NOTES ON ‘ALTERNATING CURRENT’ (Class 12) Page 01

Alternating current (A.C) is the current that changes in magnitude direction continuously with
respect to time.

It can be represented as,

The currents and voltages in a.c circuits can be expressed by the following terms:-

(i) Instantaneous (I or V) :- It is the current or voltage that in the circuit at any instant.
(ii) Peak (Im or Vm):- It is the maximum available voltage or current in the circuit.
(iii) Average (Iav or Vav):- It is the arithmetic mean of the instantaneous values of voltages or
currents in the circuit.
(iv) Room Mean Square (R.M.S) or Effective:- It is the square root of the average of the squares
of the instantaneous values of the voltages or currents in the circuit.
𝑣12+𝑣22+𝑣32+⋯.
Vrms = √
𝑛

R.M.S value is the d.c equivalent of a.c

Relation between Peak and R.M.S values

𝑇 𝐼2𝑑𝑡 𝑇 𝐼𝑚2𝑆𝑖𝑛2𝜔𝑡.𝑑𝑡
𝐼2
𝑟𝑚𝑠 =∫0 𝑇
=∫0 𝑇
𝑇 𝐼2 1−𝑐𝑜𝑠2𝜔𝑡
= ∫0 𝑚
𝑇
(
2
) 𝑑𝑡 where  = 2/T
𝐼2 𝑇 𝑇
𝐼2 = 𝑚 [∫ 𝑑𝑡 − ∫ 𝑐𝑜𝑠2𝜔𝑡 𝑑𝑡] ---------- (1)
𝑟𝑚𝑠 2𝑇 0 0

But 𝑇 𝑠𝑖𝑛2𝜔𝑡 𝑇 1 4𝜋 𝑇 1
∫0 𝑐𝑜𝑠2𝜔𝑡 𝑑𝑡 = [ 2𝜔
]0 = 2𝜔 [𝑠𝑖𝑛 𝑇
𝑡]0 = 2𝜔 [𝑠𝑖𝑛4𝜋 − sin 0] = 0

Thus equation (1) becomes, 𝐼2 =𝐼𝑚2 [ T – 0 ] So, 𝐼 =

𝐼𝑚
𝑟𝑚𝑠 2𝑇 𝑟𝑚𝑠 √2

𝑚 𝑉
Similarly, it can also be shown that , 𝑉𝑟𝑚𝑠 = √2

Note: the measurement of ac is done by the comparison of the heating effect produced by ac with that
by dc. Hence ac measuring instruments are also called ‘hot-wire instruments’.

Show that the average current in the complete cycle of a.c is zero.
𝑇 𝐼𝑚𝑠𝑖𝑛𝜔𝑡.𝑑𝑡 𝐼𝑚⁄ 𝑇
I = ∫ 𝑇 𝐼 𝑑𝑡 = ∫ == ∫ 𝑠𝑖𝑛𝜔𝑡. 𝑑𝑡 = 𝐼𝑚⁄ [−𝑐𝑜𝑠𝜔𝑡]𝑇
av 0 𝑇 0 𝑇 𝑇 0 𝑇 𝜔 0
𝐼𝑚⁄ 2 𝑇 𝐼𝑚⁄
=− [𝑐𝑜𝑠 ] = − [𝑐𝑜𝑠2 − 𝑐𝑜𝑠0] = 0 (Hence proved)
𝑇 𝑇 0 𝑇
𝑇 𝑇
𝐼 𝑑𝑡 𝐼 𝑠𝑖𝑛𝜔𝑡.𝑑𝑡 𝑇 𝐼𝑚⁄ −𝑐𝑜𝑠𝜔𝑡 𝑇/2
Note:- But in half a cycle, Iav = ∫ 2 = ∫ 2 𝑚 = = 2 𝐼𝑚⁄ ∫2 𝑠𝑖𝑛𝜔𝑡. 𝑑𝑡 = 2 [ ]
0 𝑇/2 0 𝑇/2 𝑇 0 𝑇 𝜔 0

2𝐼𝑚⁄ 2 2𝑇 𝐼𝑚 𝐼𝑚
=−  [𝑐𝑜𝑠 ] = − ⁄ [𝑐𝑜𝑠 − cos 𝑜] = − ⁄ [−1 − −1]
𝑇 𝑇 0 𝑇 𝑇

This gives Iav = 2𝐼𝑚⁄ Similary, it can be shown that Vav = 2𝑉𝑚⁄
NOTES ON ‘ALTERNATING CURRENT’ (Class 12) Page 02

Phasor diagrams

These are the vector diagrams in which the magnitudes represent the peak values and
directions represent the phase differences between voltages and currents in the a.c circuits.

A.C Circuits

Fundamental components in a.c circuits.

1. Resistor

2. Inductor

3. Capacitor

1. Circuit containing only resistor

Circuit diagram: Let V = Vm sint Current (I) = V/R = (Vm sint)/R

i.e., I = Im sint. Where Im = Vm/R. R = Vm/Im

V Hence the phase difference between voltage and current is,  = 0.

So the phasor diagram can be as shown I V

The graph showing the variation of current and voltage with time can be represented as:

Power associated with the circuit:

𝑇
𝑃 =∫
𝑇 𝑉𝐼𝑑𝑡 =1 ∫𝑇 𝑉 𝑠𝑖𝑛 𝑡. 𝐼 𝑠𝑖𝑛 𝑉𝑚𝐼𝑚 ∫ 𝑠𝑖𝑛2 𝑡. 𝑑𝑡
𝑅 0 𝑇 𝑇 0 𝑚 𝑚 𝑡𝑑𝑡 = 𝑇 0

𝑉𝑚𝐼𝑚 𝑇 1−𝑐𝑜𝑠2𝑡) 𝑉𝑚𝐼𝑚 𝑇1 𝑇 𝑐𝑜𝑠2𝑡 𝑑𝑡

= ∫0 ( 𝑑𝑡 = [∫0 2 𝑑𝑡 - ∫0 ]
𝑇 2 𝑇 2
𝑇 𝑐𝑜𝑠2𝑡 𝑑𝑡
But ∫0 2
=0 Thus we get, PR = (Vm Im) /2 Hence PR = Vrms. Irms

2. Circuit containing only inductor

Circuit diagram: Let V = Vm sint

The induced voltage in the inductor is Vind = - L(dI/dt)

V By loop rule, Vm sint - L(dI/dt) = 0

i.e., Vm sint = L(dI/dt) this gives, dI = (Vm/L) sint dt

𝑉𝑚 −𝑐𝑜𝑠𝑡
On integrating the above equation we get, Current (I) = ( )
𝐿 
𝑉𝑚
I=− 𝑐𝑜𝑠𝑡 Hence I = Im sin (t - /2) where Im = ( Vm/L)
𝐿

Here ‘L’ represents the non-resistive opposition offered by the inductor to the flow of current
in an ac circuit. It is called inductive reactance (XL) i.e., XL = L

Hence the phase difference between voltage and current is,  = /2.
NOTES ON ‘ALTERNATING CURRENT’ (Class 12) Page 03

Here the current lags behind the voltage by /2.

So the phasor diagram can be as shown V

The graph showing the variation of current and voltage with time can be represented as:

Power associated with the circuit:

𝑉𝑚𝐼𝑚 𝑇
𝑃𝐿 = ∫0
𝑇 𝑉𝐼𝑑𝑡 =1 ∫𝑇 𝑉 𝑠𝑖𝑛 𝑡. 𝐼 (𝑠𝑖𝑛𝑡 − /2)𝑑𝑡 = ∫ 𝑠𝑖𝑛 𝑡. −𝑐𝑜𝑠𝑡𝑑𝑡
𝑇 𝑇 0 𝑚 𝑚 𝑇 0

𝑉𝑚𝐼𝑚 𝑇 𝑠𝑖𝑛2𝑡 𝑉𝑚𝐼𝑚 𝑇 𝑠𝑖𝑛2𝑡 𝑑𝑡

=− ∫0 ( )𝑑𝑡 = − .0 as ∫0 =0
𝑇 2 2𝑇

Thus we get, PL = 0 i.e., an ideal inductor does not consume any power.

(Note: this is why they are used as ‘choke coils’ in ac circuits to reduce current
without power loss, in the place of resistors)

3. Circuit containing only capacitor

Circuit diagram: Let V = Vm sint

𝑑Q 𝑑𝑉
We have Q = CV Differentiating we get, 𝑑𝑡
= 𝐶 𝑑𝑡
𝑉𝑚 
i.e. I = C Vm cost .  I=1 sin(𝑡 + )
⁄𝐶 2

𝑉𝑚
Hence I = Imsin(t+/2) where Im = 1⁄
𝐶

Thus Xc = 1/C gives the non-resistive opposition to the flow of current called capacitive reactance.

Hence the phase difference between voltage and current is,  = /2.

Here the current leads the voltage by /2. So the phasor diagram I V

Is shown beside:-

The graph showing the variation of current and voltage with time can be represented as:

Power associated with the circuit:

𝑉𝑚𝐼𝑚 𝑇
𝑃𝐿 = ∫0
𝑇 𝑉𝐼𝑑𝑡 =1 ∫𝑇 𝑉 𝑠𝑖𝑛 𝑡. 𝐼 (𝑠𝑖𝑛𝑡 + /2)𝑑𝑡 = ∫ 𝑠𝑖𝑛 𝑡. 𝑐𝑜𝑠𝑡𝑑𝑡
𝑇 𝑇 0 𝑚 𝑚 𝑇 0

𝑉𝑚𝐼𝑚 𝑇 𝑠𝑖𝑛2𝑡 𝑇
= ∫ ( )𝑑𝑡 = 𝑉𝑚𝐼𝑚 .0 as ∫ 𝑠𝑖𝑛2 𝑡 = 0
𝑇 0 2 2𝑇 0

Thus we get, PC = 0 i.e., an ideal capacitor does not consume any power.

3. L-C-R series circuit

Circuit diagram: Let V = Vm sint

Since the components are in series, the current, I is the same

throughout. Let VR, VL and VC be the p.d’s across the resistor, inductor
and capacitor respectively. Assuming VL to be greater than VC, the
phasor diagram can be drawn as,

VL I

The resultant of the above vector diagram is: VR

VC
NOTES ON ‘ALTERNATING CURRENT’ (Class 12) Page 04

By parallelogram law, VL - VC V I

The resultant voltage is given VR

by, V =√𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2 But VR = IR, VL = IXL and VC = IXC

Thus V = 𝐼√𝑅2 + (𝑋𝐿 − 𝑋𝐶)2 Hence V/I = √𝑅2 + (𝑋𝐿 − 𝑋𝐶)2

This gives the total opposition to the flow of current in an a.c circuit, including resistance and
reactances. It is called the impedance (Z) of the circuit.

Thus the impedance is given by Z = √𝑅2 + (𝑋𝐿 − 𝑋𝐶)2

Also the phase difference between the voltage and current is given by
𝑉𝐿−7 (𝐼𝑋𝐿−𝐼𝑋𝐶) (𝑋𝐿−𝑋𝐶)
𝐶
tan = i.e. tan = or tan =
𝑉𝑅 𝐼𝑅 𝑅

Hence, current in the circuit can be expressed as I = Imsin(t-), the negative sign
indicates that the inductive reactance dominates capacitive. i.e. current lags behind voltage.

(Refer page 246 of Text book for ‘Analytical solution’)

Power associated with the circuit:

𝑉𝑚𝐼𝑚 𝑇
𝑃 = ∫0
𝑇 𝑉𝐼𝑑𝑡 =1 ∫𝑇 𝑉 𝑠𝑖𝑛 𝑡. 𝐼 (𝑠𝑖𝑛𝑡 − )𝑑𝑡 = ∫ 𝑠𝑖𝑛 𝑡 [𝑠𝑖𝑛𝑡. 𝑐𝑜𝑠 − 𝑐𝑜𝑠𝑡. 𝑠𝑖𝑛]𝑑𝑡
𝑇 𝑇 0 𝑚 𝑚 𝑇 0
𝑉𝑚𝐼𝑚 𝑇 𝑇
= [ 𝑐𝑜𝑠. ∫ 𝑠𝑖𝑛2𝑡. 𝑑𝑡 − 𝑠𝑖𝑛. ∫ 𝑠𝑖𝑛 𝑡. 𝑐𝑜𝑠𝑡𝑑𝑡]
𝑇 0 0
𝑉𝑚 𝐼 𝑚 𝑇 1−𝑐𝑜𝑠2𝑡 𝑇 𝑠𝑖𝑛2𝑡
= [ 𝑐𝑜𝑠. ∫ ( ) 𝑑𝑡 − 𝑠𝑖𝑛. ∫ 𝑑𝑡
𝑇 0 2 0 2

This gives P = Vrms. Irms.cos. (Refer above for the integration results)

It shows that power in the circuit depends not only the values of current and voltage, but on the value of
‘cos’ also. Hence it is called the power factor of the circuit.

i.e. power factor, cos = (VR/V) = (IR/IZ) cos = 𝑅

𝑍

𝑅
Thus power can also be expressed as P = Vrms. Irms. 𝑍

Wattless current:- It is the current in the a.c circuit in which there is a phase difference of ‘/2’ between
current and voltage.

We have P = Vrms. Irms.cos Here cos = 0

It can be represented as

Irms Thus by resolving the current into the respective components, we get the

 Vrms Wattless current = Irms.sin

RESONANCE:- It is the condition at which the inductive and capacitive reactances are equal so that the
current in the circuit is maximum.

We know Im = V/Z Current is maximum, when impedance is minimum.

So ‘Z’ is minimum when 𝑋𝐿 = 𝑋𝐶

Hence Zmin = R L = 1/C or 2 = 1/LC

NOTES ON ‘ALTERNATING CURRENT’ (Class 12) Page 05
1
Thus (Im)max = Vm/R Resonant frequency r = 1
and 𝑟 =
√𝐿𝐶 2𝜋√𝐿𝐶

Frequency response curves with current

The variation of peak current

with different frequencies

can be represented for different values

of resistances as shown beside:

VARIATION OF RESISTANCE, REACTANCE AND IMPEDANCE WITH FREQUENCY

Resistance is independent of the frequency.

Xc = 1/C So, Xc α 1/

XL = L XL α 

r

POWER AT RESONANCE
𝑅 𝑉𝑚 𝐼𝑚
We have Power P = Vrms . Irms. But Vrms = , Irms = and Im = Vm / Z
Z √2 √2

𝑉2
So 𝑅
P = 2Z𝑚2 ------------------------------- (1)

At resonance, Z is minimum. Thus power also is maximum at resonance.

𝑉𝑚2 𝑅2
Hence Pmax = 2𝑅
Eqn(1) becomes P = Pmax 𝑍2

The frequency response curve with power can be drawn as:

The frequencies at which the power in the circuit is half of

Its maximum value are called ‘half power points’.

The difference between half power points is called

Bandwidth (2)

2 = 2 - 1

[For the derivation of the expression for band width, 2 = R/L, refer the note book]

Q-FACTOR (Quality factor)

It measures the sharpness of the resonance curve. It is defined as the ratio of the resonant
frequency to the band width.
𝑟
i.e. 𝑸= Using the above equations we get,
2∆

1 𝐿 1 𝐿
Q= . so, 𝑸 = √
√𝐿𝐶 𝑅 𝑅 𝐶
NOTES ON ‘ALTERNATING CURRENT’ (Class 12) Page 06

Radio Tuner

Tuning is the process of receiving the frequency of a desired radio station. It consists of an
inductor and a variable capacitor.

During tuning, the value of capacitance is changed. At a given position the

value of capacitance is such that it satisfies for a given frequency in the

expression for resonant frequency, 𝑟 = 1 .

2𝜋√𝐿𝐶

Since the Q-factor of the circuit is adjusted to be very high, the maximum power reception
occurs only at this frequency. Thus tuning is achieved.

L-C OSCILLATIONS

A capacitor, initially charged to Q0 and is disconnected from the battery. Further it is connected
across an inductor. This causes to and fro motion of electrons in the circuit, as illustrated below:

This gives electrical oscillations of frequency  = 1/T

Mathematical Treatment

[Qn. Show that electric charges execute S.H.M in an L-C circuit. Hence derive an expression for the
frequency of oscillations. Show that the total energy in the circuit is conserved.]

------ Refer Note book---------------

Advantages of A.C

(i) Cost of production is low.

(ii) It can be transmitted to distant places with minimum power loss, with the help of
transformers.
(iii) It can easily be converted into d.c, wherever needed.
(iv) Current in the circuits can be controlled with the help of choke coils.

Disadvantages of A.C

(i) It is more dangerous.

(ii) It cannot be used in the processes like electroplating, making permanent magnets, etc.
(iii) Energy loss can occur in the case of high frequency a.c due to skin effect. [ It is the effect in
which high frequency a.c confines through the surface of a conductor, when transmitted.]
 Alternating Currents

1. The power is transmitted from a power house on high voltage ac because

(a) Electric current travels faster at higher volts
(b) It is more economical due to less power wastage
(c) It is difficult to generate power at low voltage
(d) Chances of stealing transmission lines are minimized
2. The potential difference V and the current i flowing through an instrument in an ac circuit
of frequency f are given by V = 5 cos  t volts and I = 2 sin t amperes (where  = 2f). The
power dissipated in the instrument is
(a) Zero (b) 10 W
(c) 5 W (d) 2.5 W
3. Alternating current(A.C.) can not be measured by D.C. ammeter because
(a) A.C. cannot pass through D.C. ammeter
(b) Average value of complete cycle is zero
(c) A.C. is virtual
(d) A.C. changes its direction

4. In an ac circuit, peak value of voltage is 423 volts. Its effective voltage is

(a) 400 volts (b) 323 volts
(c) 300 volts (d) 340 volts

5. In an ac circuit I = 100 sin 200 t. The time required for the current to achieve its peak
value will be
1 1
(a) sec (b) sec
100 200

1 1
(c) sec (d) sec
300 400
6. An alternating current is given by the equation i = i1 cos  t + i2 sin  t . The r.m.s. current is given
by
1 1
(a) (i + i ) (b) (i + i )2
1 2 i 2
2 2

(c) 1
(i2 + i2 )1 / 2 (d) 1 (i2 + i2 )1 / 2
1 2 1 2
2 2

7. The phase angle between e.m.f. and current in LCR series ac circuit is
(a) 0 to  / 2 (b)  / 4
(c)  / 2 (d) 
8. Current in the circuit is wattless, if
(a) Inductance in the circuit is zero
(b) Resistance in the circuit is zero
(c) Current is alternating
(d) Resistance and inductance both are zero
9. An alternating current of frequency ' f ' is flowing in a circuit containing a resistance R and

a choke L in series. The impedance of this circuit is

(a)R + 2fL (b) R 2 + 4 2 f 2 L2

(c) R 2 + L2 (d) R 2 +2fL

10. A resonant ac circuit contains a capacitor of capacitance 10 −6 F and an inductor of 10 −4 H. the

frequency of electrical oscillations will be
(a) 105 Hz (b) 10 Hz
(c) 5
10
10 (d) Hz
Hz
2 2

11. Power delivered by the source of the circuit becomes maximum when
(a) L = C (b) L = 1
C

(c) L = −  1 C 
2
(d) L = C
 
12. In a LCR circuit having L = 8.0 henry, C = 0.5 F and R = 100 ohm in series. The
resonance frequency in per second is
(a) 600 radian (b) 600 Hz
(c) 500 radian (d) 500 Hz
13. In LCR circuit, the capacitance is changed from C to 4C. For the same resonant frequency,
the inductance should be changed from L to
(a) 2L (b) L / 2
(c) L / 4 (d) 4 L
14. 4.In a series LCR circuit, operated with an ac of angular frequency  , the total impedance is
(a) [R 2 + (L − C)2 ]1 / 2

 1 2 1 / 2
(b) R 2 + L −  
  C  


 1 2 −1 / 2
(c) R 2 +  L −  
  C  


 1 2 1 / 2
(d) (R)2 +  L −  
  C  

15. For series LCR circuit, wrong statement is

(a) Applied e.m.f. and potential difference across resistance are in same phase.
(b) Applied e.m.f. and potential difference at inductor coil have phase difference of  / 2.
(c) Potential difference at capacitor and inductor has phase difference of  / 2 .
(d) Potential difference across resistance and capacitor has phase difference of  / 2.
16. In a purely resistive ac circuit, the current
(a) Lags behind the e.m.f. in phase
(b) Is in phase with the e.m.f.
(c) Leads the e.m.f. in phase
(d) Leads the e.m.f. in half the cycle and lags behind it in the other half
17. The current in series LCR circuit will be maximum when  is
(a) As large as possible
(b) Equal o natural frequency of LCR system
(c) LC

(d) 1 / LC

18. An inductor L and a capacitor C are connected in the circuit as shown in the figure. The
frequency of the power supply is equal to the resonant frequency of the circuit. Which
ammeter will read zero ampere

A1
C
A2

A3
E = E0 sin

(a) A1 (b) A2
(c) A3 (d) None of these
19. [A]: In LCR series circuit, current is maximum at resonance.
[R]: In LCR series circuit, impedance is minimum at resonance.
a) Both Assertion and Reason are true and reason is correct explanation of Assertion.
b) Both Assertion and Reason are true but reason is not the correct explanation of Assertion.
c) Assertion is true but reason is false.
d) Both assertion and reason are false.
20. [A]: If Xc < XL, the phase factor is negative.
[R]: If Xc < XL; current lags behind emf.
a) Both Assertion and Reason are true and reason is correct explanation of Assertion.
b) Both Assertion and Reason are true but reason is not the correct explanation of Assertion.
c) Assertion is true but reason is false.
d) Both assertion and reason are false.
21. [A]: Electric power is transmitted over long distances through conducting wires at high
voltage.
[R] : A Power loss is less when power is transmitted at high voltage.
a) Both Assertion and Reason are true and reason is correct explanation of Assertion.
b) Both Assertion and Reason are true but reason is not the correct explanation of Assertion.
c) Assertion is true but reason is false.
d) Both assertion and reason are false.
22. The voltage of an a.c source varies with time according to the relation E = 120 sin 100  t
cos 100 t. Then
i) The peak voltage of the source is 120 V
ii) The peak voltage = 60 V
iii) The peak voltage = 120/ 2 V
iv) The frequency of source voltage is 100 Hz
a) i) and ii) are correct b) ii) and iii) are correct
d) ii) and IV) are correct c) i) and iii) are correct
23. In the circuit shown below what will be the reading of the voltmeter and ammeter?

a) 2.2 A b) 1.2 A c) 4.2 A d) 0.5 A

 Key

1) b 2) a 3) b 4) c 5) d 6) c 7) a 8) b 9) b 10) c

11) b 12) c 13) c 14) b 15) c 16) b 17) d 18) c 19) a 20) a

21) a 22) c 23) a

Hints

1
1. Power loss 
(Voltage)2
 
2. V = 5 cos  t = 5 sin  t + and i = 2 sin  t
 
 2

Power = Vr.m.s.  ir.m .s.  cos  = 0

(Since  =  , therefore cos  = cos  = 0 )
2 2

3. In dc ammeter, a coil is free to rotate in the magnetic field of a fixed magnet.

If an alternating current is passed through such a coil, the torque will reverse its direction
each time the current changes direction and the average value of the torque will be zero.
Vo = 423 = 300 V
4. Effective voltage Vr.m.s. =
2 2

T
5. The current takes sec to reach the peak value.
4

2 1
In the given question = 200  T = sec
T 100

1
Time to reach the peak value = sec
400

1 2 21/ 2
i21 + i22
6. irms = = (i1 + i2 )
2 2

7. (a)
8. Because power = i2 R, if R = 0, then P = 0.

9. Z= R 2 + X 2L , X L = L and  = 2f

Z= R 2 + 4 2 f 2 L2

1 105
10.  = = 1 = Hz
2 LC 2 10 −6 10 −4 2

11. (b)
12. Resonance frequency in radian/second is
1 1
= = = 500 rad / sec
LC 8  0.5  10 −6

1 1 L1
13. = =  L2 =
L1 C1 L2 C 2 4

14. (b)
15. (c)
16. (b)
17. At resonant frequency current in series LCR circuit is maximum.
18. (c)
19. a
20. a
21. a
22. c
23. V= VR2+ V(−LV C

220 = VR2+ 300 − 300  VR = 220V

2

So reading in voltmeter = 220V

V = i R  220 = i 100  i = 2.2 A.
 D.C Circuits

1. Why the current does not rise immediately in a circuit containing inductance
(a) Because of induced emf
(b) Because of high voltage drop
(c) Because of low power consumption
(d) Because of Joule heating
2. The adjoining figure shows two bulbs B1 and B2 resistor R and an inductor L. When the
switch S is turned off

B1

B2

(a) Both B1 and B2 die out promptly.

(b) Both B1 and B2 die out with some delay.
(c) B1 dies out promptly but B2 with some delay.
(d) B2 dies out promptly but B1 with some delay.
3. In L-R circuit, for the case of increasing current, the magnitude of current can be
calculated by using the formula
(a) I = I0e − Rt / L (b) I = I0(1 − e−Rt / L) (c) I = I0(1 − e Rt / L ) (d) I = I0 e Rt / L

4. A capacitor is fully charged with a battery. Then the battery is removed and coil is
connected with the capacitor in parallel, current varies as
(a) Increases monotonically (b) Decreases monotonically
(c) Zero (d) Oscillates indefinite
5. The time constant of an LR circuit represents the time in which the current in the circuits
(a) Reaches a value equal to about 37% of its final value
(b) Reaches a value equal to about 63% of its final value
(c) Attains a constant value
(d) Attains 50% of the constant value
6. The resistance and inductance of series circuit are 5 and 20H respectively. At the instant
of closing the switch, the current is increasing at the rate 4A-s. The supply voltage is
(a) 20 V (b) 80 V (c) 120 V (d) 100 V
7. In inductance L and a resistance R are first connected to a battery. After some time the
battery is disconnected but L and R remain connected in a closed circuit. Then the current
reduces to 37% of its initial value in

(a) RL sec (b) R sec (c) L

sec (d) 1
sec
L R LR

8. In an LR-circuit, time constant is that time in which current grows from zero to the
value (where I0 is the steady state current)
(a) 0.63 I0 (b) 0.50 I0 (c) 0.37 I0 (d) I0

9. A solenoid has an inductance of 60 henrys and a resistance of 30 ohms. If it is connected

e−1
to a 100 volt battery, how long will it take for the current to reach  63 .2% of its final
e

value?
(a)1 second (b) 2 seconds (c) e seconds (d) 2e seconds
10. A coil of inductance 300 mH and resistance 2 is connected to a source of voltage 2V . The
current reaches half of its steady state value in
(a) 0.15 s (b) 0.3 s (c) 0.05 s (d) 0.1 s

11. Find the time constant (in  s ) for the given RC circuits in the given order respectively.
R1= 1 , R2 = 2 , C1 = 4F , C2 = 2 F

V
R1 V
R2 C1 V
C1
C2 R C1
1 R1

C2
R2
C2 R2
I) II) III)

8 8 8 8
a) 18, 4, b) 18, ,4 c) 4, 18, d) 4, , 18
9 9 9 9
12. An ideal coil of 10 Henry is joined in series with a resistance of 5 ohm and a battery of 5
volt. 2 second after joining, the current flowing in ampere in the circuit will be

1) e-1 2) (1-e–1) 3) (1-e) 4) e

13. An inductor of L = 400 mH and two resistors of R1 = 2 and R2 = 2 are connected to a

12V battery as shown in the figure. The internal resistance of the battery is negligible. The
switch is closed at time t = 0. The potential drop across ‘L’ as a function of time is

2) 6(1 − e−t / 0.2 )V

12 −3t
1) e V 3) 12e−5tV 4) 6e−5tV
t

14. The current in the given circuit is increasing with a rate 4 amp/ s. The charge on the
capacitor at an instant when the current in the circuit is 2 amp will be

1) 4 C 2) 5 C 3) 6 C 4) 3C

15. The ratio of time constants in charging and discharging in the circuit shown in figure is

1) 1:1 2)1:2 3) 3:2 4) 2:3

 Key

1) a 2) c 3) b 4) d 5) b

6) b 7) c 8) a 9) b 10) d

11) b 12) b 13) c 14) c 15) c

Hints

 di = −i − R  − RtL = i0 R .e − RtL
6. (b) i = i 1 − e − RtL    e
0
  dt 0 L L


di i0 R E E
At t = 0; = = 4=  E = 80 V
dt L L 20

7. (c) When battery disconnected current through the circuit start decreasing exponentially
according to i = i0 e − Rt / L

 0.37i0 = i0e − Rt / L  0.37 =

1
=e −Rt / L  t=L
e R
 − R t
8. (a) Current at any instant of time t after closing an L-R circuit is given by I = I0 1 − e L 

 

L
Time constant t=
R

 −R L
  1
 I = I0 1 − e L R  = I 0 (1 − e −1 ) = I0 1 − 
   e
 1 
=
I0 1 −  = 0.63 I0 = 63% Of I0
 2.718 

L 60
9. (b) t== = = 2 sec .
R 30
 − Rt 
10. (d) i = i 01 − e L   For i=
i0
, t = 0.693
L
 
 2 R


 300  10 −3
t = 0.693  = 0.1 sec
2
11. t = CR
t1 = (C1 + C2 ) ( R1 + R2 ) = 18  s
CC RR 8
t2 = C +1 C2  R + R =9
1 2
1 2 1 2

t = (C + C ) R1R2
=4
3 1 2
R1 + R2
 − t
R

12. i = i0 1− e L 
 
5 − 52 
=
i 1− e 10 
5 

. i = (1− e−1 ) A
12  − t
R

13. i0 = = 6 , p.d = v − i0 1− e L  R

2  
12 
t
2
−
p.d = 12 − 1− e 40010−3  2
2  

p.d = 12 e−5t volt.

di q
14. E =iR + L −
dt c
q
4=2+4–
3
q = 6c

15. 1 = R2 = 2R + R Or
1 3
=
2 R1 2R 2 2