𝐼1 +𝐼2 +⋯….

+𝐼𝑛
To prove 𝐼𝑚𝑒𝑎𝑛 =
ALTERNATING CURRENT (A.C.) 𝑛
:

Alternating current: The current or E.M.F. whose magnitude changes with time and Positive half
direction reverse periodically is called alternating current or E.M.F.
cycle
Different values of Sinusoidal voltage and current
Maximum
positive value
π π
I0 3π 2π
2 2
0.707I0
0.636I0 I0 or IP
I rms or Iv Negative half
I av or Im cycle
Fig. derivation of mean value of ac current
O π π 3π 2π
Let I1 , I2........In be the instantaneous of ac current through the circuit for half cycle. By
2 2
definition of mean value of currrent,
Heat produced by ac = heat produce by dc
Or, ( Iav)2 RT/2 = ( Imean )2RT/2
𝐼 +𝐼 +⋯….+𝐼𝑛
Or, Imean =Iav = 1 2
𝑛

Fig. showing Io, Irms and Iav 𝐼 +𝐼 +⋯….+𝐼

 𝐼𝑚 = 1 2 𝑛
Proved
𝑛
a) Instantaneous Value: It is the value of the alternating quantity (voltage or current) To prove I mean = 0.637 Io : Suppose instantaneous value of the alternating current is
that exists at any instant of time i.e. given by R
I  I sin ωt for instantaneous value of a.c. current I  I sin ωt … …. … (1)
E  E sin ωt for instantaneous value of a.c. voltage. The alternating current changes with time but if we
b) Peak Value or Maximum Value: It is the highest value reached by alternating assume that a constant value is given by above equation I
quantity (current or voltage). for the small time dt . Through the circuit, the small
c) Peak to Peak Value: It is the sum of the positive peak and negative peak values amount of the charge passing through the circuit is
usually written as P-P value. dq  Idt
d) Mean value or average value of the alternating current: It is steady current which
Or, dq  I sin ωt dt Fig .Ac supply for  ve half cycle

when passed through a circuit would send the same amount of charge in a half cycle
T
as is done by the alternating current in the same circuit in the same time. q 2
T
It is denoted by the symbol Imean of I m
𝐼1 +𝐼2 +⋯….+𝐼𝑛
Integrating from t  0 to t  then
2  dq   I
0 0
 sin ωt dt
𝐼𝑚𝑒𝑎𝑛 = T
𝑛
2
Imean = 0.637 Io

Or, q  I sin ωt dt
0
T

 cos ωt  2
Or, q  I 
 ω  0

Or, q    cos ωt 02

T
I
ω
I  T 
Or, q   cos ω - cosω  0
ω 2 
 T
I  2 T 2  2
Or, q    cos - cos  0 Positive half
ω T 2 T 0
cycle
Or, q    cos  - cos0
I
ω
Or, q     1  1
I
ω π π 3π 2π
2 2
2I
Or, q 
ω
2I Negative half
Or, q 
2 cycle
Fig. derivation of mean value of ac current
T
IT
q  … … … (1) Heat produced by ac = heat produce by dc
 Or, ( I2) av RT= ( Ir.m.s )2RT
If I m is the mean value of the alternating R Or, ( Ir-ms)2 = I2av
2
𝐼 2 +𝐼22 +⋯+𝐼𝑁
current then according to the definition Or, ( Ir-m-s)2 = 1
𝑁
T
 q  I m … … … (2) 𝐼12 + 𝐼22 + ⋯ + 𝐼𝑁2
2 I 𝐼𝑟.𝑚.𝑠 = √
From equation (1) and (2) we get, 𝑁
T IT
Im   𝑰
2 π Fig .Dc supply for equal to  ve half cycle To prove 𝑰𝒓.𝒎.𝒔 = 𝑶 = 𝟎. 𝟕𝟎𝟕𝑰𝑶 : Let the instantaneous value of the alternating current is
√𝟐
2I given by
Or, I m 
π I  I sin ωt
 I m  0.637I  … … … (3) If the current passed through the resistance ‘R’ for small time ‘dt’ then small amount of heat
This is the required relation between mean value and peak value of a.c current. will be produced.
2E  dH  I 2 Rdt
Similarly, E m 
π Or, dH  I 2sin 2 ωt Rdt
 E m  0.636E  The total amount of the heat in one complete cycle, t  0 to t  T .
This is the required relation between mean value and peak value of a.c voltage. Integrating both sides, we get
H T
Root mean square value (R.M.S.) or virtual value of A.C.: The root mean square value
of the current is defined as the steady current when passed through a resistance for a given 0

Or, dH  I 02 sin 2 ωt Rdt
0
R
time will produce the same amount of heat as the alternating current done in the same T
resistance at the same time. It is denoted by I r.m.s. or I e or I v .

Or, H  I 02 R sin 2ωt dt
0 I
𝐼2 + 𝐼22 + ⋯+ 𝐼𝑁2
𝐼𝑟.𝑚.𝑠 = √1 1 - cos2 ωt
T

𝑁 Or, H  I 02 R  2
dt
𝐼𝑂 0

𝐼𝑟.𝑚.𝑠 = = 0.707𝐼𝑂 I 02 R
T

 1  cos2 ωt  dt
Fig .Ac supply for ful l cycle.
√2 Or, H 
2 0
𝑰𝟐 𝟐 𝟐
𝟏 +𝑰𝟐 +⋯+𝑰𝑵
IR 
T T
To prove 𝑰𝒓.𝒎.𝒔 = : Let I1 , I2........IN be the instantaneous of ac current through 2
𝑵
the circuit for to full cycle. By definition of mean value of current,
Or, H 
2 0 0

 dt  cos2 ωt dt 
0



 I 02 R  T  sin2 ωt  T 
Or, H  t  0    
I0 or E0
2   2ω  0 
Or, E
Y’ E0
  2π 2π  
T
E  E 0 sin( ωt  )
  sin2. .T sin2. .0 
T  
2

H  0 T  0  
IR T   I  I 0 sin ωt
2   2. 2π 2.
2π  

  T T  0  ωt Y
I
I0
0  ωt  θ
   
T

2   sin4π    ωt
Or, H  0 T  0  
IR
 0  0 X X’
2   4π   Fig. Phasor diagram
  T  0 
I 02 RT
H  … … … (1) A phasor or a vector diagram ( a graph of instantneous current and voltage) represents
2
alternating voltages and currents (of same frequency) as vectors along with the phase angle
If I rms be the root mean square value from definition between them.
H  I 2rms RT … … … (2) Let us consider the instantaneous value of current and e.m.f. are
From equations (1) and (2), we get I  Isin ωt … … … (i)
Or, I 2rms .RT  0
I 2 RT
R and E  E sin  ωt  θ 
2 Then the phase diagram of Io and Eo are as shown in figure.
2
I Here, the projections on vertical axis are:
Or, I 2rms  0
2 OY= I  Isin ωt and OY’= E  E sin  ωt  θ 
I I
Or, I rms  0 While the projections on the horizontal axis are:
2 OX= I  I Cos ωt and OX’= E  E  Cos ωt  θ  .
 I rms  0.707I 0 It may be noted that the peak or r.m.s values of current and voltage (emf) in a circuit are
Fig .Dc supply for equal to full ac cycle.
E added vectorically (using parallelogram law of vector addition in the phasor diagram)
Similarly, E rms  0
2 whereas the instantaneous values of current and voltage are added algebraically. This is
known as vector or rotor diagram.
 E rms  0.707E 0
Phasor Diagram: An ideal inductor: It is an electronic device that has zero
resistance and consists of a number of turns of a conducting wire
in the form of a coil
Fig. An ideal inductor
Reactance: It is the frequency dependent resistance that arises on account of opposing e.m.f.
induced due to the change in the strength of current. It block dc but allows ac current
Inductive reactance (XL): It is the resistance which an inductor offers to current flow and
denoted by XL i.e.
V
XL  L
IL
Where VL is the PD across the inductor and IL is current through it.

 XL  ω L
 Unit: ohm Alternating current passes through the inductor: Consider a pure inductor or self
Capacitative reactance (XL): It is the frequency dependent resistance which a capacitor inductor ‘L’ connected to the alternating source of E.M.F. of E having the frequency of ‘f ’
offers to current flow and denoted by X L i.e. as shown in figure.
V V E or I E  E 0 sin ωt
X C  c  rms
Ic I rms L I  I sinω t
Where Vc is the PD across the inductor and Ic is the current through it.
E
1
XC 
ωC ωt
90 o
ωt
O π 2π 3π O
Unit: ohm I
Impedance (Z): It is total resistance of A.C. circuit as whole offers to current flow. It is
denoted by Z i.e.
E
Z Fig (a) A.C. through inductor Fig (b) Phasor diagram Fig (c) Vector diagram
I
Unit: ohm
Let instantaneous value of the alternating current is
Alternating current passes through the pure resistance: Consider a pure Ohmic resistor
I  I  sin ωt … … … (1)
of resistance R connected to the alternating current source of E.M.F. E having frequency f
as shown in fig (1). The potential difference across the terminal of the inductor
Let instantaneous value of the alternating current is d d(LI )
E 
I  I  sin ωt … … … (1) dt dt
The EMF of the source E connected across R is equal to the potential difference dI
VL  L
dt
According to the Kirchhoff’s law, E  V  0
E  E 0 sin ωt  dI 
Or, E    L   0
R I  I sinω t  dt 
dI
Or, E  L
dt
E I
ωt ωt d(I0 sin ωt)
O O Or, E  L [  I  I o sin t ]
dt
d sin ωt
Or, E  LI 0
dt
Or, E  LI 0cos ωt. ω
Fig (a) A.C. through resistor Fig (c) Vector diagram
 E  Lω I 0 cos ωt
Fig (b) Phasor diagram
E = VR
E  IR Where Lω is called inductive reactance. It is denoted by the symbol X L .
Or, E  I0 sin ωt R Or, E  X L I 0cos ωt
Or, E  E 0 cos ωt
Or, E  I 0 R sin ωt Where E 0  I 0 X L is called maximum value of the E.M.F.
 E  E 0 sin ωt … … … (2)
 π
Where E  I 0 R is the peak value of E.M.F.  E  E 0 sin ωt   … … … (2)
 2
From equation (1) and (2) it follows that the alternating current and alternating source of the
E.M.F. both are the same phase as shown in figure (2)
 From equation (1) and (2), it follows that the alternating E.M.F. leads the current by phase  cos ωt 
Or, q  I 0 

angle .  ω 
2 I cos ωt
Note : (i) In L only ac circuit, a e.m.f (E) leads the current by a phase angle π/2 if q 0 … … … (2)
ω

I = IoSinωt then E= EoSin (ωt + ) Substituting the value of q in equation (2), we get
2 I
(ii) In L only ac circuit, ac current (I) lags the e.m.f by a phase angle π/2 if Or, E   0 cos ωt
Cω

E = EoSinωt then I= IoSin (ωt - ) 1
2 Where is called capacitive reactance denoted by X C .
Cω
Alternating current passes through the capacitor: Consider a capacitor of capacitance C Or, E   X C I 0 cos ωt
connecting to the alternating source of E.M.F. of E having the frequency f as shown in the Or, E   E 0 cos ωt
figure . Let instantaneous value of the alternating current is
 π
I  Isin ωt … … … (1)  E  E 0 sin  ωt   … … … (3)
 2
Suppose E and q are the E.M.F. and the charge on the capacitor. Therefore, the potential
From equation (1) and (3) it shows that the alternating E.M.F. lags the current by the
difference across the capacitor is

phase angle .
2
 
E  E  sin ωt -  Note : (i) In C only ac circuit, ac e.m.f (E) lags the current by a phase angle π/2 if
 2

I  Isin ωt I = IoSinωt then E= EoSin (ωt - )
E or I 2
(ii) In C only ac circuit, ac current (I) leads the e.m.f by a phase angle π/2 if
C

I E = EoSinωt then I= IoSin (ωt + )
ωt 2
O π 2π 3π
90 o
VC Alternating current passes through R-L in series: Consider an inductor of inductance L
and resistor of resistance R connected in series with alternating voltage E and having

E  E  sin ωt  θ 

Fig (a ). A.C. circuit containing C only Fig ( b). Phasor diagram Fig (c). Vector diagram I  I sinω t
E = VC L R
q
E
C VL
t E
dq O π 2π 3π
Also, I  
dt
Or, dq  Idt 
I VR

Or, dq  I 0 sin ωt. dt

Integrating both sides, we get Fig (a ). A.C. through RL series circuit Fig (b). Phasor diagram Fig (C). Vector diagram

 dq   I sin t .dt
0
frequency ‘f’ shown in figure.
Or, q  I  sin ωt. dt
0
Let instantaneous value of the alternating current is
 I  I  sin ωt … … … (1) E  E  sin ωt - θ 
In LR series circuit, the phase relations between voltage and current are I  I sinω t E
(a) In pure resistance, voltage (VR) across R, and current I are in the same phase i.e. C R
voltage drop across R is I
VR
VR = IR … … …. (2)
O 

(b) In pure inductor, voltage VL leads the current I by the phase angle . i.e. VC
2 
Voltage drop across L is
VL  IX L … … … (3)
Where X L  L  2fL , is inductive reactance of the inductor Fig (b). RC ac series circuit Fig (b). Phasor diagram Fig (c). Vector diagram
Combining the above two relations (a) and (b), we draw vector diagram vector and phase
diagram as shown in figure (b and c) . From vector diagram, Let instantaneous value of the alternating current is
To find phase relation: In both diagram (c), total voltage (E) of ac circuit in LR series a.c. I  I  sin ωt … … … (1)
circuit leads the current by a phase angle θ i,e E = Eosin (ωt+θ ) The phase relations between voltage and current in RC series circuit.
From vector diagram, In RC series circuit, the phase relations between voltage and current are
V IX X (a) In pure resistor voltage VR and current I are in the same phase. The voltage drop
Tan  L  L  L
VR IR R across R is
X  VR = IR
   Tan 1  L  
 R  (b) In pure capacitor voltage VC lags the behind the current I by the phase angle . The
To determine impedance of circuit: Let Z be the impedance of RL circuit. 2
Let Z be the impedance of LR circuit, voltage drop across the capacitor is
From vector diagram, we have VC  IX C
E 2  ( VR ) 2  ( VL ) 2 1 1
Where X C   , is capacitive reactance of the capacitor
Or, E 2  ( IR ) 2  ( IX L ) 2 C 2fC
Combining the above two relations (a) and (b), we draw vector diagram vector and phase
Or, E  I R 2  X L
2
diagram as shown in figure (c)
E To find phase relation:
Z  R2  XL
2

I Further, from the figure, total e.m.f of LR circuit E leads the current I by the phase angle θ.
Where Z is the impedance of the RL series circuit. E = Eosin (ωt-θ )
Alternating current passes through the R-C circuit: Consider a resistor R and In vector diagram,
capacitance C are connecting in series with alternating voltage E having frequency f as V IX X
Tan  C  C  C
shown in figure. VR IR R
X 
   Tan 1  C 
 R 
To determine impedance of circuit: Let Z be the impedance of RC circuit .
From vector diagram, we have
E 2  ( VR ) 2  ( VC ) 2
Or, E 2  ( IR ) 2  ( IX C ) 2
Or, E  I R 2  X C
2
 E Using the Pythagoras theorem
Z
 R 2  XC
2

I In a vector diagram (c), we have

E 2  ( VR ) 2  VL  VC 
2
Where Z is the impedance of the RC series circuit.
Where Z is called the impedance which depends upon the frequency of the source as Or, E 2  I 2 R 2  IX L  IX C 
2

well as resistance and inductance of the circuit.

Or, E 2  I 2 R 2  I 2 X L  X C 
2

Alternating current through the L-C-R circuit: Consider an A.C. circuit consisting
inductor of inductance L, capacitor of capacitance C and resistance of resistor R are Or, E  I R 2  X L  XC 2
connected in series with alternating voltage E having frequency f as shown in figure.
 R 2  X L  X C   Z
E 2
E  E  sin ωt  θ 
Or,
I
I  I sinω t
 Z  R 2  X L  X C 
2

L C R E
( VL  VC )
Where Z  is called the impedance of the circuit.
E I

Special Cases:
O π 2π 3π  t
I VR Z
VC or Maximum Current I0

I0
Fig (a ). LCR series ac circuit Fig (b). Phasor diagram Fig . Vector diagram (VL  VC ) Z I0
In LCR series circuit, the phase relations between voltage and current are
(a) In pure resistance, voltage VR and current I are in the same phase.

(b) In pure inductor, voltage VL leads the current I by the phase angle .
2
(c) In pure capacitor, the voltage VC lags the current I by the phase angle. Minimum Z
We combine above three relations (a),(b) and (c) to draw phaser diagrm (a) and vector
diagram (c). In both diagram, total voltage (E) of ac circuit in LCR series a.c. circuit leads Z=R
ω
the current by a phase angle θ . ω  ωr

If VL > VC , then this resultant is VL  VC  in the direction of VL . XL =XC

Fig. Variation of Z and I0 with ω
To find phase relation:
Further, from the figure, voltage V leads the current I by the phase angle θ i.e E =
Eosin (ωt+θ ) (i) If XL > XC, Tanθ = + Ve and therefore, θ is positive. So, the e.m.f leads the
V -V current.
Tanθ  L C (ii) If XL < XC, Tanθ = - Ve and therefore, θ is negative. So, the e.m.f. lags the
VR
current.
IX L - IX C
Or, Tanθ 
IR
XL - XC
Or, Tanθ 
R
 X  XC 
 θ  Tan 1  L  … … … (i)
 R 
To determine impedance (Z):
 (iii) If XL = XC, Tanθ = 0 E (ii) Z=R
and therefore, θ = 0º. (iii) E.m.f. and current are in the same phase
So, the e.m.f and the (iv) VL and VC are opposite in phase
current are in the same (v)  fr  1/ 2 π L C
phase. Vc
LCR series Resonance. It is Quality Factor (Q - factor of the resonance circuit): Q - Factor or quality factor of a
the phenomena in which the series resonance circuit is defined as the ratio of the potential difference developed across
LCR circuit behaves as purely the inductance or capacitance at resonance to the applied e.m.f i.e. the voltage applied. It
0 t defines the sharpness of resonance in a series LCR resonance circuit i.e.
resistive circuit when the
e.m.f. and current are in the Potential across L or C
VR=V Q=
same phase at XL = XC giving Potential difference across R (  applied e.m.f)
maximum current i.e. in such 1
case, 2π  L
VL I0 X L X L ωr L 2π f r  L 2π LC 1 L
The impedance of circuit (Z) Q     
Fig. Conditions of resonance in the circuit I0 R R R R R R C
 R  X L  X C  =
2 2

 X  ω r L & ω r  2π f r 
R  X L  X L  = R
L
2 2

1 L  1 
E E Q= .  f r  
Current (Io) = o = 0 is maximum R C  2π LC 
Z R 1
Derivation of resonance frequency:
The variation of current I0 versus the angular frequency ω in a LCR series circuit is I0 X C ωr C 1 1
Again, Q = = =
shown in figure. I0R R ω r
C R 2π f r
CR
At very low frequencies, inductive reactance X L =ωL is negligible, but capacitative  X L  ω r L & ω r  2π f r 
reactance (Xc =1/ωC) is very high.
1 L  1 
When the frequency of the alternating e.m.f applied to the LCR series circuit is Q  .  f r  
increased, XL goes on increasing and XC goes on decreasing. For particular value of R C  2π LC 
X L = XC Power in A.C. circuit:
1  1  Io Consider an instantaneous value of alternating current passes through LCR ac series circuit
Or, ω r L   X L  ω r L and X C   Small R is
ωr C  ωr C 
I = Iosinωt ... ... ... (i)
1 In LCR circuit, the the e.m.f (E) leads the current by Phase angle θ
Or, ω 2r 
LC ω E = Eosin (ωt + θ) ... ... ... (ii)
1
Or, ω r 
LC The instantaneous power consumed by the circuit is
Large R P = EI
1
Or, 2 π f r  ω Let us assume that the current and e.m.f remain constant for a small interval of time dt. So,
LC the amount of electrical energy consumed in the circuit is given by
 ωr  2 π f r  dW = P dt = EIdt
ω == Eosin (ωt + θ). Iosinωt dt
fr 
1 0 ω 1 ωr ω 3
= E0.I0. sinω t sin(ω t  θ) dt
2π LC Fig. The variation of I versus ω
= E0.I0. [ sinω t . cosθ + cosω t . sinθ ]dt
This is required resonance frequency of the
LCR series circuit and is equal to the natural frequency of an LC circuit. = E0.I0. [ sin 2 ω t . cosθ + sinω t cosω t. sinθ ]
Condition of resonace: sin2ω t
= E0.I0. [ sin 2 ω t . cosθ + . sinθ ]
(i) XL = X C 2
 (1 cos2 ω t) sin2ω t  (1  cosω t)  π
= E0.I0. [ . cosθ + . sinθ  sin 2 ω t   P av = E rms .I rms cos = 0. So the current through the pure inductor L is wattles
2 2  2  2
E 0 .I 0 current.
= [ cosθ  cos2ω t.cosθ + sin2ω t . sinθ ] π
2 (ii) In case of inductor C, θ =  .
Total electrical energy consumed in the circuit in one complete cycle ( i.e. T= 0 to T = t) 2
is given by (i)  P av = E rms .I rms
T
 π
W= 0 dw cos -  = 0. So the
 2
current through the pure
E 0 .I 0 T T T 
=   cos θ.dt   cos2ω t.cos θ.dt   sin2ω t. sin θ.dt  capacitor C is wattles
2 0 
 current.
0 0

E .I  T T T 
SHORT QUESTION ANSWERS
= 0 0 cosθ  dt  cosθ  cos2ω t.dt  sinθ  sin2ω t .dt 
2  

0 0 0
1. 2076 C Q.No. 1(f), 2068 old can, 2068 can A choke coil is preferred to a resistor in an ac
E .I   cos2ω t  
T T
 sin2ω t  circuit. Why?
= 0 0 cosθ t  T0  cosθ    2ω   Ans: In pure resistor, e.m.f. and current are in same phase i.e. phase angle   0 0 , then
 sinθ
2   ω 0   0 
E .I E .I The power loss in resistor, PR  IV cos   IV cos 0 o  IV  1  IV. The power loss in
= 0 0 [ cosθ . T – 0 – 0]= 0 0 . cosθ . T. resistor is maximum
2 2
Now, the average power consumed in the A.C. circuit in one complete cycle is But, a choke coil is made by inductive coil of inductance (L). In a pure inductance,
W E 0 .I 0 π
Pav = = . cosθ e.m.f. leads the current by the phase angle . So,
T 2 2
π
E .I
 P av = 0 0 cosθ  P av = E rms .I rms cos = 0.
2 2
Again, Hence, when coil is used ,the power loss in inductor is minimum.
E I Therefore, a choke coil is preferred to a resistor in an alternating current circuit
P av = 0 . 0 cosθ
2 2
2. 2076 B Q.No. 1f, 2070 C Q.No. 1f 220 V A.C. is more dangerous than 220 V D.C.
 P av = E rms .I rms cosθ Why?
Hence, the average power consumed in the A.C. circuit is equal to the product of r.m.s. or Ans: We know that relation between root mean square voltage and peak value of voltages
virtual values of current and e.m.f. and cosθ .
is E 0  2 E rms . So, the peak value of 220 V a.c is 2 times greater than 220 V d.c.
Power factor: It is defined as the ratio of true or average power to apparent power, i.e.
Hence, a 220 V a.c. is more dangerous than 220 V
True or average power
Power factor (cosθ ) = 3. 2074 S Q.No. 1f , 2072 D, , 2071 C, 2067 2st Exam, 2058 What is meant by wattles
Apparent power
current?
Pav Ans: The current in an a.c. circuit is wattles if the average power consumed in the circuit
 cosθ 
E rms .I rms is zero. This is possible only when the phase difference between the current and the
Wattless Current: It is defined as the current through the pure inductor L or pure capacitor voltage is π/2 i.e.,
π
C which requires no power for maintenance of its circuit. It is also called idle current.  P av = E rms .I rms cos = 0
π 2
(i) In case of inductor L, θ = . π
2 In a pure inductor, e.m.f. leads the current by the phase angle .
2
 π Hence, a.c. current passes through the capacitor but d.c. current does not.
 P L = E rms .I rms cos =0
a. 2072 S Q.No. 1f How does the resonance frequency of an LCR series circuit
2
π changes if the plates of the capacitor are brought closer together?
In a pure capacitor, e.m.f. lagss the current by the phase angle . Ans: Resonance frequency of the LCR series circuit is
2
π 1
 P C = E rms .I rms cos =0 fr  ... ... ... (1)
2 2π LC
Hence, wattles current can be obtained by using an inductor or capacitor in the circuit. The capacitance of a parallel plate capacitor is given by
That is, if a circuit doesn’t have any resistance, no power is consumed in the circuit. C   0 A / d ... ... ... (2)
4. 2074 A Q.No. 11 The e.m.f. of an ac source is given by the expression, Since capacitance increases with decreasing distance between plates, so from equation
E  300Sin314t volt . Write the values of peak voltage and frequency of source. (1) , the resonance frequency of an LCR series circuit decreases with decreasing
Ans: The expression of a given e.m.f. of an ac source is distance between the plates of capacitor and vise-versa. Hence resonance frequency
E  300Sin314t volt ... ... ... (1) changes with changing distance between parallel plates of capacitor.
A sinusoidal e.m.f. (E) of an ac source is given by 7. 2072 E Q.No. 1f, 2067 Old What are the advantages of a.c. over d.c?
E  E oSin t ... ... ... (2) Ans: The following advantages of a.c. over d.c. are as follows:
Where Eo= peak value voltage and  is angular velocity i. The voltage of a.c. can be changed from lower voltage to higher and higher to
Comparing equations (1) and (2), we get lower voltage with help of step up transformer and step down transformer
Eo = 300 V and  = 314 rad/s respectively but voltage of d.c. cannot be changed.
ii. In a.c., power losses (P min) = 0 but in d.c., power loss (P max) = IV.
 314
Now, Frequency (f) =   50Hz . iii. The a.c. can be easily converted into d.c. but reverse is little difficult.
2 2 iv. Many electrical appliances like electric motor, electric fan, micro oven, etc are
Hence, the required the values of peak a.c. voltage and frequency are 300 V and 50 Hz run by only a.c. but not d.c.
respectively. 8. 2070 D Q.No. 1f, 2059 Fluorescent tube often uses an inductor to limit the
5. 2074 B Q.No. 1e Define r.m.s. value of ac. How is it related with the peak value current through the tube. Why is it better to use inductor than a resistor for
of ac? this purpose?
Ans: Root mean square or virtual or effective value of a.c.: The r.m.s. value of an a.c. is Ans: In a pure inductance, e.m.f. leads the current by the phase angle π/2. So,
defined as that value of a direct current which produces the same heating effect in π
given resistor as is produced by the alternating current when passed for the same time. P av = E rms .I rms cos = 0.
Let Irms and Io be the r.m.s and peak value of a.c. current respectively, then 2
Hence, when coil is used, the power loss in inductor is minimum.
Ieff  I rms  Io / 2  0.707 Io
But In pure resistor, e.m.f. and current are in same phase i.e. phase angle   0 0 , then
This is the relation between the r.m.s and peak value of a.c. current.
6. 2073 D Q.No. 1e Alternating current passes through a capacitor whereas direct The power loss in resistor, PR  IV cos   IV cos 0 o  IV  1  IV. The power loss in
current does not. Explain this fact on the basis of capacitive resistor is minimum. It means if we uses resistor, there is power loss in resistor. Hence,
reactance. an inductor is used in fluorescent tube to limit current through the tube without power
Ans: The capacitive reactance of the capacitor is given by the relation, losses.
1 1 9. 2071 S Q.No. 1b For a capacitor in an a.c. circuit, explain why there is a phase
XC   difference between current and voltage.
C 2f  C
1 1
For D.C. current, f  0 then X C   
2f  C 2  0  C
This relation shows that the resistance is infinity for capacitor. So, the current is block
for D.C. current. For A.C. current, f  50 Hz then
1 1
XC    certain value
2f  C 2  50  C
 X C  certain value
Ans: The phasor diagram between current and voltage for a capacitor in an a.c. circuit is 12. 2069 B Q.No. 1b Sketch the symbols of a capacitor, an inductor ,emf of a cell and
figure. E or I E  E sin ωt a resistor.
From I  Isin ωt Ans:
 -  
C E
Fig (a). capacitor Fig (b) An inductor Fig (c). E.m.f Fig (d ). Resistance
π 2π 3π ωt 90 o I VR The symbols of a capacitor, an inductor, e.m.f. of a cell and a resistor are shown in
figure.
E VC 13. 2054 Q.No. 7g What is meant by impedance of an a.c. circuit?
Fig (a ). A.C. circuit Ans: The total resistance offered by the a.c. circuit to the alternating current is called
impedance of an a.c. circuit. Hence, the impedance (Z) of a.c. series circuit containing
containing C only Fig ( b). Phasor diagram Fig (c). vector diagram
inductor of inductance(L), capacitor of capacitance (C) and resistor of resistance(R)
diagram, we have is given by
 E  E 0 cos ωt  π/2  2
 1 
Hence, it shows that the alternating E.M.F. lags the current by the phase angle  / 2 Z  R 2  (X L  XC )2  R 2   L  
 C 
10. 2069 A Q.No. 1e At high frequencies, a capacitor becomes a short-circuit and an
inductor becomes an open circuit. Explain.  1 
 X L   L and X C  C 
Ans: We know that capacitive reactance is X c 
1

1  
C 2fC This is known as the impedance of the a.c circuit. It plays the same role as resistance in the
1 D.C. circuit.
At high frequencies, f   X c  0
2    C Write application of LCR resonance?
V V Ans: Application of LCR resonance are as follows
Current through it (I ) =    and hence I is maximum.  Oscillator circuit, radio receivers, and television sets are used for the tuning purpose.
Xc 0
 The series and RLC circuit mainly involves in signal processing and
Therefore, at high frequencies, a capacitor becomes short-circuit.
communication system.
Again Inductive reactance is X L  L  2fL
 It is used to stabilize the electrical frequency of an Alternating current(AC)
At high frequencies, f   , X L  L  2    L  
Oscillator circuit
V V
Current through it (I ) =   0 and hence I is minimum.
XL  ***
Therefore, at high frequency inductor becomes an open circuit
11. 2069 A Old Q.No. 2e Define power factor in an a.c. circuit.
Ans: The power factor of an AC circuit is defined as the ratio of true or average power to
apparent power, i.e.
True or average power
Power factor (cosθ ) =
Apparent power
Pav
 cosθ 
E rms .I rms
For example, In LCR series circuit, the power factor is
i2R R R
Cos  2
 
i Z Z R 2  (X L  X C ) 2
It is dimensionless number between 0 and 1.