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CHAPTER

7 ALTERNATING CURRENT

CONTENTS NEET Syllabus

 Introduction  Alternating currents, peak and rms
 Alternating current and voltage value of alternating current/
 RMS value and Mean value voltage; reactance and impedance;
 Power in ac circuit
LC oscillations (qualitative
 ac instruments
treatment only), LCR series circuit,
 Phasor and phasor diagram
resonance; power in AC circuits,
 ac through resistor, inductor and
Capacitor wattles current.
 Impedance and Admittance  AC generator and transformer.
 ac with LR
 ac with CR
 ac with LCR
 Series resonance
 Quality factor and Sharpness of
resonance
 Transformers
 LC oscillations
 AC generator
 Parallel LCR circuit

ALTERNATING CURRENT 1
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+
I = I0 cost

+ +
I or V t
O –

V = V cost
 When a resistor is connected across the termi- – 0

nals of a battery, a current is established in the (b)

circuit. The current has a unique direction, it  The symbol used to represent the ac source in
goes from the positive terminal to the negative a given circuit is as shown in figure.
terminal via the external resistor. The magnitude
of the current also remains almost constant.
This is called direct current (dc). ac source
+ Note: Other forms of ac
I or V Saw-tooth form of ac
t

– + + +
 If the direction of the current in a resistor or in e or i
any other element changes alternately, the current V or I – – –
t
is called an alternating current (ac).
(OR)
If the current in a circuit changes its direction in Square form of ac
T +
every sec ,then current is called an alternating
2
current(ac)
sinusoidal form of a.c e or i
+ T V or I t
2
T t
I or V
T –
– 4
3T Peak value or crest value or amplitude of
4
current or voltage  I0 or V0  :
ALTERNATING CURRENT AND VOLTAGE:
 Alternating current is that current whose  The maximum value of current or voltage in an
magnitude changes with time and direction ac circuit is called its peak value or crest value
reverses for every half cycle. or amplitude.
 Similarly, the voltage whose magnitude and Instantaneous value of current or emf( I or E)
direction change with time and attains the same The value of current or emf in an ac circuit at
magnitude and direction after definite time intervals any instant of time is called its instantaneous value.
is called alternating voltage. Instantaneous current,
 If the current or voltage varies periodically as I  I0 sin  t (or) I  I 0 sin( t   )
‘sine’ or ‘cos’ function of time, the current or Instantaneous emf
voltage is said to be sinusoidal E  E 0 sin  t (or) E  E 0 sin( t  )
I = I0sin t Where ( t   ) is called phase
+
Note : Since alternating current varies continuously
I or V + +
O t with time , its effect is measured by defining either
– the mean value (Average value) of ac or by
– defining root mean square value or virtual
V = V0sin t
value of ac
(a)
2 ALTERNATING CURRENT
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Note:Average value of a function f(t) from t1 to t2 is  The value of current at any instant ‘t’ is given by
t2 t2 I  I0 sin  t .
 f .dt  f .dt  Hence the mean or average value for half cycle
t1 t1 is given by
f av  t2

defined as t 2  t1 .
 dt T /2 T /2
t1  I .dt  I 0 .sin t.dt
0 0
t2
I av  
We can also find the value of  fdt graphically..
T /2 T /2
t1

Average value is the area of f - t graph from t1 to t2.

 dt  dt
0 0
Mean (or) Average Value of AC During
T /2 T/2
Complete Cycle: 2   cos  t  2I0 T  2 
 .I 0      cos t
 The value of current at any instant ‘t’ is given T   0 T 2  T 0
by I  I0 sin  t .
I0 2I
 The average value of a sinusoidal wave over   cos   cos 0  0
one complete cycle is given by  
T 2 I0
 I av   0.637 I 0  63.7% of I 0
 Idt 

Thus, the mean or average value of ac over
0 T
I av  I 0  2  positive half cycle is + 0.637 I0 and that over
T  cos t
 T  T 0
 dt negative half cycle is – 0.637 I0.
0  Similarly, for alternating voltage or emf over
I0 I half cycle also, the same relation holds good.
 cos 2  cos 0  0 (11)  0 2
T T
i.e., Eav  E0  0.637 E0  63.7% of E0
 Thus, we see that the average value of ac over 
one complete cycle is zero.
+
 Similarly, we can prove that the average value
of alternating voltage over one complete cycle Note : a) I or E
is also zero. A B
 The reason is that alternating current or voltage t
–
during first half cycle ( 0 to T/2) is positive and
during other half cycle (T/2 to T) is negative. between A and B, Iave  0, E ave 0
Therefore, we find the mean or average value of + A
alternating current or voltage only over any half
b) I or E
cycle.
Mean (or) Average Value Of ac For Half Cycle
t
 It is an arithmetic average of all the values of – B
current in a sine wave for half cycle. (or) The between A and B, Iave  0, E ave 0
value of steady current which sends the same
amount of charge through a circuit in a certain +
time interval is sent by an alternating current c) I or E
through the same circuit in the same time interval A C B
(half the time period or half cycle) is known as t
mean or average value of alternating current over –
half cycle. (either positive half cycle or negative between A and C or C and
half cycle) B, Iave  0, E ave  0
ALTERNATING CURRENT 3
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Root Mean Square (R.M.S) Value (or) Solution:

Effective Value (or) Virtual Value of i0 5A
The rms current is irms    3.5 A
Alternating Current  I rms  : 2 2
1 1
 The instantaneous current I may be positive or The time period is T   s
 60
negative at a given instant but the quantity I2 is
always positive. So, its average is always The current takes one fourth of the time period
positive. Therefore, we calculate the average of to reach the peak value starting from zero.
I2 over a complete period and then take the Thus, the time required is
square root of it. This gives the root mean square T 1
t  s
(or) r.m.s. current. 4 240
 It is the square root of the average of squares of Illustration 2: The electric current in a circuit is given
all the instantaneous values of current over one
by i  i0 (t2 / ) for some time. Calculate the rms
complete cycle. (or)
 Root mean square value of ac is defined as that current for the period t  0 to t   .
direct current which produces the same amount Solution:
of heat in a conductor in a certain time interval 
1 2 2 i20  4 i20 5
 0 3 0
as is produced by the ac in the same conductor i 2
 i 0 (t /  )2
dt  t dt 
3 5
during the same time. It is represented by Irms.
 Root mean square value of ac is also known as i0
Thus, the rms current is irms  i  
2

effective value (Ieff) or virtual value (Iv). 5

 Let us now calculate the value of average of I2. Illustration 3:
An alternating emf is represented by
T T
2 2 2 E = 100 sin 120 t   / 4 volt. Calculate
 I .dt  I 0 .sin t.dt i) Average or mean value of emf
2
I av  0T  0 ii) RMS value of emf
T iii) Frequency of alternating emf
 dt iv) the shortest time interval after start
0 at which emf is zero.
I r.m.s  Iav 2
2E
Solution : i) E av  0 = 0.636 E0
T
I 2  1  cos 2 t  I 2  sin 2 t 
T 2 
 0  dt  0  t    I0 = 0.6365 × 100 volt = 63.65 volt
T  2  2T  2  0 2
0 E 0 100
I0 ii) E rms   = 70.7 volt
 I r.m.s   0.707 I 0 2 2
2 iii) On comparing with E = E0 sin (t  ) ,
 Clearly r.m.s. value of an alternating current is we get   120,
70.7% of its peak value.  2f  120  f  60 Hz
 Similarly, r.m.s. or virtual or effective value of (iv) Substituting zero for instantaneous value
alternating emf is given as of emf
E 0 = 100 sin (120 t   / 4)
Er.m.s.  0  0.707 E0 
2 
 sin  120 t    0
Illustration 1:  4
The peak value of an alternating current 
For value of t to be positive, 120 t   
is 5A and its frequency is 60 Hz. Find its 4
rms value. How long will the current take 3 1
 t  sec = 6.25 × 10–3 sec.
to reach the peak value starting from zero? 4 120
4 ALTERNATING CURRENT
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4) inversely proportional to the square of

frequency
8. In general in an alternating current circuit
1) the average value of current over full cycle is zero
2) the average value of square of the current over
full cycle is zero
INSTANTANEOUS, PEAK,R.M.S & AVERAGE
VALUES OF A.C AND A.V 3) average power dissipation is zero
1. In an ac circuit the current 4) the phase difference between voltage and
1) is in phase with the voltage current is zero
2) leads the voltage 3) lags the voltage 9. The r.m.s. value of potential due to super-
4) any of the above depending on the position of given two alternating potentials
circumstances E1 = E0 sin t and E2 = E0 cos t will be
2. The average e.m.f during the positive half 1) E0 2) 2E0 3) E 0 2 4) Zero
cycle of an a.c. supply of peak value E0 is
10. If the instantaneous value of current is
1) E0 /  2) E0 / 2 3) E0 / 2 4) 2 E0 /  I  2 cos(t  ) A in a circuit, the r.m.s.
3. Alternating current is transmitted to distant value of current in ampere will be
places at 1) 2 2) 2 3) 2 2 4) zero
1) high voltage and low current
2) high voltage and high current 11. The electric current in a circuit is given by
3) low voltage and low current i  2i 0 t /  for some time. What is the rms
4) low voltage and high current
current for the period t  0 to t =  ?
4. In case of a.c circuit, Ohm’s law holds good for
a) Peak values of voltage and current 1) i 0 / 2 2) i0 3) i 0 / 2 4) 2i 0 / 3
b) Effective values of voltage and current 12. If an alternating current of frequency 50 Hz
c) Instantaneous values of voltage and current
is flowing through a conducting wire; then
1) only a is true 2) only a and b are true
3) only c is true 4) a, b and c are true how many times does the current become
5. In case of AC circuits the relation V = i Z, zero in one second?
where Z is impedance, can directly applied to 1) 125 times 2) 100 times
1) peak values of voltage and current only 3) 75 times 4) 25 times
2) rms values of voltage and current only 13. The equation of alternating current is given
3)instantaneous values of voltage and current only by E = 158 sin 200  t V. The value of voltage
4) both 1 and 2 are true at time t = 1/400 sec is
6. Alternating current can not be measured by
1) 168 V 2) –79 V 3)79 V 4) 158 V
direct current meters, because
1) alternating current can not pass through an 14. A generator produces a voltage that is given
ammeter by V  240sin120t volt, where t is in second.
2) the average value of current for complete The frequency and r.m.s. voltage are
cycle is zero 1) 60 Hz and 240 volt 2) 19 Hz and 120 volt
3) some amount of alternating current is 3) 19 Hz and 170 volt 4) 754 Hz and 170 volt
destroyed in the ammeter
4) peak value of current is zero 15. The r.m.s value of I = I 1 sin t  I 2 cos t is
7. If a capacitor is connected to two different
A.C. generators, then the value of I1  I 2 I2  I2
1) 2) 1 2
capacitive reactance is 2 2
1) directly proportional to frequency
2) inversely proportional to frequency 1 2 2 I I
3) independent of frequency 3) I1  I2 4) 1 2
2 2
ALTERNATING CURRENT 5
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16. The instantaneous values of current and the peak value of current will be
voltage in an A.C. circuit are respectively I 1) 2  10 2 sec and 14.14 A
= 4 sin t and E = 100 cos  t   / 6  . The 2) 1 102 sec and 7.07 A
phase difference between voltage and
3) 5  103 sec and 7.07 A
current is
4) 5  103 sec and 14.14 A
1) 7 /1 2) 6 / 5 3) 5 / 6 4) 2 / 3
17. An a.c. source is of 120 volt, 60 Hz. The 24 The peak value of A.C. is 2 2A . It’ss
value of the voltage after 1/360 sec, from apparent value will be
the start will be 1) 1A 2) 2A 3) 4A 4) zero
1) 146.9 volt 2) 42.9 volt 25. Alternating current in circuit is given by
3) 106.8 volt 4) 20.2 volt I  I 0 sin 2 nt . Then the time taken by the
18. The voltage of an AC supply varies with current to rise from zero to r.m.s. value is
time(t) as V = 120 sin 100  t cos 100  t. equal to
The maximum voltage and frequency 1) 1/2n 2) 1/n 3) 1/4n 4) 1/8n
respectively are 26. The form factor for a sinusoidal A.C. is
120 1) 2 2 :  2)  : 2 2
1) 120 V, 100 Hz 2) V,100 Hz
2
3) 2 : 1 4) 1 : 2
3) 60 V, 200 Hz 4) 60 V, 100 Hz
19. The rms value of an AC of 50 Hz is 20 amp. 27. The voltage of an A.C. source varies with
The time taken by an alternating current in time according to the equation
reaching from zero to maximum value and V  50sin100 t cos100 t ,
the peak value of current will be where 't' is in sec and 'V' is in volt. Then
1) 2  102 sec and 14.14 amp 1) The peak voltage of the source is 100 V
2) 1  102 sec and 7.07 amp
2) The peak voltage of the source is 100 / 2V
3) 5  103 sec and 7.07 amp
4) 5  103 sec and 28.28 amp 3) The peak voltage of the source is 25 V
20. Two alternating voltage generators produce 4) The frequency of the source is 50 Hz
e.m.f of the same amplitude E0 but with a 28. The voltage over a cycle varies as
phase difference of  / 3 . The resultant emf is V  V0 sin t for
1) E0 sin t   / 3 2) E0 sin t   / 6    2
0  t  , V  V0 sin  t for t 
3) 3E0 sin t   / 6  4) 3E0 sin t   / 2    
21. A resistance of 20  is connected to a The average value of the voltage for one
source of an alternating potential V = 220 cycle is
sin(100  t). The time taken by the current to V0 V0 2V0
change from its peak value to rms value, is 1) 2) 3) zero 4)
2 2 
1) 0.2s 2) 0.25 s
3) 2.5 × 10–3s 4) 2.5 ×10–6s
CRTQ: 01) 4 02) 4 03) 1 04) 2 05) 4
22. In an AC main supply is given to be 220V. 06) 2 07) 2 08) 1 09) 1 10) 2
What would be the average e.m.f. during a 11) 4 12)2 13) 4 14) 3 15) 2
positive half cycle? 16) 4 17)1 18) 4 19) 4 20) 3
1) 198V 2) 386V 3) 256V 4) 456V 21) 3
23. The r.m.s. value of an a.c. of 50 Hz is 10 A.
The time taken by the alternating current SPQ: 22) 1 23) 4 24) 2) 25) 4 26) 2
in reaching from zero to maximum value and 27) 3 28) 4
6 ALTERNATING CURRENT
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 100sin   /2  t   /6 
 100sin  t  2 /3 
Comparing the two equations, we find that phase
1. Phase difference between AC current and
diff. between voltage and current = 2 /3 .
voltage depends on given elements in circuit.
2. Formula 17. E v  120V,v  60 Hz
3. By using stepup Transformer E  E0 sin t  E0 sin  2v  t
4. Definition 1
 120 2 sin 120  
5. Definition 360
6. For every half cycle polarity changes 3
1 1  120 2   146.9V
XC   2
7.
C C 2 f 18. V  120sin100 t cos 100 t.
8. Definition V  2  60sin100 t cos 100 t.
E 2
 E 2
 60  sin 2  100 t   60  sin 200 t 
9. E R  0 0
 E 0
2 Vmax  60
I 2 19. Iv  20A,n  50 Hz
10. I RMS  M   2A
2 2
I0  2 Iv  1.414  20  28.28A
T
 i dt ; Time taken to reach from zero to max. value
2
2
11. i  0
T I i 2
T 1 1 1
 dt
0
t  
4 4n 4  50
s 
200
s  5  103 s

 0.02 s , for 0.02s  Two times 20. E  E2  E1  E0 sin t   / 3  E0 sin t

1 1
12. T  
f 50
current becomes zero per each cycle  E0  2 sin t   / 6  cos  / 6  

 for 1s 
1
 2  100  3E0 sin t   / 6 
0.02 21. I  I0 sin100t
1
13. E  158sin 200   158V T
400 At t1   I  I0
4
V 240 120 3 T I0
14. V  m  ,   2 f  120  f  At t2  I =
2 2 2 
8 2
15. The phase angle between I1 cos t and I2 T
Hence t2  t1 
cos t is 90 . Therefore, 8
2 1
I12sin2t  I22 sin2 cos2 where T  
Irms   400
t  2I1I2 cos  t cos  t Hence time taken = 2.5 × 10–3 s
average value in one cycle 2Vm 1 T
I12  I22  22. Vavg  23. i0  2irms , T  ,t 
  2 2 1  f 4
2 for sin  t and cos t is 2 I T 1
24. I rms  0 25. t  4  4 f
I12  I22 2
I0
IR  I12  I22 ; Ir.m.s  Iv   . rms value
2 2 26. Form factor 
16. I  sin t avg value over half a cycle
E  100 cos  t   /6  27. V0  2.Vr .m.s. 28. Conceptual

ALTERNATING CURRENT 7
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1
E0 I 0 cos 
2
Form Factor And Peak Factor of AC : c) Apparent or virtual power : The
 It is defined as the ratio of r.m.s. value to the product of apparent voltage and apparent
average value during half cycle of alternating current in an electric circuit is called
current or emf apparent power. This is always positive.
I rms Erms E0 I 0
i.e., Form factor   Papp  E rms I rms 
I av Eav 2
I0 2I0  Resistance (R): It is the opposition offered by
 We know that I rms  and I av  a conductor to the flow of direct current.
2 
I0    IMPEDANCE (Z): It is the opposition offered
 Form factor     1.11 by a conductor to the flow of alternating current.
2 2I 0 2 2
| alternating emf |
 Peak Factor : It is the ratio of peak value to the Z = | alternating current |
rms value of alternating current or emf
I0 E peak value of alternating voltage
=
i.e., peak factor   0  1.414
I rms Erms
peak value of AC

RMS value of alternating voltage

POWER IN AC CIRCUITS: = RMS value of AC
 In dc circuits power is given by P = VI. But in ADMITTANCE(Y): Reciprocal of impedance
ac circuits, since there is some phase angle of a circuit is called admittance of the circuit.
between voltage and current, therefore power is 1
admittance (Y) =
defined as the product of voltage (or emf) and Z
S.I. Units ohm-1 i.e. mho or siemen.
that component of the current which is in phase
Inductive Reactance (XL)
with the voltage (or emf)
 The opposition offered by an inductor to the
Thus P = EI cos , where E and I are r.m.s.
flow of ac is called an inductive reactance.
values of emf and current.
 The quantity  L is analogous to resistance
POWER FACTOR: and is called reactance of Inductor represented
 The quantity cos is called power factor as it by X L .
determines the power consumed in the circuit  It allows D.C. but offers finite impedance to
a) Instantaneous power : Suppose in a circuit the flow of A.C.
 Its value depends on L and f.
E = E0 sin t and I  I 0 sin t    then  Inductance not only causes the current to lag
Pinstantaneous = EI = E0I0 sin t sin t    behind emf but it also limits the magnitude of
b) Average power (True power) : The current in the circuit.
average of instantaneous power in an ac circuit E0 E
over a full cycle is called average power. Its  I0   L  0  XL ,
L I0
unit is watt i.e.
E I  X L  L  2 fL  X L f ;
Pav  Erms I rms cos   0 . 0 cos 
2 2 X L  L curve

8 ALTERNATING CURRENT
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Y  X C  f curve
X C  C curve
C = const  = const
slope = tan = 2L
XL

X XC XC
f

X L  f curve  or f C
Note: Resistance, Impedance and Reactance have the
Y same units and Dimensional Formulae.
i.e. SI unit is ohm; Dimensional Formula is

slope = tan = 2f

 ML T 2 3
A2 
XL
 L - C SERIES CIRCUIT WITH
X
L
ALTERNATING VOLTAGE
 Let an alternating source of emf is connected to
 For dc, f  0  X L  0 the series combination of a pure capacitor of
capacitance (C) and an inductor of inductance
For ac, high frequencies, X L  
(L) is shown in fig.
 dc can flow easily through inductor..
L C
 Inductive reactance in terms of RMS value
VL VC
E
is X L   L  rms
Irms
~
Capacitive Reactance (XC) E

 The resistance offered by a capacitor to the flow  Let I be the rms value of current flowing in the
of ac is called capacitive reactance. circuit
1  The P.D across ‘L’ is VL  I . X L
 The quantity is analogous to resistance and
C  The current I lags VL by an angle  / 2 .
is called reactance of capacitor represented by  The P.D across capacitance is VC  I . X C .
XC
 The current I leads VC by an angle  / 2 .
E0 1 1 E E
I0   XC    0  rms  The voltage VL and VC are represented by OB
 1   C 2 fC I 0 I rms
 
 C  and OC respectively.
 It is the part of impedance in which A.C. leads Y
B

the A.V. by a phase angle of . VL
2

1 1
 Its value is Xc   . O X
C 2fC I

 Its value depends on C and f. C

VC
 It bypasses A.C. but blocks D.C. -Y
 It is produced due to pure capacitor or induced
charge. The resultant P.D of VL and VC is

ALTERNATING CURRENT 9
Active site edutech- 9844532971

V  VL  VC  I  X L  X C  At resonance Z  0 and I 0 
E0
 .
Z
 1  1
 I  L :   IZLC Resonant angular frequency 0 
 C  LC
 From the above equations, Impedance of 1
L -C circuit is Resonant frequecny f0  .
2 LC
 1  A.C THROUGH LCR SERIES CIRCUIT
ZLC  L : 
 C   A circuit containing pure inductor of inductance
(L),pure capacitor of capacitance (C) and
1
 If  L  i.e, X L  X C then resistor of resistance (R), all joined in series, is
C shown in figure.
VL  VC potential difference  Let E be the r.m.s value of the applied alternating
V  VL  VC . emf to the LCR circuit.
 Now current lags behind voltage by  / 2 .
1
 If  L  then VL  VC
C I

Resultant potential difference V   VC  VL

Now current leads emf by  / 2 .  The potential difference across L,
1 1 VL  IX L .....(i)
If  L  then Z   L  0
C C  The potential difference across C,
E
Current I    VC  IX C .....(ii)
Z
In L - C, circuit, the phase difference between  The potential difference across R,
voltage and current is always  / 2 . VR  IR .....(iii)
Power factor cos   cos  / 2  0 .
PHASOR DIAGRAM
So, power consumed in L - C circuit is
Y
P  Vrms  I rms  cos   0 B
 In L - C circuit no power is consumed. D L
VL
Note: (VL-VC)
 In L - C, circuit, the impendence or E
1 (XL-XC) 
X
Z  L  O
VR or R A I
C VC
E C
Current I  .
Z
-Y
So, the impedance and current varies with
frequency.  Since VL and VC are in opposite phase, so
1 their resultant (VL –VC) is represented by OD
 At a particular angular frequency,  L  (Here VL > VC)
C
E  The resultant of VR and (VL–VC) is given by OL.
and current I  becomes maximum The magnitude of OL is given by
Z
 I 0  and resonance occurs. OL   OA    AL 
2 2
;

10 ALTERNATING CURRENT
Active site edutech- 9844532971

 XL  XC 
 VR   VL  VC 
2 2 1
phase angle   Tan  R

 

 I R2   XL  XC  Case (ii) : If X L  X C then  is +ve. In this

2

case the current leads the emf by a phase

E
 R2   X L  XC  angle
2
Z
I
 Impedance (Z) of LCR circuit is given by  XC  X L 
  Tan 1  
 R 
Z  R2   X L  XC 
2

Case (iii): If X L  X C then  is 0. In this case

E the current and emf are in phase.
 I
R 2  X L X C  ;
2
 If X L  X C , then the circuit will be
inductive
E
  If X L  X C , then the circuit will be
2
 1 
R   L 
2
 capacitive
 C 
 If X L  X C , then the circuit will be purely
 Let  be the phase angle between E and I,
resistive.
then from Phasor diagram  The LCR circuit can be inductive or
VL  VC IX L  IX C X  XC capacitive or purely resistive depending on
tan     L
VR IR R the value of frequency of alternating source
of emf.
 1   At some frequency of alternating source,
 L  C 
tan     X L  X C and for some other frequency,,
R X L  X C . There exists a particular value of
 Current in L-C- R series circuit is given by frequency where X L  X C (This situation is
E E0 explained under resonance of LCR series
I  sin(  t   )
Z Z circuit )
Note:Relation between applied pd & potential
(or) I  I 0 .sin( t   )
differences across
 If XL and XC are equal then Z = R i.e., the components in L - C - R circuit
expression for pure resistance circuit. L C R
2
If XL = 0 then Z  R 2  X C i.e.,
expression for series RC circuit. VL VC VR

 Similarly if XC = 0 then Z  R 2  X L 2 i.e.

V
expression for series RL circuit. L C R
R
Also, cos  
Z
VL VC VR
Case (i) : If X L  X C then  is -ve. In this
case the current lags behind the emf by a ~
E

ALTERNATING CURRENT 11
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For ‘dc’ X L  2fL  2  3.14  50  50  103   15.7 

V = VR + VL + VC V  IZ The rms current in the circuit is
(only before steady state) Vrms 220V
I rms    14.01A .
For ‘ac’ X L 15.7

V  I R2   X L  X C 
2
; Illustration 5: If an input of 50 mV is applied as Vin
then Vout at 100 kHz will be
 IR 2   IX L  IX C 
2

1k
2
V 2 = VR 2 + VL - VC 
10 nF Vout
V  V   VL  VC 
2 2
R

where VL  IX L  I  L and Solution:

I V 50  159
VC  IX C  V0  in X C   7.9 mV
and VR = IR | Z| 1000 2
 159 2
C
Note: Rules to be followed for various combinations 1 1 10 3
XC   8   159 
of ac circuits Cw 10  2 p  10 5 2 p
 Compute effective resistance of the circuit as R Illustration 6: 30.0 F capacitor is connected to a
 Calculate the net reactance of the circuit as
220 V, 50 Hz AC source. Find the capacitive
1
X  X L  X C where X L   L , X C  . reactance and the current (rms and peak) in the
C
 Resistance offered by all the circuited elements circuit. If the frequency is doubled, what happens
to the flow of ac is impedance ( Z ) to the capacitive reactance and the current.
2 Solution: The capacitive reactance is
 Z  R 2  X 2  R2   X L  X C 
1
XC   106
E0 2fC
 The peak value of current is I0 
Z Vrms
The rms current is I rms   2.08A
 The phase difference between emf & current can XC
be known by constructing an ac triangle as
The peak current is I m  2Irms  2.96A
X If the frequency is doubled, the capacitive
tan  
R reactance is halved and consequently, the
X
X
Z sin  
Z
current is doubled.
Illustration 7:When an AC source of e.m.f. E = E0
cos  
R
Z

R sin (100t) is connected across a circuit, the phase
Illustration 4: difference between the e.m.f. E and the current I
A pure inductor of 50.0 mH is connected in the circuit is observed to be  / 4 , as shown in
to a source of 220 V. Find the inductive the figure. If the circuit consists possibly only of
reactance and rms current in the circuit if
R-C in series. What will be the relation between
the frequency of the source is 50 Hz.
Solution: The inductive reactance. the two elements of the circuit ?

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I E Z  R 2  ( X C  X L ) 2  R  100

[since X C  X L  3 R ]
t
V0 200
Current in the circuit I 0    2 A.
Z 100
Solution: Given E = E0 sin (100 t). Comparing this
Illustration 9: In a series resonant circuit. the ac
with E = E0 sin t , we have   100 rad s 1 .
It follows from the figure that the current leads voltage across resistance R, inductance L and
the e.m.f. which is true only for R-C circuit. capacitance C are V R = 4V, VL = 10V and VC
1 = 7V respectively. Find the voltage applied to
tan   ..... (1)
RC the circuit.
Given    / 4 . Also   100 rad s 1 . Using Solution: Voltage applied across circuit
these values in (1), we get
V  VR2   VC  VL   42  10  7  = 5V
2 2
 1 1
tan    or RC 
4
  100 RC 100
Illustration 8: An LCR circuit contains resistance
of 100  and supply of 200 V at 300 rad/
sec. If only capacitance is taken out from the
circuit and the rest of the circuit is joined,
A.C ACROSS R-L,R-C,L-C & L-C-R SERIES
current lags behind the voltage by 60°. If on CIRCUIT
the other hand, only inductor is taken out, the 1. At low frequency a condenser offers
1) high impedance 2) low impedance
current leads by 60° with applied voltage. Find
3) zero impedance
the current flowing in the circuit. 4) impedance of condenser is independent of
Solution :Resistance R = 100  frequency
2. If the frequency of alternating e.m.f. is f in
Applied voltageV0  200 volt L-C-R circuit, then the value of impedance
When the capacitance is taken out, only Z will change with log (frequency) as
1) increases
inductor and resistor will remain in circuit.
2) increases and then becomes equal to
XL resistance, then it will start decreasing
The phase tan () 
R 3) decreases and when it becomes minimum equal
XL to the resistance then it will start increasing
tan (60o )  X L  3R
R 4) go on decreasing
When the inductor is taken out, only capacitor 3. In a circuit, the coil of a choke:
and resistance will remain in circuit. 1) Always decreases the current
2) Always increases the current
XC 3) Opposes the change of current
Then phase tan ()  4) No effect with the current
R
4. With increase in frequency of an AC supply,
o X the impedance of an LCR series circuit
tan (60 )  C X C  3 R
R 1) Remains constant
2) increases 3) Decreases
Impedance 4) Decreases at first, become minimum and then
increases
ALTERNATING CURRENT 13
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5. A resistor, an inductance and a capacitance

R = 5
are connected in series to an a.c. supply.
When measured with the help of an a.c.
voltmeter, the p.d. across the resistor is i
found to be 40 V, across the inductance 30
V, and across the capacitance 60 V. What is 
100V, 50 Hz
the supply voltage?
1) 10 A 2) 20 A 3) 30 A 4) 5 A
1) 130 V 2) 50 V 3) 70 V 4)100 V 12. In series L-C-R circuit as shown in figure.
6. A 60 volt-10 watt bulb is operated at 100 The potential difference across A & B is
volt-60 Hz ac. The inductance required is R = 3

A B
1) 2.56 H 2) 0.32 H 3) 0.64 H 4)1.28 H
7. In L–R circuit, the a.c. source has voltage

220 volt. if the potential difference across 220V, 50 Hz
the inductance is 176 volt, the p.d. across 1) 100 V 2) 150 V 3) 200 V 4) Zero
the resistance will be 13. If resistance of 100  and inductance of 0.5
1) 44 volt 2) 396 volt henry and capacitance of 10 × 10–6 F are
connected in series through 50 Hertz a.c.
3) 132 volt 4) (220) × (176) volt supply, the impedance is
8. A coil having an inductance of 15/16  henry 1) 1.8765  2) 18.76 
is connected in series with a resistance of 3) 189.6  4) 101.3 
300 ohm. If 20 volt from 200 cycle source 14. The instantaneous values of current and
are impressed across the combination, the voltage in an A.C. circuit are respectively
value of the phase angle between the I  4sin t and E  100cos(t   / 3) . The
phase difference between voltage and
voltage and the current is
current is
1
1) tan (5 / 4) 1
2) tan (4 / 5) 1) 7  / t 2) 6 / 5 3) 5 / 6 4)  / 3
15. An inductance of negligible resistance, whose
3) tan 4 (3 / 4) 4) tan 1 (4 / 3) reactance is 22  at 200 Hz, is connected to
200 volt, 50 Hz power line. The value of
9. A 20 volt a.c. is applied to a circuit consisting
inductance is
of a resistance and a coil with negligible
1) 0.0175 H 2) 0.175 H
resistance. If the voltage across the
3) 1.75 H 4) 17.5 H
resistance is 12 volt, the voltage across the
16. A pure resistance and a pure inductance are
coil is
connected in series across a 100 volt A.C.
1) 16 volt 2) 10 volt line. A voltmeter gives same reading whether
3) 8 volt 4) 6 volt connected across resistance or inductance.
What does it read?
10. Alternating voltage V  400sin(500 t) is 1) 50 V 2) 70.7 V 3) 88.2 V 4) 100 V
applied across a resistance of 0.2 k  . The 17. An alternating voltage E(in volt)
r.m.s. value of current will be equal to
= 200 2 sin 100t  is connected to a
1) 14.14 A 2)1.414 A
1 F capacitor through an ac ammeter..
3) 0.1414 A 4) 2.0 A
The reading of the ammeter shall be:
11. A series L-C-R circuit is given in figure. The 1) 10 mA 2) 20 mA
current through the circuit is 3) 40 mA 4) 80 mA
14 ALTERNATING CURRENT
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18. A direct current of 10A is superimposed on connected across this combination. Ideal
an alternating current I = 40 cos t amperes ac volt meters v1 and v2 show 120 volt
flowing through a wire. The effective value and 160 volt respectively. What is the
of the resulting current will be phase difference between V1 and V2.
L R
1) 10 2A 2) 20 2A 3) 10A 4) 30A
19. In an A.C. circuit, resistance, inductance and V 1 V 2

capacitance are connected in series. The

values of potential differences across three 
are 70 V, 90 V and 65 V respectively. The
value of potential difference of A.C.source is   
1) 0 2) 3) 4)
1) 225 V 2) 95 V 6 4 2
3) 85 V 4) 74.3 V 26. In an LR circuit, the AC source has voltage
20. A 10 F capacitor, is connected across a 200 220 volt. The potential difference across
V, 50 Hz AC supply. The peak current the inductance is 176 volt. The potential
difference across the resistance will be
through the circuit is
1) 0.6 amp 2) 0.6 2 amp 1)  220  176  volt 2)  220  176  volt

3) (0.6 / 2)amp 4) (0.6 / 2)amp 3) 220  176volt

21. An inductor, a capacitor and a resistor are 4) [(220)2  (176)2 ] volt
connected in series to an a.c. supply. When
measured with an a.c. voltmeter, the 27. A 50W, 100V lamp is to be connected to an
potential difference across the inductor, ac mains of 200V, 50Hz. The capacitance
capacitor and resistor are respectively 90 of the capacitor essential to be put in series
volt, 60 volt and 40 volt. Then the supply with the lamp is:
voltage is
1) 4.6  F 2) 2.9  F
1) 190 volt 2) 100 volt
3) 130 volt 4) 50 volt 3) 1.5 F 4) 9.2  F
22. The magnetic field energy in an inductor 28. A coil of inductance 5.0 mH and negligible
changes from maximum value to minimum resistance is connected to an alternating
value in 5.0 ms when connected to an AC voltage V = 10 sin (100 t). The peak current
source. The frequency of the source is in the circuit will be
1) 20 Hz 2) 50 Hz 3) 200 Hz 4)500 Hz 1) 2 amp 2) 5 amp 3) 10 amp 4) 20
23. In an L-R circuit, an inductance of amp
0.1 H and a resistance of 1  are connected
29. An alternating e.m.f. 100 cos 100 t volt is
in series with an ac source of voltage V = 5
connected in series to a resistance of 10
sin 10 t. The phase difference between the
ohm and inductance 100 mH. The phase
current and applied voltage will be
difference between the current in the circuit
1)  / 2 2)  / 4 3)  / 6 4) 0 and the e.m.f. is
24. The reactance of a capacitor of capacitance
C is X. If both the frequency and capacitance 1)  / 4 2) zero 3)  4)  /2
be doubled, then new reactance will be30. An emf E = 4 cos (1000 t) is applied to an
X LR circuit of inductance 3 mH and
1) X 2) 2X 3) 4X 4) resistance 4  . The amplitude of current
4
25. A pure inductor and a pure resistor are in circuit is
connected in series and an ac supply is 1) 0.8 A 2) 0.6 A 3) 0.4 A 4) 0.2 A

ALTERNATING CURRENT 15
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1) 60o 2) 30o 3) 45o 4) 90o

39. In an L-C-R series circuit,
AC ACROSS LR, CR, LC AND LCR SERIES
CIRCUIT R  5, X L  9, X C  7 . If applied
31. In an LCR series circuit the rms voltages voltage in the circuit is 50V then impedance
across R, L and C are found to be 10 V, 10 of the circuit in ohm will be
V and 20 V respectively. The rms voltage 1) 2 2) 3 3) 2 5 4) 3 5
across the entire combination is
40. In the following circuit, the values of cur-
1) 30 V 2) 1 V 3) 20V 4) 10 2 V
rent flowing in the circuit at f = 0 and
32. An alternating voltage of f =  will respectively be
E  200 2 sin(100t)V is connected to a conden- –5
0.01H 10 F 25
ser of 1  F through an A.C. ammeter. The
reading of the ammeter will be
1) 10 mA 2) 40 mA 3) 80 mA 4) 20 mA
33. In a circuit, the frequency is f  1000 Hz and
2
200V
the inductance is 2 henry, then the reactance
1) 8A and 0A 2) 0A and 0A
will be
3) 8A and 8A 4) 0A and 8A
1) 200 2) 200
41. Radio receiver receives a message at 1MHz
3) 2000 4) 2000 band, If the available inductance is 1mH,
34. A condenser of capacity 1pF is connected to then calculate required capacitance
an A.C source of 220V and 50Hz frequency. 1)2.5pF 2)25 pF 3) 1 pF 4)50 pF
The current flowing in the circuit will be 42. A pure inductor of 25mH is connected to an
1) 6.9 x 10-8A 2) 6.9A ac source of 220V. Given the frequency of the
3) 6.9 x 10-6A 4) zero source as 50Hz, the rms current in the circuit is
35. The frequency at which the inductive reac- 1) 7A 2) 14A 3) 28A 4) 42A
tance of 2H inductance will be equal to the 43. In a series LCR circuit the voltage across the
capacitive reactance of 2 F capacitance resistance, capacitance and inductance is
(nearly) 10V each, If the capacitance is short circuited,
1) 80Hz 2) 40 Hz 3) 60Hz 4) 20Hz the voltage across the inductance will be
36. If the instantaneous current in a circuit is
 10 
given by I  4 cos t    A, the rms value 1) 10V 2) 10 2V 3)  V 4) 20V
of the current is  2
44. In a series LCR circuit, the potential differ-
1) 2 A 2) 2 A 3) 2 2 A 4) zero ence between the terminals of the induc-
37. A 100 km telegraph wire has capacity of tance is 60V. between the terminals of the
0.02 F / km , if it carries an alternating capacitor is 30 V and that between the ter-
current of frequency 5 kHZ. The value of minals of resistance is 40V. The supply volt-
an inductance required to be connected in age will be equal to
series so that the impedance is minimum. 1) 130V 2) 10V 3) 50V 4) 70V
1) 50.7mH 2) 5.07mH 45. An inductor of inductance 2H and a resis-
3) 0.507mH 4) 507mH tance of 10 are connected in series to an
ac source of 110V, 60Hz. The current in the
38. In a series LCR circuit R  10 and the
circuit will be
impedance Z  20  . Then the phase diffe-
1) 0.32A 2) 0.15A 3) 0.48A 4) 0.80A
rence between the current and the voltage is
16 ALTERNATING CURRENT
Active site edutech- 9844532971

46. A coil has an inductance of 0.7 henry and

is joined in series with a resistance of
220  . When the alternating emf of 220 CRTQ: 01) 1 02) 3 03) 3 04) 4 05) 2
V at 50Hz is applied to it then the phase 06) 4 07) 3 08) 1 09) 1 10) 2
through which current lags behind the 11) 2 12) 4 13) 3 14) 3 15) 1
applied emf and the wattles component of 16) 2 17) 2 18) 4 19) 4 20) 2
current will be respectively
21) 4 22) 2 23) 2 24) 4 25) 4
1) 300 , 1 A 2) 450 , 0.5 A 26) 4 27) 4 28) 4 29) 1 30) 1
3) 60 , 1.5 A
0
4) none of these
SPQ: 31) 4 32) 4 33) 3 34) 1 35) 1
47. If value of R is changed then
36) 3 37) 3 38) 1 39) 2 40) 2
10V 10V R 41) 2 42) 3 43) 3 44) 3 45) 2
46) 2 47) 3 48) 3 49) 3 50) 4
51) 3

Supply

1) Voltage across L remains same 1

1. XC  2. Z  R 2   X L  X C 
2
2) Voltage across C remains same C 2 f
3) Voltage across LC combination remains same V
3. I=
4)Voltage across LC combination changes R  (L)2
2

Inductance decreases the current.

48. A current in circuit is given by i =3 + 4 sin
2
t. then the effective value of current is :  1 
4. Z  R 2   L  
  C
1) 5 2) 7 3) 17 4) 10
With increase in frequency of a.c. supply
49. A coil of inductive reactance 31  has a (  2n) , Z decreases first, becomes mini-
resistance of 8  . It is placed in series with 1
a condenser of capacitative reactance 25  mum when   and then increases.
LC
the combination is connected to an a.c source
V  VR2 VC VL   402  60 30  50V
2 2
of 110volt. The power factor of the circuit is: 5.
1) 0.56 2) 0.64 3) 0.80 4) 0.33 V 2 60  60
6. R   360
50. Which of the following device in alternating P 10
circuit provides maximum power : P 10 1 V V
I   A 
1) only capacitor 2) capacitor V 60 6 Z R  X 22
2

3) only inductor 4) only resistor X L  LW

51. The self inductance of a choke coil is
V  VL2 VR2 VR  V2 VL2   220 176
2 2
7.
10mH.When it is connected with a 10 volt
D.C. source. Then the loss of power is 20 X L L 2 f 15 20  200
8. Tan    
watt. When it is connected with 10 volt R R 16 3000
A.C. source loss of power is 10 watt. The 5
   Tan 1  
frequency of A.C. source will be: 4
1) 50 Hz 2) 60 Hz 3) 80Hz 4) 100 Hz 9. VL  V 2  VR2  202  122

ALTERNATING CURRENT 17
Active site edutech- 9844532971

Vm 400  L 10  0.1 
10. I    1.414 A 23. tan    1 
2R 2  0.2  103 R 1 4
V 100 1
11. I  At resonance, I    20 24. X C  . if  '  2 and C ' = 2C,
R 5 C
X
12. At resonance, VL  VC   VL  VC  0 Then, X 'C = C
4
25. As current is common through both the
13. Z  R 2   X C  X L 
2
elements, so in the inductor, voltage (V1) is
1 1 ahead of current I while, in the resistor,
XC  
CW C 2 f voltage (V2) is in same phase as current.
X L  LW  L 2 f Hence phase difference between V1 and V2 is 90º.
  26. In an LR circuit ; VL2  VR2  V 2
14. I  u sin t E  100 cos  t  
 3
VR  V 2  VL2   220 
2
 176 
2

 
 100 sin   t   V 2 100 
2
2 3 27. R    200
5  P 50
 5
 100 sin  t      P 50
 6  6 I   0.5
V 100
15. Here, X L  22, n  200 Hz
V V 200
X L  L  2nL , I  AC  Z  AC   400
Z I 0.5
XL 22
L   0.0175 H X C  Z 2  R2
2n 2  200
V 28. XL  L  100  5  103  0.5
16. V  VR  VL ;
2 2 VR  VL 
2 E0 10
I0    20A
CV0 XL 0.5
17. I  CV  29. E  100 cos 100 t volt
2
E0  100 V,   100 rad /s
18. I net  I d2.c  I rms
2

R  10,L  100 mH  101 H

19. V  VR2   VL  VC  2
X L L 100  10 1
tan     1
 702   90  65   5525  74.3V
2 R R 10

E0 
20. i 0  4
Z E0
1 1 1000 30. Amplitude of current in circuit is I0 
XC    Z
6
C 50  2  10  10  E0 = 4
 2 Z  R 2   L 
2
  1000
E0  200 2, i  200 2 
1000 5
 
2
0.6 2 amp.  42  1000  3  103 =5

21. V  VR2  (VL  VC ) 2  1600  900  50 volt 4

I0   0.8A
5
T
22. t  31. V  VR2  VL  VC 
2
4

18 ALTERNATING CURRENT
Active site edutech- 9844532971

rmsE E 0 C 220
32. Irms  X  I L  I sin   sin 450  0.5 A
C 2 2  220
1000
33. X L   L  2 fl  2   2  2000 47. L-C-R Circuit
2
Erms 1 48. I DC  3, I AC  4 ; I RMS  I DC
2
 I RMS
2
34. irms  35. f 
Xc 2 LC
2
I0  4 
I RMS  32     17 A
36. Irms   2
2
1 1 1 49. L C R Circuit
37.   L 2 
LC  C (2 n) 2 C
Z  R2  XL  XC   82  31 25 10
2 2

R
38. cos  
Z R 8
cos     0.8
39. Impedance, Z  R 2   X L  X C 
2 Z 10
50. Resistor
40. I
E

E
51. L-R circuit
Z 2
 1 
R 2   2 f L  20
 2  f C  DC : P  VI  I   2A
10
41. 1 10
f   1 106 V  IR R   5
2 LC 2
1 V R
C   25 pF AC : P  VI cos   V 
4  103  1012
2
Z Z
Vrms
42. I rms  Z  R 2  X L2  Z  50
XL
43. Here, R  X L  X C
(  voltage across them is same)
Total voltage in the circuit
2 1/ 2 RESONANT FREQUENCY
V  I  R 2   X L  X C    IR  10 volt
  Electrical resonance is said to take place in a
When capacitor is short circuited series LCR circuit, when the circuit allows
10 10 maximum current for a given frequency of
I'  
R  XL 
2 2 1/2 2R alternating supply, at which capacitive reactance
becomes equal to the inductive reactance.
 Potential drop across inductance The current (I) in a series LCR circuit is given by
10
 I ' X L  I 'R  V
2 I
E

E
Z 2

44. In series LCR circuit, V  VR2  VL  VC   1  .....(i)

2
R 2   L 
 C 
Vrms
45. Impendence Z  R 2  X L2 , I rms  From the above equation (i), it is clear that current
Z I will be maximum if the impedance (Z) of the
46. L-R circuit, X L  L  L2 f circuit is minimum.
22 At low frequencies, L   L  2  f is very small
 0.7  2   50  220 1 1
7 and C   C  2  f is very large.
V 220 220 At high frequencies, L is very large and
I  ; Tan   1 ;   450
Z 2R 220 1
is very small.
C
ALTERNATING CURRENT 19
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1 c) Impedance Z = 0
For a particular frequency (f0), L   i.e. E0
C d) peak value of current I0  
X L  X C and the impedance (Z) of LCR Z
1
circuit is minimum and is given by Z = R. e) Resonant frequency f 0 
2 LC
Therefore, at the particular frequency ( f 0 ), the 
f) Voltage and current differ in phase by
current in LCR circuit becomes maximum. The 2
frequency ( f ) is known as the resonant g) Power factor cos   0
0
RESONANCE IN L - C - R CIRCUIT :
frequency and the phenomenon is called
At resonance,
electrical resonance.
a) Net reactance X = 0
Again, for electrical resonance (XL–Xc) = 0. b) X L  X C
i.e. XL = XC c) Impedance Z = R ( minimum )
1 1 E0 E0
or L   2  d) peak value of current I0  
C LC Z R
1 1
or     2f 0   ( maximum but not infinity )
LC LC 1
1 e) Resonant frequency f 0 
or f 0  .....(ii) 2 LC
2 LC f) Voltage and current will be in phase
This is the value of resonant frequency. g) power factor cos   1
The resonant frequency is independent of the h) Resonant frequency is independent of value of R.
resistance R in the circuit. However, the i) A series L - C - R circuit behaves like a pure
sharpness of resonance decreases with the resistive circuit at resonance.
increase in R. HALF POWER FREQUENCIES AND
Series LCR circuit is more selective when BAND WIDTH.
resistance of this circuit is small.  The frequencies at which the power in the circuit
Y
is half of the maximum power (The power at
R3>R2>R1 resonance) are called half power frequencies.
I0 Pmax

R1
R2
I P Pmax
R3 P=
2

O X
f0 1 2 3 V
f
.Note: Series LCR circuit at resonance admit  The current in the circuit at half power frequencies
maximum current at particular frequencies, so (HPF) is 1 2 or 0.707 or 70.7% of maximum
they can be used to tune the desired frequency current (current at resonance).
or filter unwanted frequencies. They are used in  There are two half power frequencies
transmitters and receivers of radio, television and
telephone carrier equipment etc. 1  called lower half power frequency. At this
RESONANCE IN L - C CIRCUIT : frequency the circuit is capacitive.
At resonance , 3  called upper half power frequency. It is
a) Net reactance X = 0
greater than 2 . At this frequency the circuit is
b) X L  XC
inductive.
20 ALTERNATING CURRENT
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 Band width    : The difference of half power The Q-factor of series resonance circuit is
defined as the ratio of voltage developed
frequencies 1 and 3 is called band width across the inductance or capacitance at
   and   3  1. resonance to the applied voltage (which is
 For series resonant circuit it can be proved the voltage across R).
voltage across L or C
   R / L  Q
Applied Voltage( voltage acrossR)
QUALITY FACTOR (Q) OF SERIES
IX L X L  L 1 L
RESONANT CIRCUIT : Q   
(i) The characteristic of a series resonant circuit IR R R LC R
is determined by the quality factor (Q - factor)
1 L
of the circuit. Q=
(ii) It defines sharpness of i - v curve at resonance. R C
When Q - factor is large, the sharpness of VL VC 0 L 1
resonance curve is more and viceversa. Q - factor  V or V  R or  CR
R R 0
Re sonant frequency 
(iii) Q  Band width
 0

(v) The quality factor (Q) is also defined as 2
times the ratio of the energy stored in L (or C)
For series resonant circuit, it can be proved to the average energy loss per period.
R  max imum energy stored in circuit 
that,    Q  2  
L  energy loss per period 
0 L The maximum energy stored in inductor,
Q = 0 
 R
1
U  LI 20
1 2
But at resonance, 0 
LC The energy dissipated per second,

1 L 1 L 2 I 20 R
U R  I .R 
Q   rms
2
LC R R C
Energy dissipated per time period,
I0
I 20 R
UR  .T
R = very low 2
Q - Factor = Large Substituting these values, we get
R = low
Q - Factor = Normal  
 1 2 
R = High  LI 0  2 L
V0 V Q - Factor = Low Q  2  22  
 I 0 R   T R
  
Resonance curve
  2  T 
 
(iv) Q - Factor tells the relation between voltage
across the inductor or capacitor and peak L 1 L 1 L
 0    
value of voltage. R LC R R C
The sharpness or selectivity of a resonance Different forms of Q factor
circuit is measured by Q-factor, called 0 1 L
quality factor. a) Q 
 R C

ALTERNATING CURRENT 21
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maximum P.D across C E rms R

Q Pave  E rms  
b) maximum P.Dacross R  1 
2
 1 
2
R 2   L   R 2  L  
VC X C 1  C   C 
  
VR R 0 CR
maximum P.D across L E2rms R
c) Q  2
maximum P.D across R  1 
R 2  L  
VL X L 0 L  C 
  
VR R R Average power is also known as true power.
2maximum energy stored The quantity E rms I rms is called the apparent
d) Q power or virtual power.
loss of energy in one time period
It is customary to express true power in kW and
2maximum energy stored
e) Q apparent power in kVA.
T average energy exp ended
POWER IN LCR CIRCUIT cos  is called the power factor of LCR circuit.
Instantaneous power, P = EI Its value varies from zero to 1.
 E0 sin t I0 sin t   Power factor is defined as the ratio of true
 E0 I0 sin t sin  t    power to apparent power.
In L – C – R series circuit,
 E 0 I0 sin t  sin t cos   cos t sin  
R R
 E 0 I0 cos  sin2 t+E0 I 0 sin  sin t cos t Power factorcos 
Z  1 
2
If the instantaneous power is assumed to remain 2
R L 
 C 
constant for a small time dt, then work done over
a complete cycle is given by POWER FACTOR :
T (a) It may be defined as cosine of the angle of
W   P.dt lag or lead (i.e. cos )
0 (b) It is also defined as the ratio of resistance
T
 E I  R
W    E 0 I 0 cos  sin2 t  0 0 sin   2sin t cos t  dt
 2  and impedance (i.e. )
0 Z
T T
E I c) The ratio
WE0 I0 cos  sin  t dt 0 0 sin  sin 2 t dt
2

0
2 0 True power W kW
T    cos 
 W  E 0 I0 cos  Apparent power VA kVA
2 Special Cases :
Average power over complete cycle,
Case I : If the ac circuit contains only pure
W E 0 I0 resistance, then   0 0 .
Pave   cos
T 2 E 2rms
Pave E rms I rms cos 00 E rms I rms 
E0 I0 R
  cosE rms I rms cos Case II : If the ac circuit contains only pure
2 2
inductance, then   90 0
E 0 I0
 Pave  cos   E rms .I rms cos   Pave  E rms .I rms cos 90  0
2
R Case III : If the ac circuit contains only pure
Also, Pave  E rms I rms or
2 capacitance, then
 1 
R 2  L     900.  Pave  0
 C 
22 ALTERNATING CURRENT
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Case IV : If the ac circuit contains L and R, then Iron core

R R
cos  Starter
Z R 2 L2  2
Choke
Coil of Cu wire ~ coil
R
Pave E rms I rms ~ L, R
Application of choke coil
R 2 L2  2 Choke coil
E 2rms R  It consists of a copper coil wound over a soft

R 2 L2  2 iron laminated core. This coil is put in series
Case V : If the ac circuit contains C and R, with the circuit in which current is to be
then reduced.
R R  Soft iron is used to improve inductance (L) of
cos  the circuit.
Z 1 .
R2   The inductive reactance or effective opposition
C 2
2
of the choke coil is given by
2
Erms R X L   L  2 vL
 Pav 
1
R2  2 2 .  For an ideal choke coil r  0 , no electric
C energy is wasted, i.e., average power P = 0.
Wattless current or idle current  In actual practice choke coil is equivalent to a
If the voltage and current differ in phase by  / 2 , R  L circuit.
then Power factor, cos   cos90 0  0 .  Choke coil for different frequencies are made
In this case, as there is no resistance to the flow by using different substances in their core.
of current, the average power consumed by the  For low frequency L should be large thus iron
current is zero. Such a current is, therefore, called core choke coil is used. For high frequency ac
wattless current. Since this current does not circuit, L Should be small, so air cored choke
perform any work, therefore, this current may coil is used.
also be called idle current. Such a current flows  The choke coil can be used only in ac circuits
only in purely inductive, purely capacitive or L– not in dc circuits, because for dc frequency
C circuits.
The average power consumed per cycle in a pure v  0 . Hence X L  2 vL  0.
inductive or capacitive or L - C circuit is zero.  Choke coil is based on the principle of
Such current for zero power consumption is Wattless current.
called Wattless current. E
CHOKE COIL:  The current in the circuit I 
Z
 Choke coil (or ballast) is a device having high
inductance and negligible resistance. with
 It is used to control current in ac circuits and is
 R  r    L 
2 2
used in fluorescent tubes. Z .
 The power loss in a circuit containing choke coil  The power loss in the choke
is least.
pav  Vrms I rms cos   0
 In a dc circuit current is reduced by means of a
rheostat.This results in a loss of electrical energy r r r
I 2 R per sec. as cos   Z  2 
L
0
r  L
2 2

ALTERNATING CURRENT 23
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Illustration 10: Solution :

A 750 Hz., 20 V source is connected to a
resistance of 100 ohm, an inductance of In the figure voltages across source, inductor
0.1803 henry and a capacitance of 10 and capacitor are given as V = 100 V, VL =
microfarad all in series. Calculate the time 400 V and VC = 400 V respectively.
in which the resistance (thermal capacity
2 J/0C) will get heated by 100C. From the formula V  VR2  (VC  VL ) 2 ... (i)
TC  mc  thermal capcity  Substituting value in (i)
Solution: 100  VR2  (400  400) 2  VR  100 V
X L   L  2 fL  2  750  0.1803  849.2
So reading of voltmeter
and,
(reading across R) is 100 V.
1 1 1 Reading of ammeter (i.e., current passing
XC     21.2
C 2fC 2 750 105 through it).
So, X  X L  X C  849.2  21.2  828
V
and hence, I [since the circuit is in resonance, so
Z
Z  R 2  X 2  (100)2  (828) 2  834 Z = R = 50 
But as in case of ac, 100
I  2A
Vrms R 50
Pav  Vrms I rms cos   Vrms   Voltage across inductor (VL )
Z Z Quality factor=
i.e., Applied voltage (V)
400
2 2   4.
V   20  100
Pav   rms   R    100  0.0575W
 Z   834  Illustration 12:
and as, An inductance of 2.0 H, a capacitance of
U  P  t  mc  (TC)  18 F and a resistance of 10 k are
( TC)   2  10 connected to an ac source of 20 V with
t   348s
P 0.0575 adjustable frequency.
Illustration 11: (i) At what frequency, will the current be
In the LCR series circuit find the voltmeter maximum in the circuit ?
and ammeter reading in the figure shown (ii) What is this maximum current.
below. Also find the quality factor of
Solution:
circuit.(Given R=50  )
(i) Current will be maximum at resonant
frequency fr.
400 V 400 V
r 1 1
fr     26.5Hz
V VL VC 2 2 LC 2 218106

(ii) At resonance R = Z.
At resonant frequency,
A
 I max 
E max E max
 
20 2
 2.83mA
Z R 10000 
100V, 500 Hz

24 ALTERNATING CURRENT
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Illustration 13: circuit are given by V= 5 cos ( t)volts

and I = 2sin( t) Amp. Find the power
A 0  electric iron is connected to an ac
dissipated in the instrument.
supply of 200 V, 50 Hz. Calculate (i) average
power delivered to iron (ii) peak power and Solution:
(iii) energy spent in one minute. Current I = 2 sin ( t)
Solution: Root mean square value of current
(i) Average power is based on rms value of I0 2
emf. I rms    2
2 2
(E V )2 2002
Average power =
R
=
50
= 800watt Voltage V  2cos  t   2sin  t  
 2
(ii) Peak power = Root mean square value of voltage
V0 2
E 02 (200 2 ) 2 Vrms    2
 1600 watt 2 2
R 50 Power dissipated P  Vrms Irms cos  = 0
(iii)Energy spent is based on rms power.   
(since   and cos   =0
So energy spent 2 2
= P × t = 800 watt × 60 sec Illustration 16:
= 48000 Joule. A capacitor, an inductor and an electric bulb
Illustration 14: are connected in series to an AC supply of
variable frequency. If the frequency of the
An L-C-R series circuit is connected to supply is increased gradually, what will
an external emf E= 200 sin (100  t). The happen to brightness of bulb.
values of the capacitance and resistance
Solution:
in the circuit are 1µF and 100 
The brightness of bulb is proportional to
respectively. Find the inductance for which square of current passing through bulb.
current in the circuit is maximum. Current through the bulb or circuit
Solution:
2
V0
where Z  R 2   
Current in circuit I 1
 L 
Z  C 
V0
I0  where Z  R 2   X C  X L 2
Z
For current I0 to be maximum, impedance Z As frequency is increased. Impedance Z will
must be minimum Z will be minimum, when XC first decrease, reaches minimum value of R
= XL i.e.
at resonance and will again increase.
1 1
 L L 2
C C Correspondingly, the current will first
(  =100  from E = 200 sin (100  t)  1
1 1 100
increase  I   , reaches maximum value of
L   2 H 10H Z  
100 10 100 10 
2 6 2 6 I0 = V0/R at resonance and will then
decrease
Illustration 15:
The potential difference V and the current So the intensity of bulb will first increase,
I flowing through an instrument in an ac reaches a maximum and then decreases.

ALTERNATING CURRENT 25
Active site edutech- 9844532971

6. An inductance L and capacitance C and

resistance R are connected in series
across an AC source of angular frequency
1
 . If  2  LC then
1) emf leads the current
RESONANCE, POWER IN AC CIRCUIT 2) both the emf and the current are in phase
1. An LCR circuit is connected to a source of 3) current leads the emf
alternating current. At resonance, the
applied voltage and the current flowing 4) emf lags behind the current
through the circuit will have a phase 7. Power factor is one for
difference of 1) Pure inductor
1)  / 4 2) zero 3)  4)  / 2
2) Pure capacitor
2. In series L - C - R resonant circuit, to
3) Pure resistor
increase the resonant frequency
4) Either an inductor or a capacitor
1) L will have to be increased
8. Choose the wrong statement of the
2) C will have to be increased
following.
3) LC will have to be decreased
1) The peak voltage across the inductor can
4) LC will have to be increased be less than the peak voltage of the
3. If in a series L - C - R ac circuit, the voltages source in an LCR circuit
acr oss R, L , C ar e V 1 ,V2 ,V3 respectively.
2) In a circuit containing a capacitor and an
Then the voltage of applied AC source is
always equal to ac source the current is zero at the instant
source voltage is maximum
1) V1 +V2 +V3
3) When an AC source is connected to a
2
2) V1  (V2  V3 ) 2 capacitor,then the rms current in the circuit
gets increased if a dielectric slab is
3) V1 -V2 -V3 inserted into the capacitor.
2 4) In a pure inductive circuit emf will be in
4) V1  (V2  V3 )2
phase with the current.
4. The phase difference between voltage and 9. The value of current in two series L C R
current in an LCR series circuit is circuits at resonance is same when
1) zero always 2)  / 4 always connected across a sinusoidal voltage
3)  4) between0 and  / 2 source. Then
5. In an LCR a.c circuit at resonance, the 1) both circuits must be having same value of
current capacitance and inductance
1) Is always in phase with the voltage 2) in both circuits ratio of L and C will be
same
2) Always leads the voltage
3) for both the circuits R must be same
3) Always lags behind the voltage
4) both circuits must have same impedance at
4) May lead or lag behind the voltage
all frequencies

26 ALTERNATING CURRENT
Active site edutech- 9844532971

10. The power factor of a.c. circuit having L and 16. The average power transferred in series L-
R connected in series to an a.c. source of C-R circuit is given by (where symbol have
angular frequency  is given by their usual meaning)

R 1) Pav = Vrms.Irms.sin 
R 2  2 L2
1) 2)
R R 2  2 L2 R
2) Pav = Vrms.Irms.
Z
L R
3) 4) 3) Pav = V0.I0.cos 
R L
11. The capacitor offers zero resistance to 4) Pav = V0.I0.sin 
1) D.C. only 2) A.C. & D.C. 17. The amplitude of current oscillations in LCR
3) A.C. only 4) neither A.C. nor D.C. circuit will be maximum when  is
12. Power factor is defined as
1) As large as possible
1) apparent power/true power
2) Equal to natural frequency of LCR system
2) true power/apparent power
3) true power (apparent power)2 1
3) LC 4)
4) true power x apparent power LC

13. A 200V, 50Hz A.C. source is applied across 18. A series LCR circuit is tuned to resonance.
the LCR series circuit . If XC > XL then The impedance of the circuit now is
1) Current lags the applied voltage  1/2
 2  1  
2) Current leads the applied voltage 1) R   L   
   C 
3) Current and voltage are in same phase 1/2
 2  1  
2

    
2
4) Phase relationship can’t be determined 2)  R   L 
  C  
14. The graph between current and frequency
1/2
for LCR series circuit is drawn for three  2  1  
2

different resistor R1, R2 & R3. Which one 3)  R    L  

  C  
is correct regarding resistance
4) R
R1
R2 19. At resonance, in a series LCR circuit, which
i0 R3 relation does not hold?
1 1
1)  2)  
 LC LC
1) R1 > R2 < R3 2) R3 > R2 > R1 1 1
3) L  4) C 
3) R1 = R2 = R3 4) R1 > R2 > R3 C L

15. In series L-C-R circuit the quality factor is 20. In an AC circuit, with voltage V and current I
the power dissipated is
1 R
1) 2) 1 1
R LC LC 1) VI 2) VI 3) VI
2 2
1 L RC
3) 4) 4) Depends on the phase angle between V and I
R C L

ALTERNATING CURRENT 27
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21. The figure shows variation of R, XL and XC 26. Power factor of the A.C. circuit given in
with frequency f in a series L, C, R circuit. figure is 0.6. The value of R is
Then for what frequency point, the circuit is
X = 10
inductive C

XC XL

200V, 50 Hz

R
1) 3  2) 7.5  3) 4  4) 10 
A B C 27. A 30 mH pure inductor is connected to 220V,
f

1) A 2) B 3) C 4) All points 50 Hz A.C. supply. Net power absorbed by

22. For a series LCR circuit the power loss at the circuit over a complete cycle is
resonance is 1) Zero 2)10W 3) 12W 4) 5W
V2
28. A resistor, a capacitor of 100µF capacitance
1) L  1  2) I2 L
 C  and an inductor are in series with on AC
V 2
source of frequency 50 Hz. If the current in
3) I2R 4)
C the circuit is in phase with the applied
23. The power loss in an AC circuit will be voltage. The inductance of the inductor is
minimum when:
1) 0.1 H 2) 0.3 H
1) Resistance is high, inductance is high
2) Resistance is high, inductance is low 3) 0.5 H 4) 0.7H
3) Resistance is low, inductance is low 29. In series L-C-R circuit as shown in figure
4) Resistance is low, inductance is high the voltage of the source is
24. In the given A.C. circuit the average power
VR= 10V VC= 10V
consumed in the circuit is
R = 5 VL= 10V

I = 2A


200V, 50 Hz 10
1) 10 2 volt 2) volt
1) 20 W 2) 30 W 3) 40 W 4) 100 W 2

25. The circuit shown in the diagram is in 3) 10 3 volt 4)10 volt

resonance. Power factor of the circuit is
R=10 L=5H 30. In an LCR circuit having L = 8 henry,
C=2µF C = 0.5 F and R  100 ohm in series, the
resonance frequency in hertz is

220V,50Hz 1) 600 2) 600
1) 0.3 2) 0.6 3) 1.0 4) 0.9
3) 250 /  4) 5000

28 ALTERNATING CURRENT
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31. An A.C. circuit using an inductor and a 3) 2.5 watt 4) 5 watt.

capacitor in series has a max. current. If 37. Rms voltage of 110 V is applied across a series
L = 0.5 henry & C  8 F, then the angular circuit having a resistance 11  and
frequency of A.C. voltage will be (in rad/s) impedance 22  . The power consumed is
1) 500 2) 5  10 5 3) 4000 4) 5000 1) 275 W 2) 366 W
32. The resonant frequency of a circuit of 3) 550 W 4) 1100 W
negligible resistance containing an
38. A coil of inductive reactance 31  has a
inductance of 50 mH and a capacitance of
resistance of 8  . It is placed in series with
500 pf is
a condenser of capacitive reactance 25  .
5
10 1 The combination is connected to an a.c.
1) Hz 2)
  source of 110 volt. The power factor of the
100 1000 circuit is
3) Hz 4) Hz
 
1) 0.33 2) 0.56 3) 0.64 4) 0.80
33. If a power of 100 W is being supplied across
39. An ac ammeter is used to measure current
a potential difference of 200 V, current
in a circuit. When a given direct current
flowing is
passes through the circuit, the ac ammeter
1) 2A 2) 0.5A 3)1A 4) 20A reads 3A. When an AC passes through it,
34. In an A.C. circuit, voltage applied is then it reads 4A. Then the reading of this
V = 220 sin 100 t. If the impedance is 110  ammeter, if dc and ac flow through the circuit
and phase difference between current and simultaneously is
voltage is 60 , the power consumption is 1) 3A 2) 4A 3) 7A 4) 5A
equal to
40. In a circuit L, C and R are connected in
1) 55 W 2) 110W 3) 220 W 4) 330 W series with an alternating voltage source of
35. An alternating voltage is applied across the frequency f. The current leads the voltage
R-L combination V  220 sin 120t and the by 45°. The value of C is
current I = 4 sin 120t  60 develops. The 1 1
power consumption is 1) 2f  2fL  R  2) f  2fL  R 

1) zero 2) 110 W 3) 220 W 4) 440 W

1 1
36. In an A.C. circuit, V and I are given by 3) 2f  2fL  R  4) f  2fL  R 

V = 100 sin (100t) volt. 41. In an LCR circuit, the capacitance is made
I = 100 sin (100t +  /3) mA. one-third , then what should happen to
inductance, so that the circuit remains in
The power dissipated in the circuit is resonance?
1) 104 watt 2) 10 watt 1) 8times 2) 1/3times 3) 2times 4) 3 times

ALTERNATING CURRENT 29
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42. A lamp consumes only 50% of power in an a frequency that is 10% lower than the
A.C. circuit. What is the phase difference resonant frequency is
between the applied voltage and the circuit 1) 0.5 A 2) 0.7 A
current
3) 0.9 A 4) 1.1 A
   
1)
6
2)
3
3)
4
4)
2 48. When an AC source of emf e  E0 sin 100t 
is connected across a circuit, the current
43. Radio waves with wavelength 300  metre
leads emf by a phase of 450 , if the circuit
are transmitted from a transmitter. An
consists possibly of R  C or R  L or L  C
inductive coil connected in series with a
in series, find the relationship between the
capacitor of 1.0 F to receive these waves.
two elements:
The inductance of coil in henry is
1) R  1k , C  10  F
1) 2.5 × 10–9 2) 2.5 ×10–8
3) 1 × 10–7 4) 2.5 × 10–7 2) R  1k , C  1 F

44. The potential difference V across and the 3) R  1k , L  10 H

current I flowing through an instrument in 4) R  1k , L  1H
an AC circuit are given by V  5 cos t V 49. In LC circuit the capacitance is changed
and I  2sin t A .The power dissipated in from C to 4C. For the same resonant
the instrument is frequency, the inductance should be changed
1) zero watt 2) 5 watt from L to

3) 10 watt 4) 2.5 watt 1) 2L 2) L/2 3) L/4 4) 4L

45. A direct current of 5 2 amp. is

superimposed on an alternating current I
= 10 sin t flowing through a wire. The 50. In an ac circuit V and I are given by
effective value of the resulting current will V  200sin100t volt and
be I  200sin 100t   / 3 milliampere. The
power dissipated in the circuit is:
1) (15/2) amp 2) 10 amp
1) 10 4 watt 2) 10 watt
3) 5 5 amp 4) 15 amp
3) 2.5 watt 4) 5 watt
46. A 60V and10 watts bulb is operated at 100
V – 60 Hz ac. The inductance required to 51. A voltage 10V and frequency 10 Hz is
3

1
be connected in series with the bulb is applied to  F capacitor in series with a

1) 2.28 H 2) 1.28 H 3) 2H 4) 1 H resistor of 500 . Find the power dissipated

47. An LCR circuit has L = 10 mH, R = 3 , 1) 500 / 2W 2) 1/100W

and C = 1  F connected in series to a source
3) 10 2W 4) 1/10W
of 15 cos  t volt. The current amplitude at
30 ALTERNATING CURRENT
Active site edutech- 9844532971

1 10mC. At resonance
52. If power factor of a R- L series circuit is
2 a) What is the total energy stored initially?
when applied voltage is V  100sin100 t
b) What is the natural frequency of the
volt and resistance of circuit is 200 then circuit?
the inductance of the circuit is
1) 10J, 103 Hz 2) 1J ,103 Hz
1) 2 3H 2) 3 / H
3) 1J ,159 HZ 4) 10 J ,15.92 J
2 3 2
3) H 4) H 57. For the series LCR circuit shown in figure,
 
the resonating frequency and current
53. A circuit consisting of an inductance and a
amplitude at resonance respectively are
resistance joined to a 200 volt supply (A.C).
8 mH
It draws a current of 10 ampere. If the power
used in the circuit is 1500 watt, then the
wattless current component is 220V 20F

10 7
1) 10 7A 2) A 44 
4
3) 5 / 4A 4) 4A 1) 2500 rad s-1 and 5 2A
54. The power factor for the circuit shown below 2) 2500 rad s-1 and 5A
is
5
3) 2500 rad s-1 and A
XL = 100 R = 60 XC = 20 2

L R C 5
4) 250 rad s-1 and A
2

58. In the circuit shown in figure, what will be

220 V, 50 Hz the reading of the voltmeter?

1) 3 / 5 2) 1/ 2 3) 4 / 5 4) 3 / 2 100V 100V
V
55. A series LCR circuit is connected across an L

ac source E  10sin 100 t   . Current R C
 6

  200 V, 100 Hz
from the supply is I  2sin 100 t   . The
 12 
1) 300V 2) 900V 3) 200V 4) 400V
average power dissipated is
59. A series LCR circuit has R  5 , L  40
1) 10 2W 2) 5W 3) 5 2W 4) 5 / 2W
mH and C  1 F , the bandwidth of the
56. An A.C circuit contains a 20mH inductor and circuit is
a 50  F capacitor with an initial charge of 1) 10Hz 2) 20Hz 3) 30Hz 4) 40Hz

ALTERNATING CURRENT 31
Active site edutech- 9844532971

60. A pure resistive circuit element X when particular frequency

connected to an ac supply of peak voltage
200V gives a peak current of 5A which is in 4) remains constant
phase with the voltage. A second circuit
element Y, when connected to the same 64. In the series LCR circuit as shown in the
supply also gives the same value of peak figure, the voltmeter V and ammeter A read-
current but the current lags behind by 90 .
0
ings are
If the series combination of X and Y is
400 V 400 V
connected to the same supply, what will be
V
the rms value of current ?

10 5
1) A 2) A 3)  5 / 2  A 4) 5A
R = 50 L C

2 2 A

61. In a series resonance LCR circuit, the volt-

age across R is 100V and
100 V, 50 Hz
R = 1 k , C  2  F . The resonant fre-
quency is 200 rad/s. At resonance, the volt- 1) V = 100V, I = 2A
age across L is 2) V = 100V, I = 5A
1) 40V 2) 250V 3) V = 400V, I = 2A
3) 4  10 V
3 4) 2.5  10 V
2
4) V = 300 V, I = 2A
62. An ideal resistance R, ideal inductance L, 65. In a series resonant LCR circuit, the volt-
ideal capacitance C and ac voltmeters V1, age across R is 100V and the value of
V2, V3 and V4 are connected to an ac source
R  1000 . The capacitance of the capacitor
as shown. At resonance.
is 5  10 6 F ; angular frequency of ac is
V4
R C 200 rad s 1 . Then the potential difference
L
V1 V2 V3 across the inductance coil is
1) 100V 2) 40V
3) 250V 4) 400V
1) reading in V3 = reading in V1
66. The figure shows a LCR network connected
2) reading in V1 = reading in V2 to 300V ac supply. The circuit elements are
3) reading in V2 = reading in V4 such that R  X L  X C  10 V1, V2 and V3
4) reading in V2 = reading in V3 are three ac voltmeters connected as shown
in the figure. Which of the following repre-
63. A series LCR circuit is connected to an ac sents the correct set of readings of the volt-
source of variable frequency. When the fre- meters ?
quency is increased continuously, starting V1 V2 V3
from a small value, the power factor
1) goes on increasing continuously
R L C
2) goes on decreasing continuously
3) becomes maximum at a 300 V
32 ALTERNATING CURRENT
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1) V1  100V , V2  100V , V3  100V 70. An LC circuit contains a 40mH inductor and

a 25  F capacitor. The resistance of the
2) V1  150V , V2  0, V3  150V
circuit is negligible. The time is measured
3) V1  300V , V2  100V , V3  100V from the instant the circuit is closed. The
4) V1  300V , V2  300V , V3  300V energy stored in the circuit is completely
magnetic at times ( in milliseconds)
67. As given in the figure, a series circuit
1) 0, 3.14, 6.28
connected across a 200V, 60Hz line consists
of a capacitor of capacitive reactance 30 , 2) 0, 1.57, 4.71
a non-inductive resistor of 44 , and a coil 3) 1.57, 4.71, 7.85
of inductive reactance 90 and resistance 4) 1.57, 3.14, 4.71
36 . The power dissipated in the coil is 71. An alternating supply of 220V is applied
XC = 30 across a circuit with resistance 22 and
impedance of 44 . The power dissipated
200V R1 = 44  in the circuit is
60 Hz
1) 1100W 2) 550W
Coil with 3) 2200W 4) (2200/3)W
XL = 90
R2 = 36 
72. In an ac circuit, the current is
1) 320 W 2) 176W  
I  5sin 100t   ampere and the poten-
3) 144W 4) 0W  2
RESONANCE & POWER IN AC CIRCUIT tial difference is V  200sin 100t  volt.
Then the power consumed is
68. In an ac circuit the reactance is equal to the
1) 200 watt 2) 500watt
resistance. The power factor of the circuit
3) 1000watt 4) zero
will be
73. A capacitor of capacitance 1 F is charged
1 1 to a potential of 1V. It is connected in par-
1) 1 2) 3) 4) zero
2 2 allel to an inductor of inductance 103 H .
69. An inductance L, a capacitance C and a re- The maximum current that will flow in the
circuit has the value
sistance R may be connected to an ac source
of angular frequency  , in three different 1) 1000 mA 2) 1 mA
combinations of RC, RL and LC in series. 3) 10 mA 4) 1000 mA
1 74. A coil of inductance 0.1H is connected to
Assume that  L  . The power drawn
C 50V, 100Hz generator and current is found
by the three combination are P1 , P2 , P3 re- to be 0.5A. The potential difference across
spectively. Then, resistance of the coil is
1) P1  P2  P3 2) P1  P2  P3 1) 15V 2) 20V
3) P1  P2  P3 4) P1  P2  P3 3) 25V 4) 39V

ALTERNATING CURRENT 33
Active site edutech- 9844532971

75. An alternating voltage ( in volts) given by

V  200 2 sin 100t  is connected to 1 F
capacitor through an ideal ac ammeter in CRTQ: 01) 2 02) 3 03) 4 04) 4
series. The reading of the ammeter and the 05) 1 06) 1 07) 3 08) 4
average power consumed in the circuit shall 09) 3 10) 2 11) 4 12) 2
be 13) 2 14) 2 15) 3 16) 2
1) 20 mA, 0W 17) 4 18) 4 19) 1 20) 4
2) 20mA, 4W 21) 3 22) 3 23) 4 24) 1
3) 20 2mA,8W 25) 3 26) 2 27) 1 28) 1

4) 20 2mA, 4 2W 29) 4 30) 3 31) 1 32) 1

33) 2 34) 2 35) 3 36) 3
76. In an ac circuit, V and I are given by
37) 1 38) 4 39) 4 40) 1
V  150sin 150t  volt and
41) 4 42) 2 43) 4 44) 1
  45) 2 46) 2 47) 2 48) 1
I  150sin  150t  
 3 49) 3
ampere. the power dissipated in
SPQ: 50) 2 51) 4 52) 3 53) 2
the circuit is
54) 1 55) 3 56) 3 57) 1
1) 106W 2) 150W
58) 3 59) 2 60) 3 61) 2
3) 5625 W 4) zero
62) 4 63) 3 64) 1 65) 1
77. A series resonant LCR circuit has a quality
66) 4 67) 1 68) 3 69) 3
factor (Q- factor) 0.4. If R  2k  ,
70) 3 71) 2 72) 4 73) 1
C = 0.1  F , then the value of inductance is
74) 4 75) 1 76) 3 77) 2
1) 0.1H 2) 0.064H
78) 1
3) 2H 4) 5H
78. A resistor of 500 , an inductance of 0.5H
are in series with an ac which is given by
V  100 2 sin 1000t  . The power factor of 1. At resonance   00
the combination is
1
1 1 2. f 
2 LC
1) 2)
2 3 3 Formula
3) 0.5 4) 0.6 4. In LCR circuit, phase difference is between
0 to 900
5.   00

34 ALTERNATING CURRENT
Active site edutech- 9844532971

6. X L  X C , emf leads current  2R

23. Since P  ; For the power loss to be
7. For pure resistor I and V lies in same phase. Z2
Hence cos   1 . less, R should be less. Also for small values
of R, Z  X L . So X L should be high.
8. In pure inductor, phase difference between
 V 200
emf and current = 24. P  VI cos  I Z  100
2 Z 22
XL V R V2
9.  same, then X L  XC V   R
XC Z Z Z2
2
R R  200 
10. cos    P   5  20W
Z R 2   L 
2  100 

11. capacitor does not offer zero resistance for R R

24. cos    ,
both AC and DC Z R2   X L  X C 
2

12. Formula 1
X L  L , X C 
13. Current lags voltage C

V 1 and resonance, X L = X C  cos   1

14. At resonance I  I
R R R VR
26. cos   =
R1  R2  R3 Z V

VL 1 L
15. Q-factor=  VZ   200   120 
2 2
 160V  IX L  10 I
VR R C
16. formula VR  IR I  16 A
17. Amplitude of current oscillations will be max.
at resonance; 120  16R
30
L 
1
, 
1 R  7.5
C LC
4

18. At resonance, L  I / C  Z  R .  
27.   , P  VI cos 0
2 2
1
19.   is not true (dimensionally). 1
LC 28. f 
20. In a.c. circuit, average power dissipated 2 LC

P  E r Ir cos  , where  is phase angle 1 1

L    0.1H
between V and I. 4 fc 4 50104 2
2 2

Depends on phase between V and I.

29. At resonance V  VR  10V
21. At A : XC > XL ; At B : XC = XL
; At C : XC < XL 1 1
30.   2 LC 
22. At resonance V = VR (as VX = 0) 2 8  0.5  10 6
1 250
VR2 V 2 I 2 R 2  hertz
P=    I 2R 2  2  10 3

R R R

ALTERNATING CURRENT 35
Active site edutech- 9844532971

1 I 2 R 25 R
31.   ; or I = 5A
LC J J

1 1 1 1
L  2 f L 
32.   2 LC   c  tan 45  2 fc
2 50  103  500  1012 40. tan  
R R
1 105
  Hz . 1 1
2  5  10 6   R  2 fL  C 
2 fc 2 f (2 fL  R)
33. I  P / V  100 /200  0.5A.
1 C
E I 41.   , c  ,   const  L '  3L
34. P  0 0 cos  LC 3
2
V0 I 0
EI 42. Pavg  cos 
35. P  Er Ir cos   0 0 cos  2
2
1  P0  P0 1 

220  4
cos 60  220W .    cos   cos   ,  
2 2  2 2 3
2
43. Resonance must occur
36. P  Vrms Irms cos 
3  108 106
  300 , f   ;
 100   100      2  10
6

 
3
 cos   10 W = 2.5W 300
 2  2   3 
1 1
37. P  Vrms Irms cos   2  106  L   2.5  10 7 H
LC C 4  1012

Vrms R 44. V  5 cos t volt = 5sin(t  90) volt

 Vrms  110  Irms   cos  
Z Z
I  2sin t amp
R R
38. cos    Phase difference   90
Z (XL  X C )2  R 2
P  E v Iv cos   E v Iv cos 90 = zero.
8
  0.8 45. The effective value of resulting current is
(31  25)2  82 calculated from total heat produced in a given
resistance
39. Quantity of heat liberated in the ammeter of
resistance R  10 
2

 
2
I2vR  5 2 R    R
(i) due to direct current of 3A =  2
 3  2 R / J ) 
  I2v  50  50  100
(ii) due to alternating current of Iv  10 A
4 A   4  R / J 
2
  46. Power of bulb Ps = 10W
Total heat produced per second Maximum voltage across bulb= VS = 60V
Maximum current passing through bulb
 3  4
2 2
R 25R R PS 10 1
   I   A
J J J VS 60 6
Let the equivalent alternating current be I Resistance of bulb
virtual A. then

36 ALTERNATING CURRENT
Active site edutech- 9844532971

VS2 60  60 R 1 R
R   360 52. cos    
PS 10 Z 2 2
Let inductance required is L  Z  2 R  R 2  X L2  2 R
Total impedance of circuit  X L  3R
Applied voltage 100
Z   600 3R
current 1/ 6  L  3R  L 

Also, Z2  R 2   L 
2

3  200 2 3
 2  60  120   H
100 
 600    360   620 L2
2 2
53. Apparent power = 200 10  2000W
Solving we get L = 1.28H  Power factor

90 90 1 True power 1500 3

47.  0  x =9000 rad/s cos    
100 100 LC Apparent power 2000 4
Wattless current =
E0
i0  2
2
 1  3 10 7
R  L 
2
I rms sin   10 1     A
 C  4 4
48. CR circuit 54. Power factor
XC  1 R 60
Tan  , Tan  Cos  
Z 602   X L  X C 
2
R 4 C R
1 60 60 3
R   
,verify the option 60  100  20 
2 100 5
C
2

49. 1 55. Phase difference between voltage and current

L  L1C 1  L 2 C 2
C
L    
L C  L2 4C  L2     
4 12  6  4

50. P  Vm I m cos  1
 Power factor = cos  
2
 
P  200  200   cos   103  10 Average power dissipated
 3
Vmlm 10  2 1
 cos     5 2 walt
 500   500  500 2
2 2
51. Z  R2  XC2  2 2 2

1q 2 1 10 10 
R 500 1 3 2
Power factor cos     , 56. a) U E     1.0 J
Z 500 2 2
Power dissipated 2C 2 50  106

10 
2
1 1 1 1
= Vrms I rms cos     W b)   
LC 20  10  50  10 6
3
500 2 2 10
 10 3 rad / sec  f  159 HZ

ALTERNATING CURRENT 37
Active site edutech- 9844532971

57. Here, R  44, V0 200V

XL    40
L = 8mH = 8 103 H I0 5A
In the series combination of X and Y
C  20 F  20  106 F
1 1 Z  R 2  X L2  402  402  40 2
r  
LC 8 10 3  20  106 Vrms V
I rms   0
1 104 Z 2Z
r  4
  2500 rad s 1
4 10 4 V
61. Current, I 
Z
V0 2Vrms 2  220
I0     5 2A
R R 44 VR   VL  VC 
2 2
where V 
58. Here, VR  ? , VL  VC  100V , V  200V
Z  R2   X L  X C 
2

As V  VR2  VL  Vc 
2

At resonance, X L  X C
 200  VR  100  100   VR
2 2
Z  R
1 VL  VC V  VR
59. Resonant angular frequency , r 
LC VR 100V
I   3  101 A
Quality factor, R 10 
Re sonant angular frequency At resonance,
Q
Band width 1
VL  VC  IX C 
r c
Bandwidth =  2  1 
Q 62. At resonance
Where r  resonant frequency,, Voltage across L = Voltage across C
Q  quality factor
 Reading in V2  reading in V3
, L
 Also, Q  63. Power factor becomes equal to one at the
R
resonant frequency
R R
2  1   V
2 LC r L 2 L 64. Current I  ;
Z
R 5
2  1    20 Hz As V  VR2  VL  VC 
2
2 L 2  40 103
60. As current is in phase with the applied voltage VR
X must be R, 65. The current in the circuit is I 
R
V0 200V At resonance
R   40
I0 5A
I
As current lags behind the voltage by 900, y VL  VC  IX C 
must be an inductor C

38 ALTERNATING CURRENT
Active site edutech- 9844532971

66. The impedance of the circuit

71. Current, I  V
Z
Z  R2   X C  X L 
2

Power dissipated in the circuit, P  I 2 R

Voltage across R is VR  IR
72. Power consumed, P  Vrms I rms cos 
Voltage across L is VL  IX L
73. Charge on the capacitor
Voltage across C is VC  IX C
q0  CV  1 10 6 1  106 C  1 C
67. Impedance of the circuit
Here, q  q0 sin t
 R1  R2    X L  X C 
2 2
Z
Maximum current, I 0   q0
X 1
68. tan   1 ;
Now,  
R LC
  tan 1 1  45 E 50
74. I  ; 0.5= Z  100
1 Z Z
Power factor = cos   cos 450 
2 Z 2  R 2   2 L2 ,
69. The LC circuit draws no power. then R = 78
1
When  L  Now VR  VLR2  VL2  39V ;
C
the impedance of the RC and LR circuits VR2  VL2  VLR2 

are equal, and hence they draw the same V0

75. Vrms 
power. 2
1
 P1  P2  P3 The capacitive reactance is X C 
C
1 3 Vrms
70.   = 10 Hz I rms 
2 LC 2 XC ;
The average power consumed in the circuit
1
T   2 ms P  I rmsVrms cos 

76. The power dissipated in ac circuit is
Energy stored is completely magnetic at times
1
T 3 5 P  V0 I 0 cos 
t  , T , T ,........ 2
4 4 4
1 L L
or   QR 
2
 77. Q 
Hence, t  ms, R C C
2
78. Impedance of the RL circuit is
3 5
ms, ms Z  R 2  X L2
2 2
 1.57 ms, 4.71ms, 7.85ms R
Power factor cos  
Z

ALTERNATING CURRENT 39
Active site edutech- 9844532971

+ q0
t
– q0
LC OSCILLATIONS

 A capacitor (C) and an inductor (L) are  However in an actual LC circuit, some resistance
connected as shown in the figure. Initially the is always present due to which energy is
charge on the capacitor is Q dissipated in the form of heat. So LC oscillation
i will not continue infinitely with same amplitude
as shown.
+ +
C L + q0
– –
t

– q0
 Energy stored in the capacitor
Q2 Let q be the charge on the capacitor at any time
UE 
2C
di
 The energy stored in the inductor, UB = 0. t and be the rate of change of current. Since
dt
The capacitor now begins to discharge through no battery is connected in the circuit,
the inductor and current begins to flow in the
circuit. As the charge on the capacitor decreases, q di
 L.  0
1 2 c dt
U E decreases but the energy U B  LI in the
2 dq
magnetic field of the inductor increases. Energy but i  
dt
is thus transferred from capacitor to inductor.
from the above equations, we get
When the whole of the charge on the capacitor
disappears, the total energy stored in the electric q d2q d2q 1
 L 2  0,  q  0 The above
field in the capacitor gets converted into C dt dt 2 LC
equation is analogous to
magnetic field energy in the inductor. At this stage,
there is maximum current in the inductor. d2r
2
 2 r=0 (differential equation of S.H.M)
dt
 Energy now flows from inductor to the capacitor
Hence on comparing
except that the capacitor is charged oppositely.
This process of energy transfer continues at a 1 1
2  ; 
definite frequency (v). Energy is continuously LC LC
shuttled back and forth between the electric field 1 1
2  ; 
in the capacitor and the magnetic field in the LC 2 LC
inductor.
The charge therefore oscillates with a frequency
If no resistance is present in the LC circuit, the 1
LC oscillation will continue infinitely as shown.  and varies sinusoidally with time.
2 LC
40 ALTERNATING CURRENT
Active site edutech- 9844532971

COMPARISION OF L - C
OSCILLATIONS WITH SHM :
 The L - C oscillations can be compared to S.H.M of a block attached to a spring
1
 In L - C oscillations  0 
LC
K
 In Mechanical oscillations  0  where K is the spring constant
m
1  V
 In L - C oscillations  tells us the
C  q 
potential difference required to store a unit charge
 F
 In a mechanical oscillation K    tells us the external force required to produce a unit displacement
x
of mass
 In L - C oscillations current is the analogous quantity for velocity of the mass in mechanical oscillations
 In L - C oscillations energy stored in capacitor is analogous to potential energy in mechanical oscillations
 In L - C oscillations energy stored in inductor is analogous to kinetic energy of the mass in mechanical
oscillations
 In L - C oscillations maximum charge on capacitor q0 is analogous to amplitude in mechanical oscilla-
tions
  As Vmax = A  in mechanical oscillations,
I 0  q0 0 in L- C oscillations
ENERGY OF LC OSCILLATIONS:
 Let q0 be the initial charge on a capacitor. Let the charged capacitor be connected to an inductor of
inductance L. LC circuit will sustain an oscillations with frequency At an instant t, charge q on the
capacitor and the current i are given by; q(t )  q0 cos t ; i  q0 sin t
Energy stored in the capacitor at time t is

1 1 q 2 q02
U E  CV 
2
 cos 2 (t )
2 2 C 2C
1 2
Energy stored in the inductor at time t is U M  Li
2
1 2 2 2 q2 1
 Lq0  sin t   0 sin 2 (t )(  2  ) Sum of energies
2 2C LC

q02 q02
UE UM  (cos t  sin t ) 
2 2

2C 2C
 As q0 and C, both are time independent, this sum of energies stored in capacitor and inductor is
constant in time. Note that it is equal to the initial energy of the capacitor.

ALTERNATING CURRENT 41
Active site edutech- 9844532971

i im i

++++ + + + + ––––
C L
–– –– – – – – ++++

q 2m 1 Lim2
UE = UB = 0 UE = 0 UB =
2c 2c

1 1 1
U= kA 2 , K= 0 U  0, K  0 U = 0, K = mv 2max U  0, K  0 U= kA 2 , K= 0
2 2 2

(a) (b) (c) (d) (e)

 Fig : The oscillations in an LC circuit are depicts the corres-ponding stage of a

analogous to the oscillation of a block at the mechanical system (a block attached to a
end of a spring. The figure depicts one-half spring). As noted earlier, for a block of a
of a cycle. mass m oscillating with frequency 0 , the
equation is
 At t = 0, the switch is closed and the
capacitor starts to discharge. As the current d2 x
 02 x  0
increases, it sets up a magnetic field in the dt 2

inductor and thereby, some energy gets

stored in the inductor in the form of  Here, 0  k / m , and k is the spring
2
magnetic energy: UB = (1/2) Li . As the constant. So, x corresponds to q. In case of
a mechanical system
current reaches its maximum value im, (at t =
T/4) as in Fig. (c), all the energy is stored in  dv   d2x 
F  ma  m    m  2 
the magnetic field: UB = (1/2) Li2m.  dt   dt 
You can easily check that the maximum For an electrical system,
electrical energy equals the maximum
di d 2q
magnetic energy. The capacitor now has no e  L  L 2
dt dt
charge and hence no energy. The current
now starts charging the capacitor, as in Fig.  Comparing these two equations, we see that
(d). This process continues till the capacitor L is analogous to mass m: L is a measure of
is fully charged (at t = T/2) Fig. (e). But it resistance to change in current.
is charged with a polarity opposite to its
1
initial state in Fig. (a). The whole process In case of LC circuit, 0  and for
LC
just described will now repeat itself till the
mass on a spring, 0  k / m . So, 1/C is
system reverts to its original state. Thus, the
analogous to k. The constant k (=F/x) tells
energy in the system oscillates between the
capacitor and the inductor. us the (external) force required to produce a
unit displacement whereas 1/C (=V/q) tells
 The LC oscillation is similar to the us the potential difference required to store a
mechanical oscillation of a block attached to unit charge. Table gives the analogy between
a spring. The lower part of each figure
mechanical and electrical quantities.

42 ALTERNATING CURRENT
Active site edutech- 9844532971

ANALOGIES BETWEEN MECHANICAL AND primary as well as secondary, iP = current in

ELECTRICAL QUANTITIES primary; iS = current in secondary (or load
current)
Mechanical System Electrical System
As in an ideal transformer there is no loss of
Mass m Inductance L
power i.e. Pout = Pin so VSiS - VPiP and VP
Force constant k Reciprocal
capacitance l/C  eP and VS  eS. Hence
Displacement x Charge q
Velocity v = dx/dt Current I = dq/dt eS NS VS i P
    k; [k
Mechanical energy Electromagnetic e P N P VP iS
energy
1 2 1 2 1 q2 1 2 = Transformation ratio (or turn ratio)]
E  kx  mv U  L
2 2 2C 2 Step up transformer Step down transformer
TRANSFORMER It increases voltage It decreases voltage
and decreases current and increases current
 It is a device which raises or lowers the
voltage in ac circuits through mutual P S P S

induction.
VS > VP VS < V P
 It consists of two coils wound on the same
N S > NP N S < NP
core. The alternating current passing through
ES > EP ES < EP
the primary creates a continuously changing
iS > iP iS < iP
flux through the core. This changing flux
RS > RP RS < RP
induces an alternating emf in the secondary.
tS > t P tS < tP
(a) Transformer works on ac only and never on k>l k<l
dc.
(b) It can increase or decrease either voltage or
(h)Efficiency of transformer (  ) : Efficiency
current but not both simultaneously.
is defined as the ratio of output power and
(c) Transformer does not change the frequency of input power
input ac.
out P S S Vi
(d) There is no electrical connection between the i.e. %  P  100  V i  100
in P P
winding but they are linked magnetically.
(e) Effective resistance between primary and For an ideal transformer Pout = Pin so
secondary winding is infinite.  =100% (But efficiency of practical
transformer lies between 70%–90%)
(f) The flux per turn of each coil must be same
dS d For practical transformer Pin = Pout + Plosses
i.e. S  P :   P
dt dt Pout Pout  P P 
so  100  100  in L 100
(g) If NP = Number of turns in primary, NS = Pin Pout PL  Pin
number of turns in secondary, VP = applied
(input) voltage to primary, VS = Voltage (i) Losses in transformer : In transformers
across secondary (load voltage or output), some power is always lost due to heating
eP = induced emf in primary; eS = induced effect, flux leakage, eddy currents, hysteresis
emf in secondary,  = flux linked with and humming.

ALTERNATING CURRENT 43
Active site edutech- 9844532971

(i) Cu loss (i2R) : When current flows Thus, some part (may be very small) of
through the transformer windings some the electrical energy is wasted in the
power is wasted in the form of heat (H = form of humming sounds produced by
i2Rt). To minimize this loss windings are the vibrating core of the transformer.
made of thick Cu wires (To reduce (j) Uses of transformer : A transformer is
resistance) used in almost all ac operations e.g.
(ii) Eddy current loss : Some electrical (i) In voltage regulators for TV, refrigerator,
power is wasted in the form of heat due computer, air conditioner etc.
to eddy currents, induced in core. To (ii) In the induction furnaces.
minimize this loss transformers core are (iii) Step down transformer is used for
laminated and silicon is added to the core welding purposes.
material as it increases the resistivity. The
(iv) In the transmission of ac over long
material of the core is then called silicon-
distance.
iron (steel).
Transmission lines
(iii) Hysteresis loss : The alternating
current flowing through the coils
G  Low
V
High
V
High
V
Low
V
Load
House or
factory

magnetises and demagnetises the iron Power Step up Step down

Station Transformer Transformer
core again and again. Therefore, during (v) Step down and step up transformers are
each cycle of magnetisation, some energy used in electrical power distribution.
is lost due to hysteresis. However, the
(vi) Audio frequency transformers are used
loss of energy can be minimised by
in radiography, television, radio,
selecting the material of core, which has
telephone etc.
a narrow hysteresis loop. Therefore core
(vii)Radio frequency transformers are used
of transformer is made of soft iron. Now
in radio communication.
a days it is made of “Permalloy” (Fe-
22%, Ni-78%) (viii) Transformers are also used in impedance
matching.
(iv) Magnetic flux leakage : Magnetic flux
SKIN EFFECT:
produced in the primary winding is not
completely linked with secondary  A direct current flows uniformly throughout the
because few magnetic lines of force cross section of the conductor.
complete their path in air only. To  An alternating current, on the other hand,
minimize this loss secondary winding is flows mainly along the surface of the
kept inside the primary winding. conductor. This effect is known as skin effect.
(v) Humming losses : When alternating  When alternating current flows through a
current passes, the core of transformer conductor, the flux changes in the inner part of
starts vibrating and produces humming. the conductor are higher.

44 ALTERNATING CURRENT
Active site edutech- 9844532971

 Therefore, the inductance of the inner part is i2 N1 103 1

then i  N  2104  20
higher than that of the outer part. Higher the 1 2

frequency of alternating current, more is the  1 25

skin effect.  i 2  i1   20   20 1.25A or
 
 The depth upto which ac current flows through Alternatively, e2i 2  5103 W
a wire is called skin depth.  
VR2 5103
R  RVR2  i2   1.25A
PR 4103
( VR  Rated voltage, PR  Rated power) Illustration 18: A 1µF condenser is charged to
The working principle of metal 50V. The charging battery is then
detector disconnected and a 10 mH coil is
Metal detector works on the principle of connected across capacitor so that the LC
resonance in ac circuits. When a person walks oscillations occur. Find the maximum
through a metal detector, in fact he walks current in the coil?
through a coil of many turns connected to a Solution:
capacitor tuned suitably. If he carries some
Maximum charge stored on capacitor
metal, impedance of the circuit changes,
bringing significant changes in the current. This q0 = CV = 1 × 10–6 × 50 = 5 × 10–5C
change in current is detected and the electronic Maximum current in the inductor
circuit causes the alarm.
q0 5  105
IIllustration 17: A power transformer is used I0    0.5A
LC 10  103  106
to step up an alternating emf of 200 V to 4
KV and to transmits 5 KW power. If the Illustration 19: An inductor of 2H inductance
carries a current of 2A. To prevent
primary is of 1000 turns, calculate,
sparking when the circuit is broke
assuming the transformer to be ideal.
capacitor of 4 µF is connected across
(i) The number of turns in the inductor. Find the voltage rating of
secondary capacitor.
(ii) The current rating of the secondary Solution:To prevent sparking, the capacitor
Solution: must be such that, it can hold all the
energy contained by the inductor
e2 N 2
(i) For an ideal transformer  1
e1 N1 Energy stored in capacitor UC  CV2
2
4000 Ns (where V is voltage of capacitor)
  N s  2 104 turns
200 1000
1
(ii) P = (power to be transferred) = e1i1 Energy stored in inductor U L  LI 2
2
P 5103
 i1   = 25 A 1 1 2 2
e1 200 CV 2  LI2  V  LI  2  26 = 1414 V
2 2 C 4 10

ALTERNATING CURRENT 45
Active site edutech- 9844532971

2) the bulb will get fused

3) the bulb will glow, but with less brightness
4) the bulb will not glow
5. The ratio of primary voltage to secondary
voltage in a transformer is ‘n’. The ratio
TRANSFORMER AND LC OSCILLATIONS
of the primary current to secondary
1. The core of a transformer is laminated so
current in the transformer is
that
1) n 2) 1/n 3) n2 4) 1/n2
1) energy loss due to eddy currents may be
6. In a step down transformer, the number of
reduced
turns in the primary is always
2) rusting of the core may be prevented
1) greater than the number of turns in the
3) change in flux may be increased secondary
4) ratio of voltage in the primary to that in the 2) less than the number of turns in the
secondary may be increased secondary
2. A step up transformer is used to 3) equal to the number of turns in the
1) increase the current and increase the secondary
voltage 4) either greater than or less than the number
of turns in the secondary
2) decrease the current and increase the
voltage 7. The transformer ratio of a step up
transformer is
3) increase the current and decrease the
1) greater than one
voltage
2) less than one
4) decrease the current and decrease the
voltage 3) less than one and some times greater than
one
3. A transformer changes the voltage
4) greater than one and some times less than
1) without changing the current and frequency one
2) without changing the current but changes 8. A transformer develops 400V in
secondary coil for an input of 200V A.C.
the frequency
Then the type of transformer is
3) without changing thefrequencybut changes the 1) Step down
current 2) Step up
4) without changing the frequency as well as the 3) Same
current 4) Same but with reversed direction
4. A step up transformer is connected on the 9. Transformers are used in
primary side to a rechargable battery which 1) d.c circuits only
can deliver a large current. If a bulb is 2) a.c. circuits only
connected in the secondary, then
3) Both a.c and d.c circuits
1) the bulb will glow very bright 4) Integrated circuits.

46 ALTERNATING CURRENT
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10. The magnitude of the e.m.f. across the secondary primary and 200 windings in the secondary.
of a transformer does not depend on The primary is connected to a.c. supply of
1) The number of the turns in the primary 120 volts at 10 amperes. Check the correct
situation for this transformer out of the
2) The number of the turns in the secondary
following: (Assume that transformer is ideal)
3) The magnitude of the e.m.f applied across
1) The secondary voltage 240 volts and current
the primary
10 amperes
4) The resistance of the primary and the
2) The secondary voltage 240 volts and current
secondary
5 amperes
11. For an ideal transformer ratio of output to
3) The secondary voltage 60 volts and current
the input power is always
10 amperes
1) greater than one 2) equal to one
4) The secondary voltage 240 volts and current
3) less than one 4) zero 20 amperes
12. To reduce the iron losses in a transformer, 16. A transformer has an efficiency of 80% and
the core must be made of a material having works at 100 volt and 4 kw. If the secondary
1) low permeability and high resistivity voltage is 240V. The current in secondary is

2) high permeability and high resistivity 1) 13.3 A 2) 20 A

3) low permeability and low resistivity 3) 10 A 4) 5 A

4) high permeability and low resistivity 17. A small town with a demand of 800 kw of
electric power at 220 V is situated 15 km away
13. In a transformer as the voltage increases :
from an electric plant generating power at
1) Current also increases 440 V. The resistance of the two wire line
2) Current decreases carrying power is 0.5  per km . The town
gets power from the line through a 4000/220
3) Current will remain unchanged
step down transformer at a substation in the
4) we cant predict about current town. The power loss in the line is
14. A transformer is used to illuminate a bulb of 1) 300 k w 2) 500 k w
(36 W and 12V) with the help of 220 volt
3) 600 k w 4) 700 k w
mains. If the efficiency of the transformer
is 75 %, then current in primary coil is 18. In step down transformer 220/110V the
primary is connected to 10 V battery. The
1) 0.3 A 2) 0.42 A
out put voltage is
3) 0.22 A 4) 0.5 A
1) 5 V 2) 10 V
15. A transformer has 100 windings in the
3) 110 V 4) Zero

ALTERNATING CURRENT 47
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19. A transformer steps up an A.C. voltage from 

1) LC 2) 2 LC
4
230 V to 2300 V. If the number of turns in
the secondary coil is 1000, the number of 3) LC 4)  LC
turns in the primary coil will be 25. An AC source of angular frequency  is fed
across a resistor R and a capacitor C in
1) 100 2) 10,000 3) 500 4) 1000
series. The current registered is I. If now
20. The transformer ratio of a transformer is 5. the frequency of source is changed to  / 3
If the primary voltage of the transformer is (but maintaining the same voltage), the
400 V, 50 Hz, the secondary voltage will be current in the circuit is found to be halved.
1) 2000 V, 250 Hz 2) 80 V, 50 Hz The ratio of reactance to resistance at the
original frequency  is
3) 80 V, 10 Hz 4) 2000 V, 50 Hz
21. A step-up transformer works on 220V and 3 5 3 5
1) 2) 3) 4)
gives 2 A to an external resistor. The turn 5 3 5 3
ratio between the primary and secondary coils 26. Turn ratio in a step up transformer is 1:2 if
is 2:25. Assuming 100% efficiency, find the a Lechlanche cell of 1.5V is connected
secondary voltage, primary current and power across the input then the voltage across the
delivered respectively output will be:
1) 2750 V, 25 A, 5500 W 1) 0.1V 2) 1.5V
2) 2750 V, 20 A, 5000 W 3) 0.75V 4) zero
3) 2570 V, 25 A, 550 W 27. The efficiency of a transformer is 98%. The
4) 2750 V, 20 A, 55 W primary voltage and current are 200 V and
22. The transformer ratio of a transformer is 10:1. 6A. If the secondary voltage is 100 V, the
The current in the primary circuit if the secondary current is
secondary current required is 100 A assuming 1) 11.76 A 2) 12.25 A
the transformer be ideal, is
3) 3.06 A 4) 2.94 A
1) 500 A 2) 200 A
28. A condenser of capacity C is charged to a
3) 1000 A 4) 2000 A
potential difference of V1. The plates of the
23. The transformer ratio of a transformer is
condenser are then connected to an ideal induc-
10:1. If the primary voltage is 440V,
tor of inductance L. The current through the
secondary emf is
inductor, when the potential difference
1) 44 V 2) 440V
across the condenser reduces to V2 is?
3) 4400 V 4) 44000 V
C 
1) V V 
2 2

2) V 1 V 2
C 
2 2

24. A fully charged capacitor C with initial 1 2

L L
charge q0 is connected to a coil of self
   C V12 V22  
1/2
2 2 1/ 2
inductance L at t  0 . The time at which the C V 1 V 2
energy is stored equally between the electric 3) 4)  L


L  
and the magnetic fields is
48 ALTERNATING CURRENT
Active site edutech- 9844532971

1) 22V 2) 2200V
3) 220V 4) zero
29. The ratio of the secondary to the primary
35. The number of turns in primary and
turns in a transformer is 3:2 and the output
power is P. neglecting all power losses, the secondary coils of a transformer is 50 and
input power must be :- 200. If the current in the primary coil is 4A,
P 2P 3P then the current in the secondary coil is
1) 2) P 3) 4) (assume ideal transformer)
2 3 2
30. The primary and secondary coils of a 1) 1 A 2) 2 A 3) 4 A 4) 5 A
transformer have 50 and 1500 turns
36. A transformer has 1500 turns in the primary
respectively. If the magnetic flux “ linked
coil and 1125 turns in the secondary coil. If
with the primary coil is given by   0  4t ,
the voltage in the primary coil is 200 V, then
where  is in Webbers, t is time in seconds
and 0 is a constant, the output voltage the voltage in the secondary coil is
across the secondary coil is 1) 100 V 2) 150 V
1) 30 volts 2) 90 volts 3) 200 V 4) 250 V

3) 120 volts 4) 220 volts 37. In the primary coil of transformer current
and voltage are 5A and 220 volts. In the
31. If 2.2 kW power transmits 22000 volts in a
secondary coil 2200V voltage produces.
line of 10  resistance, the value of power
Then ratio of number of turns in secondary
loss will be:-
coil and primary coil will be (assume ideal
1) 0.1 watt 2) 14 watts transformer)
3) 100 watts 4) 1000 watts 1) 1:10 2) 10 :1
32. In a step up transformer, the ratio of turns
3) 1 :1 4) 11:1
of primary to secondary is 1:10 and primary
38. The ratio of number of turns in primary to
voltage is 230 V. If the load current is 2A,
then current in the primary is (assume ideal secondary coils of transformer is given as
transformer) 2:3. If the current through the primary coil
1) 20A 2) 10A 3) 2A 4) 1A is 3A, then the current through load
resistance is (assume ideal transformer)
33. In a transformer, number of turns in the
primary are 140 and that in the secondary 1) 1A 2) 4.5 A 3) 2 A 4) 1.5 A
are 280. If current in primary is 4A then 39. In a electrical circuit consisting of an
that in the secondary is (assume ideal inductance ‘L’ and a capacitance ‘C’ at
transformer) resonance. The time period of oscillations
1) 4A 2) 2A 3) 6A 4) 10A of charge is
34. The number of turns in primary and
L
secondary windings of a transformer are 1) T  2 2) T  2 LC
C
1000 and 100 respectively. If 200V dc
voltage is impressed across the primary
2 C
terminals, the voltage across secondary 3) T  4) T  2
LC L
terminals is
ALTERNATING CURRENT 49
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40. An A.C circuit contains a resistor ‘R’ an 44. A step down transformer is used on a 1000V
inductor ‘L’ and a capacitor ‘C’ connected line to deliver 20A at 120V at the second-
in series. When it is connected to an A.C ary coil. If the efficiency of the transformer
generator of fixed output voltage and is 80%, the current drawn from the line is
variable frequency, the current in the circuit 1) 3A 2) 30A
is found to be leading the applied voltage 3) 0.3A 4) 2.4A

rad, when the frequency is f1. When the 45. A power transmission line feeds power at
4 2300V to a step down transformer with its
frequency of the generator increased to ‘f2’ primary windings having 4000 turns. What
the current is found to be lagging behind the
should be the number of turns in the sec-
 ondary windings in order to get output power
applied voltage by rad. The resonant
4 at 230V ?
frequency of the circuit is
1) 200 2) 400
1) f  f1 f 2 2) f  f1 f 2 3) 600 4) 800
46. If i1 = 3 sin t and i2  4 cos t , then i3
f1 f 2 f1 is
3) f  4) f  f2 i1 i2
2

41. A transformer is used to light 140 watt 24

i3
volt lamp from 240 volt ac mains, the
current in the main cable is 0.7 amp. The
1) 5sin t  530  2) 5sin t  600 
efficiency of the transformer is
3) 5sin t  450  4) 5sin t
1) 63.8% 2) 74%
3) 83.3% 4) 48%
42. A transformer has 100 turns in the primary
CRTQ: 01) 1 02) 2 03) 3 04) 4
coil and carries 8A current. If input power is
one kilowatt, the number of turns to secondary 05)2 06) 1 07) 1 08) 2 09) 2
coil to have 500V output will be (assume 10) 4 11) 2 12) 2 13) 2 14) 3
ideal transformer) 15) 2 16) 1 17) 3 18) 4 19) 1
1) 100 2) 200 3) 400 4) 300 20) 4 21) 1 22) 3 23) 3 24) 1
43. A transmitting station transmits radio waves
25) 1 26) 4 27) 1 28) 4
of wavelength 360m. The inductance of coil
required with a condenser of capacity SPQ: 29) 2 30) 3 31)1 32) 1
1.20  F in the resonant circuit to receive
33) 2 34) 4 35) 1 36) 2 37) 2
them is
1) 3.04 10 8 H 2) 2.04 10 8 H 38) 3 39) 2 40) 1 41) 3 42) 3
3) 4.04 108 H 4) 6.04  10 8 H 43) 1 44) 1 45) 2 46) 1

50 ALTERNATING CURRENT
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18. V0=0, Transformer does not works on DC

source
n s Vs
19. 
n p Vp
1. To minimise eddy currents
2. definition Vs
20. Frequency remains same. V  5
3. definition p

4. Transformer does not work on DC

21. Es N s i p , P  E i
VP V I I 1   s s
n P  S n  P  E p N p is
5. VS VS I P IS n
6. T<1 NS I p
22. 
7. T>1 N p Is

8. VS  VP , then it is called step up N S Vs

23. 
Transformer N p Vp
9. Only AC circuits 24. As initially charge is maximum
10. emf does not depends on resistance
q  q0 cos t  i  dq   q0 sin t
11. output power=input power, for an ideal dt
transformer
1 2 q2
12. The material must be high permeability and Given Li 
2 2C
high resistivity
 q cos t 
2
1
13. Definition  L  q0 sin t   0
2

2 2C
VS I S 36 48
14.    IP   0.22 A 1
VP I P 220 I P 220 But,    tan t  1
LC
VS N S I P VS 200
15.   ;   VS  240V
VP N P I S 120 100   
t  t   LC
4 4 4
VI 8
16.   S S  ; 8  240  I3S 
VP I P 10 10 4  10 25. i ........ 1
R  XC2
2

8  4 10
IS   13.3 A
24 i 
 ........  2 
17. VP  4000V ; R  15  15  0.5  15 2 R2  9 X C 2

power loss P '  I 2 R XC 3

by solving 
2 2 R 5
 P  800  103 
P'    R     15  600kW
 VP   4000  26. Transformer doesn’t work on dc.

ALTERNATING CURRENT 51
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Vs I s 98 100  is where 1  2 f1 , 2  2 f 2
27.   
Vp I p 100 200  6
equating f  f1 f 2
28. In an LC circuit, by law of conservation of
energy 41. Outpur power = 140 walt
1 1 1 Input power = 240  0.7  168watt
CV12  CV22  Li 2
2 2 2
Output power
29. Transformer, out put power=InputPower Efficiency = Input power 100
30. Transformer
140
N S VS 1500 VS  100  83.3%
   168
N P VP 50 VP
42. Here, N p  100, I p  8 A
d
VP   4  VS  120V Input power, V p I p  1kW  1000W
dt
31. P  VI 1000 1000
Vp    125V ;
2.2  1000  22000 I ; I  0.1A Ip 8

N s Vs
Power loss P '  I 2 R   0.1 10 = 0.1 W
2
As 
N p Vp
N S Vs I P
32. N  V  I c
43.  
P P S

N S Vs I P
33. N  V  I and  
1
P P S
2 LC
34. Transformer does not works on DC
44. Here VP  1000V , I s  20 A, Vs  120V
N S Vs I P
35. N  V  I Vs I s
P P S 
Vp I p
N S Vs I P
36. N  V  I
P P S 45. Vs  N s
Vp N p
N S Vs I P
37. N  V  I 46. From Kirchhoff’s current law,
P P S

N S Vs I P i3  i1  i2  3sin  t  4 sin  t  900 

38. N  V  I
P P S  32  42  5
1 4sin 900 4
39.   Where tan   
LC 3  4 cos 90 0
3

40.
 1

 
  1 L1     2 L 2 
1 
 i3  5sin t  530 
 c
 1 1    2c2 

52 ALTERNATING CURRENT
Active site edutech- 9844532971

and E2, respectively.

R L,r

Resistor Coil

1. If the rms current in a 50 Hz AC circuit is e = E0 sint

5A, the value of the current 1/300 s after erms = E

its value becomes zero is The power (thermal) developed across the
coil is
1) 5 2 2) 5 3 / 2A

3) 5 / 6A 4) 5 / 2A E  E12 E  E12  E22

1) 2)
2R 2R
2. When a voltage measuring device is
 E  E1 
2 2
E
connected to Ac mains, the meter shows 3) 4)
the steady input voltage of 220V. This 2R 2R
means 6. In LR circuit current resonant frequency is
600 Hz and half power points are at 650 and
1) input voltage cannot be AC voltage, but a
550Hz. The quality factor is
DC voltage
1 1
2) Maximum input voltage is 220V 1) 2) 3) 6 4) 3
6 3
3) the meter reads not v but < v2 > and is 7. In an L-R circuit, the value of L is  0.4/  
calibrated to read  V 2 henry and the value of R is 30 . If in the
circuit, and alternating emf of 200V at 50
4) the pointer of the meter is struck by some
cycle per second is connected the impedance
mechanical defect
of the circuit and current will be
3. Which of the following combinations should
be selected for better tuning of an L-C-R 1) 11.4,17.5A 2) 30.7, 6.5A
circuit used for communication ? 3) 40.4,5A 4) 50, 4A
1) R  20 , L = 15H, C  35 F 8. The self inductance of the motor of an
2) R  25, , L  2.5 H , C  45  F electric fan is 10H. in order to impart
3) R  15, L  35 H , C  30  F maximum power at 50Hz, it should be
connected to a capacitance of
4) R  15, L  15 H , C  45  F
4. The output of a step- down transformer is 1) 1 F 2) 2  F
measured to be 24V when connected to a 12W 3) 4  F 4) 8 F
light bulb. The value of the peak current is
9. The value of L, C and R for a circuit are
1) 1/ 2A 2) 2A 1H, 9F and 3 What is the quality factor
3) 2 A 4) 2 2A for the circuit at resonance ?
5. For the circuit shown in the figure the rms 1 1
1) 1 2) 9 3) 4)
value of voltages across R and coil are E1 9 3

ALTERNATING CURRENT 53
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10.The voltage time (V - t) graph for triangular wave 1) zero 2) X g 3) - Xg 4) Rg

having peak value V 0 is as shown in figure. 14. To reduce the resonant frequency in an
L-C-R series circuit with a generator
1) the generator frequency should be reduced
2) another capacitor should be added in
parallel to the first

The rms value of V in time interval from 3) the iron core of the inductor should be
T removed
t  0 to is
4 4) dielectric in the capacitor should be
V0 V0 V0 removed
1) 2) 3) 4) 2Vo
3 2 2
15. An inductor of reactance 1 and a resis-
11. A circuit containing resistance R1 , tor of 2 are connected in series of the
Inductance L1 and capacitance C1 terminals of a 6V(rms) AC source. The
connected in series resonates at the same power dissipated in the circuit is
frequency 'n' as a second combination of
R2 , L2 and C2 . If the two are connected in 1) 8W 2) 12W
series. Then the circuit will resonates at 3) 14.4W 4) 18W

L2C2 16. The potential difference across a 2H

L1C1
1) n 2) 2n 3) 4) inductor as a function of time is shown in
L1C1 L2C2
figure. At time t = 0, current is zero.
12. An LCR circuit has L = 10 mH, R = 3 , Current at t = 2 second is
and C = 1  F connected in series to a source V (volt) L

of 15 cos  t volt. The current amplitude at 10

a frequency that is 10% lower than the
resonant frequency is t(s)
2 4
1) 0.5 A 2) 0.7 A
1) 1A 2) 3A 3) 4A 4) 5A
3) 0.9 A 4) 1.1 A
17. A bulb is rated at 100 V, 100 W, it can be
trea-ted as a resistor. Find out the inductance
of an inductor (called choke coil) that should
13. An alternating current generator has an in-
ternal resistance Rg and an internal reac- be connected in series with the bulb to
tance Xg. It is used to supply power to a pas- operate the bulb at its rated power with the
sive load consisting of a resistance Rg and help of an ac source of 200 V and 50 Hz
a reactance XL. For maximum power to be 
delivered from the generator to the load, the 1) H 2) 100 H
3
value of xL. For maximum power to be de-
livered from the generator to the load, the 2 3
3) H 4) H
value of XL is equal to  

54 ALTERNATING CURRENT
Active site edutech- 9844532971

18. A group of electric lamps having a total frequency double the resonance frequency.
power rating of 1000W is supplied by an AC The impedance is 10 times the minimum
voltage E= 200 sin ( 310 t + 600). Then the impedance. The inductive reactance is
current in a circuit is 1) R 2) 2R 3) 3R 4) 4R
1) 10A 2) 10 2A 24. A 100 V a.c source of frequency 500 Hz is
connected to a LCR circuit with L = 8.1 mH,
3) 20A 4) 20 2A
C  12.5 F and R  1 0  , all connected
19. In a transformer the output current and in series. The potential difference across the
voltage are respectively 4A and 20V. If the resistance is
ratio number of turns in the primary to
1) 100 V 2) 200 V
secondary is 2:1. What is the input current
and voltage ?(assume ideal transformer) 3) 300 V 4) 450 V

1) 2A and 40V 2) 1 A and 20V

3) 4A and 10V 4) 8A and 40V
CRTQ: 01) 2 02) 3 03) 3 04) 1
20. A current of 5A is flowing at 220V in the
primary coil of a transformer. If the voltage 05) 2 06) 3 07) 4 08) 1
produced in the secondary coil is 2200V and 09) 3 10) 1 11)1 12) 2
50% of power is lost, then the current in
secondary will be
S.P.Q: 13) 3 14) 2 15) 3 16) 4

1) 2.5A 2) 5A 3) 0.25A 4) 0.5A 17) 4 18) 2 19) 1 20) 3

21. The rms value of a semi-circular current 21) 3 22) 1 23) 4 24) 1
wave which has a maximum value of ‘a’ is


1) 1 / 2 a  2)  3 / 2 a
1
3)  2 / 3 a 4) 1/ 3a 1: Given, V = 50 HZ, Irms = 5A ; t  s
300

22. Two series resonant circuits with component We have to find I  t 

values L1C1 and L2C2, respectively have the
I 0 = Peak value = 2, Irms
same resonant frequency, They are then
connected in series, so that the combination = 2  5  5 2A
has the same resonant frequency
1 2 I  I 0 sin t  5 2
1)   2)  
LC LC
1
5 sin 2 vt  5 2 sin 2  50 
3)   1 4)   300
2 LC 2 LC
23. An AC source of variable frequency is  3
 5 2 sin 5 2  5 3 / 2A
applied across a series L-C-R circuit. At a 3 2

ALTERNATING CURRENT 55
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2. The voltmeter connected to AC mains reads mean 7. Z 2  R 2   2 fL  ; I  V / Z

2

value (<v2>) and is calibrated in such a way that

1
it gives value of < V2>. Which is multiplied by 8. Z  R2   X L  X C  ;  
2
2 LC
form factor to give rms value.
3. Quality factor (Q) of an L-C-R circuit is given by 1 L
9. Quality factor Q 
R C
1 L
Q Vt 4V t
R C 10. V  T0  0 Vrms  V2
T
4
Where R is resistance, L is inductance and C is
capacitance of the circuit. To make Q high, T 4 
 

2
 t d t 
R should be low, L should be high and C should 
4V 0  0
 T
 V0

T  4  3
be low 
 dt 

 0 
These conditions are best satisfied by the values
given in option 3
1 1
11. n  
4. Secondary voltage VS = 24V 2 L1C1 2 L2C2
Power associated with secondary Ps = 12W
C1C2
L1C1  L2C2 ; Lnet  L1  L2 ; Cnet 
P 12 1 C1  C2
IS  s   A  0.5 A
vs 24 2  CC 
Lnet Cnet  ( L1  L2 )  1 2 ;
Peak value of the current in the secondary  C  C 
1 2

1 Lnet Cnet  L2C2

I0  Is 2 = (0.5) (1.414) = 0707 = A
2 90 1
90
12. cv  cv0  x =9000 rad/s
5. Draw the phasor diagram. 100 100 LC

E 2  E12  E 22  2E1E 2 cos . E0

i0 
2
 1 
E2
R2    L  
E2   C
E

13. For delivering maximum power from the


I E1 generator to the load, total internal reactance
Thermal power developed in coil is must be equal to conjugate of total external
P  E 2 cos  I reactance
Hence,
E EE E 2  E12  E 22
I  1  P  1 2 cos  
R R 2R X int  X ext  X g   X L    X L
f0
6. Quality factor  f  f  Xg   XL   XL  XL  X g
2 1

56 ALTERNATING CURRENT
Active site edutech- 9844532971

14. We know that resonant frequency in an L-C-R 18. Power = E0i0 cos  ;
circuit is given by V0  1
2 LC i0
 I rms 
2
Now to reduce V0 either we can increase L or
we can increase C. vs ns
19. v  n ;
p p
To increase capacitance. We must connect
another capacitor parallel to the first Since the input power = output power
15. Given, X L  1, R  2 Pout Vs I s
20.  %   100
Pin V p I p
Erms  6V , Pav  ?
Average power dissipated in the circuit 21. The equation of a semi circular wave is

I E x 2  y 2  a 2 or y 2  a 2  x 2
Pav  Erms I rms cos  ; I rms  0  rms
2 Z
1 a 2
2a  a
I rms  y dx
Z  R 2  X L2  4  1  5
1 a 2 2a 2
I rms 
6 R
A ; cos   
2 2
I rms 
2a  a
 a  x 2
 dx 
3
5 Z 5
1 1
6 2 72 72 22. By problem   
Pav  6      14.4W L1C1 L2C2
5 5 5 5 5
When the combination L1C1 and L2C2 are
di
16. e  L  e dt  L  i2  i1  connected in series, the combination will have
dt
inductance L and capacitance C given by
e dt  area of  le for t  0 to 2 sec.
L  L1  L2 ; C  C1C2
100  100 C1  C2
17. Resistance of bulb is R   100
100
C1C2
Rated current is
100
 1A Now LC   L1  L2 
100 C1  C2

Vrms L1C1
In ac, I rms  ; Z  200 From (1) we have L2  ;
Z C2
2
1002   L   200  2 L2  30000 and Substituting (5) in (4) we get on simplification
1
30000 3 LC = L1C1  .
L 2
 henry.. 2
100  
1
It follows that  
LC

ALTERNATING CURRENT 57
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23. Z 2  R 2  ( L  1/ C ) 2

10 R 2  R 2  (2o L  1/ 2oC ) 2
minimum impedance Zmin = R

o2 LC  1 ------- (1)

1
2 o L   3R ------- (2)
2 o C

1
from(1) R  XC  R
2 o C

from(2) X C  2 o L  3 R  R  4 R

24. Here, E  100V , v  500 Hz

L  8.1 103 H , C  12.5 106 F , R  10 

X L   L  2 vL  1000  8.1 10 3  25.4

 1
XC  
C 1000  12.5  106

103
   25.4
12.5

Z  R2   X L  X C 
2

 102  25.425.4 10

2

Ev 100
Iv    10 A
Z 10
Potential difference across
R  I v R  10 10  100V

58 ALTERNATING CURRENT
Active site edutech- 9844532971

NEET - PHYSICS

XII CLASS - NEET - SAMPLE COPY

NEET
2016 - 2022

XII CLASS

PHYSICS

KGN SS  2
Active site edutech- 9844532971

NEET - PHYSICS

7 ALTERNATING CURRENT
1. An inductor 20mH, a capacitor 50  F and 6. Figure shows a circuit that contains three
identical resistors with resistance R =
a resistor 40 are connected in series
9.0  each, two identical inductors with
across a source of emf V  10sin 340t the
inductance L = 2.0 mH each, and an ideal
power loss in A.C circuit is [NEET 2016]
battery with emf  = 18V. The current
1) 0.51W 2) 0.67W
through the battery just after the switch
3) 0.76 W 4) 0.89W closed is [NEET 2017]
2. A small signal voltage V(t) =V0 sin t is
applied across an ideal capacitor C
L R
[NEET 2016]
+ R
-
1) Current I(t), lags voltageV(t) by 900C
L V
2) Over a full cycle the capacitor C does not R

consume any energy from the voltage source

3) Current I(t) in in phase with voltage V(t) 1) 0.2 A 2) 2A
4) Current I(t) leads voltage V(t) by1800 3) 0 ampere 4) 2 mA
3. The potential differences across the 7. An inductor 20 mH, a capacitor 100 F
resistance, capacitance and inductance are and a resistor 50  are connected in series
80V, 40V and 100V respectively in an L-C- across a source of emf, V = 10 sin 314t.
R circuit. The power factor of this circuit The power loss in the circuit is
is [NEET 2016] [NEET 2018]
1) 0.8 2) 1.0 3) 0.4 4) 0.5 1) 0.79 W 2) 0.43 W
4. A 100 resistance and a capacitor of
3) 2.74 W 4) 1.13 W
100 reactance are connected in series
across a 220V source. When the capacitor 8. A circuit when connected to an AC source
is 50% charged, the peak value of the of 12V gives a current of 0.2 A. The same
displacement current is : [NEET 2016] circuit when connected to a DC source of
12 V, gives a current of 0.4 A. The circuit
1) 4.4A 2) 11 2
is [Odisha NEET 2019]
3) 2.2A 4) 11A
1) series LR 2) series RC
5. Which of the following combinations
should be selected for better tuning of an 3) series LC 4) series LCR
L-C-R circuit used for communication ? 9. A 40  F capacitor is connected to a 200
[NEET-II 2016] V, 50 Hz ac supply. The rms value of the
1) R = 20 , L = 1.5 H, C = 35 F current in the circuit is, nearly :
2) R = 25 , L = 2.5 H, C = 45 F [NEET 2020]
3) R = 15 , L = 3.5 H, C = 30 F 1) 25.1 A 2) 1.7 A
4) R = 25 , L = 1.5 H, C = 45 F 3) 2.05 A 4) 2.5 A

KGN SS  33
Active site edutech- 9844532971

NEET - PHYSICS
10. A series LCR circuit is connected to an 13. An inductor of inductance L, a capacitor of
ac voltage source. When L is removed capacitance C and a resistor of resistance
from the circuit, the phase difference ‘R’ are connected in series to an ac source
 of potential difference ‘V’ volts as shown
between current and voltage is . If in figure.
3
instead C is removed from the circuit, the Potential difference across L, C and R is
 40V, 10V and 40V, respectively. The
phase difference is again between amplitude of current flowing through LCR
3
current and voltage. The power factor of series circuit is 10 2 A. The impedance
of the circuit is [NEET 2021]
the circuit is : [NEET 2020]
1) –1.0 2) zero 3) 0.5 4) 1.0
10 V 40 V
11. A light bulb and an inductor coil are 40 V

connected to an ac source through a key

S
V
as shown in the figure below. The key is
closed and after sometime an iron rod is 1) 5  2) 4 2 
inserted into the interior of the inductor.
3) 5 2  4) 4 
The glow of the light bulb. [NEET 2020]
14. A step down transformer connected to an
ac mains supply of 220 V is made to
operate at 11V, 44 W lamp. Ignoring power
losses in the transformer, what is the
current in the primary circuit ? [NEET 2021]
1) 4A 2) 0.2 A 3) 0.4 A 4) 2 A
15. Given below the two statements:
Statement -I:
In an ac circuit the current through a
1) decreases 2) remains
capacitor leads the voltage across it.
unchanged
Statement -II:
3) will fluctuate 4) increases In a.c. circuits containing pure capaci-
12. A series LCR circuit containing 5.0 H tance only, the phase difference between
inductor 8 F capacitor and 40  resistor the current and the voltage is  :
is connected to 230 V variable frequency In the light of the above statements ,
ac source. The angular frequencies of the choose the most appropriate answer from
source at which power transferred to the the options given below: [NEET 2022]
circuit half the power at the resonance 1) both statement-I and statement-II
angular frequency are likely to be[NEET 2021] are correct
1) 42 rad/s and 58 rad/s 2) both statement -I and statement -
II are incorrect
2) 25 rad/s and 75 rad/s
3) Statement -I is correct but statement-
3) 50 rad/s and 25 rad/s II is incorrect
4) 46 rad/s and 54 rad/s 4) Statement-I is incorrect but statement-
II is correct

KGN SS  34
Active site edutech- 9844532971

NEET - PHYSICS
16. An inductor of inductance 2 mH is con- 1 1.5 3
nected to a 220 V, 50 Hz a.c. source. Let Q1  6
 50   10.35
20 35 10 70
the inductive reactance in the circuit is
X1. If a 220 V dc source replaces the ac 1 2.5 5
Q2  6
 40   9.43
source in the circuit, then the inductive 25 45 10 90
reactance in the circuit is X 2 , X1 and X 2 1 3.5 100 35
respectively are : [NEET 2022] Q3  6
  22.77
15 30 10 15 3
1) 6.28 , zero 2) 6.28 , infinity
3) 0.628 , zero 4) 0.628 , infinity 1 1.5 40
Q4  6
  7.30
17. A standard filament lamp consumes 100 W 25 45  10 30
when connected to 200 V ac mains supply. Clearly Q3 is maximum of Q1, Q2, Q3, and Q4.
The peak current through the dulb will be : Hence, option (c) should be selected for better
tuning of an L-C-R circuit.
[NEET 2022]
6. At time, t = 0 i.e., when switch is closed,
1) 0.707 A 2) 1 A 3) 1.414 A 4) 2 A inductor in the circuit provides very high
resistance (open circuit) while capacitor starts
KEY charging with maximum current (low
resistance).
01) 1 02) 2 03) 1 04) 3 05) 3 Equivalent circuit of the given circuit
06) * 07) 1 08) 1 09) 4 10) 4 L R i

11) 1 12) 4 13) 1 14) 2 15) 3 +


-
R +
-
R R

C
R
16) 3 17) 1 L

HINTS & SOLUTIONS Current drawn from battery,

 2 2 18
i    4A
1. Z  R   Xc  XL 
2 2
 R / 2 R 9
2 * None of the given options is correct.
V  7. Impedance Z in an ac circuit is
P  I R   rms  R =0.47W..
2
rms
 Z 
Z  R 2   X C  X L  ; where X C =
2
Nearest value in the options is 0.51W
2. Power = Vrms . Irms cos  capacitive reactance and XL = inductive
reactance.
VL  Vc 100  40 3 
3. tan     or   37 0
VR 80 4 Also XL = and XL = L
C
Power factor 2
 1 
 50 
2
Z    314  20  103 
4  314  100  10
6

 cos   cos 37 0  or 0.8
5 or Z = 56 
0
4.  id max   ic max  i0   Vrms 
2

Z The power loss in the circuit is Pav =   R

So, current decreased.  Z 
5. Quality factor of an L-C-R circuit is given by,  
2

10
1 L  Pav     50  0.79W
Q
R C

   2 56 


KGN SS  35
Active site edutech- 9844532971

NEET - PHYSICS
8. When circuit is connected to an AC source
Vrms  VR2  VL  VC 
2
of 12 V, gives a current of 0.2 A.
12
 Impedance, Z =  60  Vrms  402   40  10   50V
2
0.2
When the same circuit is connected to a DC I 0  10 2 A
source of 12V, gives a current of 0.4 A.
12 I0
 30  I rms   10 A
 Resistance, R = 2
0.4
As, power factors, Vrms 50
Impedance, Z = I  10  5 
R 30 1 rms
cos      cos 60 14. Here P = 44 W, VP = 220V, Vs = 11 V
Z 60 2
Power, P = VsIs
   60, i.e., current lags behind the emf.
44 = Is  Is = 4A
So, we can conclude that the circuit is a Now, VpIp = VsIs
series L.R.
220  I p  11 4
V V
9. I   VC 11 4 4
X C 1/ C Ip    0.2 A
220 20
 200  40  106  2  50  2.5 A
XC i
10. When L removed tan  
R
XL
When L removed tan  
R
15.
XC X L 900
  Resonance,
R R Vc
R R
cos     1
Z R
2mH
11. z  R  X ; X L , Z , I 
2 2
L
1 1
12. r    50rad / s
LC 5  80  106 16.
Power transferred = half of power at 
resonance 200V, 50Hz
So, frequencies at which power transferred is
half =  r   For AC X L = L For DC, =0
R 40 X 1 =100 × 2×10 -3 XL = L
    4 rad / s
2L 2  5 X 1 = 0.2 X2 = 0
So, range is r   = 50  4
X 1 = 0.628
= 54 rad/s and 46 rad/s
13. Given, VL = 40 V 17. I rms Vrms  P , I rms  200  100
VC = 10V, VR = 40 V 1
I rms  , So, I peak  I rms 2
40 V 10 V 40 V
2
1 1
  2   0.707 A
2 2
S

KGN SS  36