P hysi cs | 23.25
Solved Examples
JEE Main/Boards
Example1: A resistance R, inductance L and a capacitor
C all are connected ina series with an AC supply. The
resistance of R is 16 Ω, and for a given frequency, the
inductive reactance of L is 24 Ω,and capacitive reactance
of C is 12 Ω. If the current in the circuit is 5 A, find
(a) The potential difference across R, L and C
(b) The impedance of the circuit
(d) Phase angle
R 2 + ( X C − XL ) where XC and XL are the capacitive
2
and inductive reactances respectively. Phase difference
 X − XC 
between voltage and current is φ =tan−1  L
.
 R



Potential drop across resistance is IR and that across
reactance is IX.
R 2 + ω2L2
. When capacitor is connected
( )
I (1 / ωC ) =5 × 12 =60 V
(iii) Capacitor VC =×
(ii) Inductor VL = I × ωL = 5 × 24 = 120 V
(b) The impedance of the circuit
2

1 
Z
R +  ωL −
=
=
ωC 

= 20 Ω
2
(16 ) + ( 24 − 12)
2
2
(c) The voltage of AC supply is given by
E =I × Z =5 × 20 =100 V
(d) Phase angle between voltage & current is
 ωL − ( 1 / ωC ) 
 24 − 12 
−1

 tan−1 
tan
=

R


 16 
= tan−1 0.75 = 360 52'
R 2 + ( X C − XL )
2
and the power factor of the
R
circuit is cos φ =
.
2
R 2 + ( XL − X C ) 


Z=
We want to find the value of the capacitor to make the
circuit’s power factor 1.0
(A) Find the value resistance and inductive reactance.
The power in AC circuit is given as
(i) Resistance VR =I × R =5 × 16 =80 V
)
R
For a LR circuit, current lags behind voltage in phase.
(a) Potential difference across
(
voltage by angle φ and the power factor of circuit is
in series in the circuit, the impedance of the circuit is
Sol: In a series LCR circuit, the impedance of circuit is
φ
Sol: In series LR circuit, the current lags the applied
cos φ =
(c) The voltage of AC supply
Z=
Example 2: A circuit draws a power of 550 W from a
source of 220 V, 50Hz. The power factor of the circuit is
0.8 and the current lags in phase behind the potential
difference. To make the power factor of circuit as 1.0,
what capacitance will be connected in the circuit?
P=
V 2rms × cos φ ⇒ Z=
...(1)
Z
V 2rms
( 220 )
2
× 0.8
= 70.4 Ω
550
× cos φ
=
P
Power factor cos φ =
… (i)
R
, so we get value of resistance as
Z
R = Z × cos φ = 70.4 × 0.8 = 56.32 Ω
Inductive Reactance is
ω
=
L
(Z
2
)
2
− R=
( 70.4 ) − (56.32)
2
2
ω=
L 42.2 Ω
(B) Capacitance needed to be connected in circuit to
make power factor = 1.0
When the capacitor is connected in the circuit.
Impedance
Z
=
2

1  
R 2 +   ωL −
 
ωC  


 ...(2)
…(ii)
2 3 . 2 6 | Alternating Current
and power factor is given by
cos φ =
as, U= P × t= mc∆θ=
R
=
t
2

1  
R 2 +   ωL −
 
ωC  



1
When cos φ= 1, ωL=
…(iii)
ωC
From (iii) we get
=
C
=
1
1
=
ω ( ωL ) 2πf ( ωL )
2 × 10
min
= 348
348 sec
secs= �5.8
5.8min.
0.0575
Sol: For LRC circuit, total potential difference is
V=
VR2 + ( VC − VL ) .
2
Inductive reactance,
= 75 µF.
Therefore to make a circuit with power factor = 1, 75 µF
capacitor is to be connected in a series with resistance
and inductor.
Example 3: A 750 Hz, 20 V source is connected to
a resistance of 100 ohm, an inductance of 0.1803
Henry and a capacitance of 10 microfarad all in series.
Calculate the time in which the resistance (thermal
capacity 2J/°C) will get heated by 10°C.
Sol: For an LCR circuit, the average power dissipated as
heat is =
Pav
2
Vrms
circuit.
Z
2
× R , where Z is the impedance of the
Product of power and time equals the heat generated.
= 849.2 Ω and
XL = ωL = 2πfL = 2π × 750 × 0.1803
=
P
Example 4: A 100 V ac source of frequency 500 Hz
is connected to a series LCR circuit with L=8.1 mH,
C = 12.5 µ F and R= 10 Ω . Find the potential different
across the resistance.
1
= 75 × 10−6 F
×
×
×
2
3.14
50
42.2
(
) ( )
X
=
C
( TC=
) × ∆θ
( TC ) ∆θ ;
1
1
=
ωC 2πfC
Capacitive reactance,
=
XC
= 21.2Ω
2π × 750 × 10−5
⇒ XL =
XC
This is the condition of resonance. This means that total
potential drop occurs across the resistance only.
(
∴ V = VR2 + VL − VC
2
= VR = 100 V
The total potential difference across resistance is the
same as the applied voltage across circuit.
Example 5: A 0.21 H inductor and a 12 Ω resistor
are connected ina series to a 20 V, 50 Hz ac source.
Calculate the current in the circuit and the phase angle
between the current and the source voltage.
Z
834Ω
R2 + X2 =
(100 ) + ( 828 ) =
2
2
R 2 + (ωL)2 ;
Impedance =
Z
But as in case of ac,
122 + ( 2 * 3.14 * 50 * 0.21 )
2
Vrms
R
P=
Vrms Irms cos=
φ Vrms ×
×
av
Z
Z
2
=
 Vrms 
 20 
×R 
100 0.00575W And
 =
 ×=
 834 
 Z 
Pav 
i.e.=
)
 ωL 
angle φ =tan−1 
 . And RMS value of the current is
 R 
V
Irms = rms where Z is impedance of the circuit.
So X = XL − X C = 849.2 − 21.2 = 828 Ω
2
106
= 25.45Ω
2π × 500 × 12.5
Sol: In series LR circuit, the current lags voltage by phase
1
Z
And hence=
XL = 2π × 500 × 8.1 × 10−3 = 25.45 Ω
(12 ) + (65.94 )=
2
2
V
rms
Current Irms = =
Z
Phase angle φ
67Ω
200
= 3.28A
67
P hysi cs | 23.27
 ωL 
−1  65.94 
tan−1 
 = tan 
;
R


 12 
VR
l
Example 6: A current of 4 A flows in a coil when
connected to a 12 V dc source. If the same coil is
connected to a 22 V, 50 rad/sec ac source, a current of
2.4 A flows in the circuit. Determine the inductance of
the coil. Also, find the power developed in the circuit if
a 2500 µ F condenser is connected in a series with the
coil.
Sol: For dc supply, the coil is purely resistive; inductance
does not come into picture. For AC voltage source,the
reactance of the inductor is non-zero. When a capacitor
is connected in a series in a circuit, the impedance of
circuit is Z =
R + ( XL − X C ) .
2
V2
Z2
R.
12
= 3Ω
4
Impedance of coil, Z=
12
= 5Ω ;
2.4
2
Now, Z=
R 2 + ω2L2 ;
2
Z −R
4
= = 0.08 H
50
ω
(c) Voltage of AC supply and
(d) Phase angle
Sol: For the LCR circuit, impedance is
Z=
R 2 + (X C − XL )2 .
The phase angle between voltage and current is given

 .

VL =I XL =5 × 24 =120 V,
VC =I X C =5 × 12 =60 V
(b) Using the formula of Impedance
Z=
R 2 + ( XL − X C )
Z=
(16 ) + ( 24 − 12)
2
2
2
=20 Ω
(c) Voltage of AC source is
Reactance of the capacitor
X C=
(b) Impedance of circuit
VR =5 × 16 =80 V
( Reactance of inductor in dc circuit is zero)
or=
L
(a) P.D. across R,L and C
(a) P.D. across each component is found below
Resistance of the coil, R=
2
VS
 X − XC
by φ =tan−1  L
 R

The real power in the circuit is
=
P I2=
R
VL
20
tan = (5.495 ) = 78.69 °
2
V
E = IZ = 5 × 20 =
100 V
1
1
=
= 8Ω
ωL 50 × 2500 × 10−6
(d) Phase angle is
X − XC )
 24 − 12 
−1 ( L
tan−1 
=
Φ tan
=

R
 16 
∴ When the capacitor is connected in series,
Z = R 2 + ( XL − X C ) = 32 + ( 8 − 4 ) =5 Ω
2
2
( 2.4 )
2
)
Example 8: A coil of resistance 20 Ω and inductance
0.5H is switched to dc 200 V supply. Calculate the rate
of increase of current:
R 3
= ;
Power factor, cos φ =
Z 5
2
Z cos
=
φ
Power developed P= Irms
(
= tan−1 0.75 = 360 87'
× 3 =17.28 W.
Example 7: A resistance R, an inductance L, and
capacitor C are connected in series with an AC supply
24 Ω and
where R=16 Ω . Inductive reactance X=
L
capacitive reactance X= 12 Ω . If the current in the
C
circuit is 5 A, find
(a) At the instant of closing the switch
(b) After one time constant
(c) Find the steady state current in the circuit
Sol: The current in the LR circuit attains constant value
over a long period of time. Generally, the current in the
2 3 . 2 8 | Alternating Current
(
)
circuit is given by
=i i0 1 − e− t/ τ where τ is one time
constant.
(a) Current at any time is given by:
Rt

− 
=i i0  1 − e L  



… (i)
...(1)
Differentiating above equation w.r.t. t, we get
Rt
−
dI  V R  − RtL 
=
dI  V . R  e=
L
∴ i0
=
dt  R . L  e=
∴ i0
dt  R L 

dI V 200
At =
t 0, dI= V= 200=
At =
t 0, dt= L= 0.5=
dt L
0.5
V
V  ...(2)
R  ...(2)
R
400 A / s
400 A / s
(b) Current after one time constant τ =
From equation (ii)
(c) The quality factor Q of the circuit
(d) The amplitude of the voltage across the inductor at
the resonant frequency.
Sol: When the LCR circuit is set to resonance, the
resonant frequency is f =
Quality factor is=
Q
… (ii)
1
1
.
2π LC
ω0L 1 L
.
=
R
R C
(a) Using formula of resonant frequency
The resonant frequency, for the circuit is given by
L
R
=
dI
= 400
=
e−1 147.15 A / s
dt
ω0
1
1
=
2π 2π LC
=
f
1
1
= 5033Hz
2π 10 × 10−3 H 100 × 10−9 F
(
)(
)
(b) At resonance current is Maximum i.e. I0
(c) For steady state t = ∞
V0
200
=
= 10.0 A
R 20.0Ω
So from (i) we get i(∞ ) = i0 = 400 A
=
I0
Example 9: What is average and RMS current over
half cycle if instantaneous current is given by i=4
sin ωt + 3cos ωt. ?
(c) The quality factor Q of the circuit is given by
Sol: Reduce the given expression of current in standard
=i i0 sin ωt + φ , where i0 is the maximum current
form
in the circuit.
(
)
Given i = 4 sin ωt + 3cos ωt.
4

3
t  5sin ( ωt + α )
= 5  sin ωt + cos ω=
5
5

where cos α =
4
3
and sin α = ;
5
5
(
=i i0 sin ωt + φ
Comparing with
i0 = 5 A ; ⇒
irms
)
 5 
 10 
=
 A ; iavg =   A
 π 
 2
JEE Advanced/Boards
Example 1: A sinusoidal voltage V(t) = (200 V) sin ωt is
applied to a series LCR circuit with L=10.0 mH, C=100
nF and R=20.0 Ω . Find the following quantities:
(a) The resonant frequency
(b) The amplitude of current at resonance
=
Q
(
)(
)
−1
10.0 × 10−3 H
ω0L 2π 5033s
=
R
( 20.0Ω )
= 15.8
(d) At resonance, the amplitude of the voltage across
the inductor is
VL = I0 XL= I0 ω0L
0
(
)(
)
=
(10.0A ) 2π 5033 s−1 10.0 × 10−3 H
= 3.16 × 103 V
Example 2: Consider the circuit shown in figure. The
sinusoidal voltage source is V (t) = V0 sinωt . If both
switches s1 and s2 are closed initially, find the following
quantities, ignoring the transient effect and assuming
that R, L, V0 and w are known:
(a) The current I(t)as a function of time
(b) The average power delivered to the circuit
(c) The current as a function of time, a long time after
only S1 is opened
P hysi cs | 23.29
Thus,
V
the
current
I(t) I0 sin(ωt −=
=
φ)
R₀
A
C
B
R
as
a
V0
function
of
time

ωL 
sin  ωt − tan−1

R 

R 2 + ω2L2
is
Note that in the limit of vanishing resistance R=0,
φ = π / 2 , and we recover the expected result for a
(d) The capacitance C if both s1 and s2 are opened for a
long time, with the current and voltage in phase.
(e) The impedance of circuit when both s1 and s2 are
opened.
(f) The maximum energy stored in the capacitor during
oscillations.
purely inductive circuit.
(d) If both the switches are opened, then this would be
a driven RLC circuit, with the phase angle φ given by tan
φ=
XL − X C
R
=
ωL −
R
1
ωc
(g) The maximum energy stored in the inductor during
oscillations.
If the current and voltage are in phase, lthen= φ ,
Sol: In LCR circuit explained above, when the switches
are closed, the current follows path of least resistance
(e) From (d), we see that both switches are opened; the
circuit is at resonance with XL = XC. Thus, the impedance
of the circuit becomes
i.e., L and C are short-circuited. Impedance of series
Z=
implying tan φ =0. Let the corresponding angular
1
frequency be ω0 ; we then obtain. ω0L = And the
ω0 c
1
(i) The frequency at which the inductive reactance XL is
capacitance is C =
And the capacitance
equal to half the capacitive reactance X C .
ω20L
(h) The phase difference between the current and the
voltage if the frequency of V (t) is doubled.
R 2 + ( XL − X C ) = R
2
R 2 + ( X C − XL ) . The energy stored
(e) The electric energy stored in the capacitor is
1 2
2
1 2 1
in inductor is UL = LI and that stored in capacitor is =
UE =
CVC
C ( IX C ) It attains maximum when the
2
2
2
1
UC = CVc2 .
current in at its maximum I0:
2
2
(a) When both switches s1 and s2 are closed, the current
V02L
1 2 2 1  V0  1
U
CI
X
C
=
=
=


C,max
goes through only the generator and the resistor, so
2 0 C 2  R  ω2 C2 2R 2
0
the total impedance of the circuit is R and the current
2
Where we have used ω0 =
1 / LC.
V0
IR (t)
sin ωt
is =
R
(g) The maximum energy stored in the inductor is given
(b) The average power is given by:
by.
2
2
V 20
V
1 2 LV0
2
0
=
P(t)
IR (t)V(t)
=
sin=
ωt
is given by.UL,max
LI
= =
R
2R
2 0 2R 2
(c) If only S1 is opened, after a long time a current
(h) If the frequency of the voltage source is double, i.e.,
will pass through the generator, the resistor and the
inductor. For this RL circuit, the impedance becomes
ω = 2ω = 1 / LC , then the phase becomes
2
LCR circuit is Z =
=
Z
1
=
R 2 + XL2
1
R 2 + ω2L2
 ωL 
And the phase angle φ is φ =tan−1 

 R 
0
 ωL − 1 / ωC 
φ =tan−1 

R


 2 / LC L − LC / 2C
= tan−1 
R


 3 L
= tan−1 

 2π C 


(
) (
) 



φ =tan 

R


 2 / LC L − LC / 2C
2=3tan
. 3 0−1| Alternating Current

R


 3 L
= tan−1 

 2π C 


(
) (
) 



IrmsinL
=
2
(i) If the inductive reactance in one-half the capacitive
reactance,
XL =
1
X ;
2 C
Then
=
ω
1 1 
;
2  ωC 
⇒ ωL = 
ω0
1
=
2LC
2
Example 3: Two inductances of 5.0 H and 10.0 H
are connected in parallel circuit. Find the equivalent
inductance and RMScurrent in each inductor and in
mains circuit when connected to source of 10 V AC.
I1
5.0 H
I
10 V AC
applied voltage, then RMScurrent through inductor is
V
.
I=
XL
=
E E0 sin ωt , then current drawn from supply is,
Let


π  E0
π
I I0 sin  ωt −=
sin  ωt −  (Since current lags
=

2
L
2
ω




π
by )
2
Where L is equivalent inductance of circuit.

π
=
I =
I1 + I2
sin  ωt − 
2
ωL1

=


π E
π
sin  ωt −  + 0 sin  ωt − 
2
L
2
ωL1
ω




2
⇒
E0
Example 4: A series LCR circuit containing a resistance
of 120 Ω has angular frequency 4 × 105 rads–1. At
resonance, the voltage across resistance and inductance
are 60 V and 40 V respectively. Find the value of L and
C. At what frequency does the current lag the voltage
by 45o?
Sol: At resonance, XL = XC. The phase angle by which
 X − XC
the current lags the voltage is φ =tan−1  L
 R

For resistance VR = IrmsR;
VR
60
=
= 0.5 A
R 120
40 = 0.5 × 4 × 105 × L ⇒ L = 2 × 10−4 H
1
At resonance, XL = XC i.e. ω0L =
ω0 C
=
C
1
=
ω20L
1
( 4 × 10 )
5
10
H
3
V
10
1
IrmsinL
=
=
−
=
;
1
ωL1
2π × 50 × 5 50π
2
1
µF
32
=
× 2 × 10−4
When the current lags behind the voltage by = 45o,
using tan φ =
XL − X C
R
, gives
1
ωL −
 ω2 L 
1
C
ω
1=
⇒ R =ωL −
=ωL −  o 
 ω 
R
ωC


∴ωR = ω2L − ω2oL
(
120 ω = 2 × 10−4 ω2 − (4 × 105 )2
)
VL
Source Voltage
VL - VC
I
I
I
1 1 15
3
= +
= +
= = ;
L L1 L2 5 10 50 10
⇒L=



Irms ω0L
For inductor V=
L
V
Sol: When two inductors are connected in parallel, the
LL
net inductance is L = 1 2 . If V is the RMS value of
L1 + L2
E0
1
1
3
Irmsincircuit =
+
=
50π 100π 100π
or I=
rms
I2 10.0 H
V
10
1
=
−
=
;
ωL 2
2π × 50 × 10 100π
o
45
VR
VC
i
P hysi cs | 23.31
On solving the above equation,we get
ω = 8 × 105
ω = −2 × 105
or
∵ Frequency can’t be negative
∴ Ignoring negative root we have ω = 8 × 105 Hz
V
10
=
I V00 sin ( ω=
t + φ ) 10 sin ( 314 t + π / 2 )
=
I Z sin ( ω=
t + φ ) 19.3 sin ( 314 t + π / 2 )
Z cos 314t 19.3
= 0.52
= 0.52 cos 314t
Example 5: An inductor of 20mH, a capacitor 100 µF
and a resistor 50 Ω are connected in a series across
a source of e.m.f. V=10 sin (314t). Find the energy
dissipated in the circuit in 20 minutes. If resistance is
removed from the circuit and the value of inductance is
doubled, then find the variation of current with time in
the new circuit.
Example 6: A choke coil is needed to operate an arc
lamp at 160 V (rms) and 50 Hz. The lamp has an effective
resistance of 5 Ω when running at 10 A (RMS). Calculate
the inductance of the choke coil. If the same arc lamp is
to be operated on 160 V (dc), what additional resistance
is required? Compare the power losses in both cases.
L
Sol: For the LCR circuit, the energy dissipated over a
=
Vrms Irms cos φ t . When resistance is
long
time is U
removed,the circuit becomes LC circuit, the impedance
and hence current changes.
(
2π
ω
T= =
2π
= 0.02s. So, we have to calculate the
314
average energy at time t>>T.
L
R
C
10 sin 314t
Energy dissipated in time t
 I
V R
U = ( Vrms Irms cos φ ) t =  0 × 0 ×  t


2 Z
 2
=
∴U

V02R
V0 
t  I0
=

2
Z 
2Z

102 × 50 × 20 × 60
=
∴U
= 864.2 J
2 × 3153.7
When resistance is removed,and inductance is doubled,
then cos φ = 0 ⇒ φ = π / 2
Value of impedance is
1
1
=
− ωL'
=
− 314 × 40 × 10−3 Ω
Z'
−4
ωC
314 × 10
=19.3 Ω
And the current in the circuit is found to be
Lamp
Choke
VL
)
The circuit is as shown in figure. One time cycle
R
VR
V = V0 sin t
Sol: Choke coil has large inductance and low internal
resistance, sothere is no power loss in the choke coil.
Hence, when alamp of some resistance is connected
in series with the coil, the net RMS voltage in circuit
is
Vrms )
(=
( Vrms )R + ( Vrms )L .
2
2
2
When the same lamp
is operated on dc, additional resistance in a series is
required to limit the current in the lamp to 10 A.
Voltage drop across the lamp is
( Vrms )R = ( Irms )(R ) =
10 × 5 = 50 V Voltage drop across
choke coil is
∴ ( Vrms ) =
L
=
( Vrms ) − ( Vrms )R
2
(160 ) − (50 ) =
2
2
As (=
Vrms )
L
∴L =
2
152 V
=
(irms
) XL (irms )( 2πfL ) ;
( Vrms )L
( 2πf )(irms )
Substituting the values
L=
152
= 4.84 × 10−2 H
2
π
50
10
( )( )( )
When lamp is operated on DC supply with a resistance
R’ in series, then voltage drop across the circuit is
=
V i (R + R' ) or 160=10(5+R’);
∴ R=' 11Ω
2 3 . 3 2 | Alternating Current
The capacitive reactance
Choke coil has no resistance.Therefore,for ac circuit
power loss in choke coil is zero, while in case of dc, the
1
1
loss due to additional resistance R’ is
X =
=
=
P= i2R'
10 ) (11 )
(=
2
C
1100 W
Example 7: A series AC circuit contains an inductor
(20 mH), a capacitor (100 µF ) and resistance (50 Ω). AC
source of 12 V (RMS), 50 Hz is applied across the circuit.
Find the energy dissipated in the circuit in 1000 s.
Sol: The average power dissipated in series LCR circuit
The net reactance is X=
and t = 1000 s � T
)(
2
50Ω 12V
Vrms R RVrms
=
Pav V=
=
rms
Z Z
Z2
Z2
7200
Z2
...(i)
100
Ω − 2π=
Ω 25.5 Ω
π
=
Pav
The average power dissipated is
Pav =
1
− ωL
ωC
(50 Ω ) + ( 25.5 Ω ) =
2
2
3150 Ω2
From (i), average power
T=1/f=20 ms.
(
100
Ω
π
XL = ωL = 2π × 50 × 20 × 10 −3 Ω = 2π Ω.
Thus, Z 2 =
The time period of the source is,
2π × 50 × 100 × 10
=
Ω
−6
The inductive reactance
=
is Pav Vrms Irms cos φ . For time t � T , the energy =
dissipated is U = Pavt.
ωC
7200
= 2.3 W
3150
∴ The energy dissipated int = 1000s is
)
2
s 2.3 × 103 J
U = Pav × 1000 =
… (i)
JEE Main/Boards
Exercise 1
Q.5 What is the relation between peak value and root
mean square value of alternating e.m.f?
Q.1 The resistance of coil for direct current (dc)is 10 Ω .
When alternating current (ac) is sent through it; will its
resistance increase, decrease or remain the same?
Q.6 Is there any device which may control the direct
current without dissipation of energy?
Q.2 Prove that an ideal inductor does not dissipate
power in an A.C. circuit.
Q.3 What is impedance? Derive a relation for it in an
A.C. Series LCR circuit. Show it by a vector.
Q.4 An A.C. supply E = E0 sinω t is connected to a series
combination of L, C and R. Calculate the impedance
of the circuit and discuss the phase relation between
voltage and current.
Q.7 What is the phase relationship between current
and voltage in an inductor?
Q.8 Find the reactance of a capacitance C at f Hz.
Q.9 Prove that an ideal capacitor connected to an A.C.
source does not dissipate power.
Q.10 State the principle of an A.C. generator.
Q.11 How are the energy losses reduced in a
transformer?
P hysi cs | 23.33
Q.12 Discusses the principle, working and use of a
transformer for long distance transmission of electrical
energy.
Q.13(a) What will be instantaneous voltage for A.C.
supply of 220 V and 50 Hz?
(b) In an A.C. circuit, the rms voltage is 100 2 V , find
the peak value of voltage and its mean value during a
positive half cycle.
Q.14 What should be the frequency of alternating 200
V so as to pass a maximum current of 0.9 A through an
inductance of 1 H?
Q.15 An alternating e.m.f of 100 V (r.m.s), 50 Hz is
applied across a capacitor of 10 µF and a resistor of 100
W in series.Calculate (a) The reactance of the capacitor;
(b) The current flowing (c) the average power supplied.
Q.16 The effective value of current in a 50 cycle A.C.
circuit 5.0 A. What is the value of current 1/300s after
it is zero?
Q.17 A pure capacitor is connected to an ac source of
220 V, 50 Hz, what will be the phase difference between
the current and applied emf in the circuit?
Q.18 A 100 Ω resistance is connected to a 220 V, 50 Hz
A.C. supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
Q.19 A pure inductance of 1 H is connected across
a 110V, 70 Hz source, find (a) reactance (b) current
(c) peak value of current.
Q.20 A series circuit contains a resistor of 10 Ω , a
capacitor, an ammeter of negligible resistance. It is
connected to a source 220V-50 Hz, if the reading of
an ammeter is 2.0 A, calculate the reactance of the
capacitor.
Q.21 A series LCR circuit connected to a variable
frequency 230V source and L=5.0 H,C=80 µF , R=40 Ω .
(a) Determine the source frequency which drives the
circuit in resonance.
(b) Obtain the impedance of the circuit and the
amplitude of the current at the resonating frequency.
(c) Determine the rms potential drops across the three
elements of the circuit. Show that the potential drop
across the LC combination is zero at the resonating
frequency.
Q.22 A circuit containing a 80 mH inductor and a 60 µF
capacitor in series is connected to 230 V, 50Hz supply.
The resistance of the circuit is negligible. (a) Obtain
the current amplitude and rms values. (b) Obtain the
rms value of potential drops across each element, (c)
What is the average transferred to the inductor? (d)
What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the
circuit? [‘average’ ‘implies’ averaged over one cycle;].
Q.23 Answer the following questions: (a) in any A.C.
circuit, is the applied instantaneous voltage equal to the
algebraic sum of the instantaneous voltage across the
series element of the circuit? Is the same true for rms
voltage? (b) A capacitor is used in the primary circuit of
an inductor coil. (c) A supplied voltage signal consists
of a super position of a D.C voltage and A.C. voltage
of high frequency. The circuit consists of an inductor
and a capacitor in series. Show that the D.C. signal will
appear across C and the A.C. signal across L. (c) An
applied voltage signal consists of a superposition of a
D.C. voltage and an A.C. Voltage of high frequency. The
circuit consists of an inductor and a capacitor in series.
Show that the D.C. signal will appear across C and the
A.C. signal across L. (e) Why is choke coil needed in the
use of florescent tubes with A.C. mains? Why can we
not use an ordinary resistor instead of the choke coil?
Q.24 An inductance of negligible resistance, whose
reactance is 22 Ω at 200 Hz is connected to a 220 V, 50
hertz power line, what is the value of the inductance
and reactance?
Q.25 An electric lamp market 220 V D.C. consumes a
current of 10 A. It is connected to 250 V-50 Hz A.C.
main through a choke. Calculate the inductance of the
choke required.
Q.26 A 2 µF capacitor, 100 Ω resistor and 8H inductor
are connected in series with an A.C. source. What
should be the frequency of this A.C source, for which
the current drawn in the circuit is maximum? If the peak
value of e.m.f of the source is 200 V, find for maximum
current, (i) The inductive and capacitive reactance of
the circuit; (ii) Total impedance of the circuit; (iii) Peak
value of current in the circuit ; (iv) The phase relation
between voltages across inductor and resistor; (v) The
2 3 . 3 4 | Alternating Current
phase difference between voltage across inductor and
capacitor.
Q.27 A step-down transformer converts a voltage of
2200 V into 220 V in the transmission line. Number
of turns in primary coil is 5000. Efficiency of the
transformer is 90% and its output power is 8 kW.
Calculate (i) Number of turns in the secondary coil (ii)
input power.
Q.28 What will be the effect on inductive reactance XL
and capacitive XC, if frequency of ac source is increased?
Q.29 The frequency of ac is doubled, what happens to
(i) Inductive reactance (ii) Capacitive reactance?
and minimum current that can flow in the circuit is
10V
.S
( )
0.1 H
10
10
(A) 2 Amp
(B) 3 Amp
(C) 1 Amp
(D) Nothing can be concluded
Q.4 The ratio of time constant in build-up and decay in
the circuit shown in figure is
R
2R
Exersice 2
Single Correct Choice Type
V
Q.1 A rectangular loop with a sliding connector of
length 10 cm is situated in uniform magnetic field
perpendicular to plane of loop. The magnetic induction
is 0.1 tesla and resistance of connecter (R) is 1 Ω. The
sides AB and CD have resistance 2 Ω and 3 Ωrespectively.
Find the current in the connecter during its motion with
constant velocity of 1 meter/sec.
A
D
3
2
L
(A) 1:1
(B) 3:2
(C) 2:3
(D) 1:3
Q.5 A current of 2A is increased at a rate of 4 A/s
through a coil of inductance 2H. The energy stored in
the inductor per unit time is
(A) 2 J/s
(B) 1 J/s
(C) 16 J/s
(D) 4 J/s
Q.6 The current in the given circuit is increased with a
rate a=4 A/s. The charge on the capacitor at an instant
when the current in the circuit is 2 amp will be:
E=4V
R
B
1
A
( A ) 110
C
1
A
(B ) 220
1
A
( C ) 55
R=1
1
A
(D ) 440
Q.2 For L-R circuit, the time constant is equal to (A)
Twice the ratio of the energy stored in the magnetic
field to the rate of dissipation of energy in the resistance.
L=1H
(A) 4 µC
)
C=3F
(B) 5 µC
(C) 6 µC
(D) None of these
(B) Ratio of the energy stored in the magnetic field to
the rate of dissipation of energy in the resistance.
Q.7 A coil of inductance 5H is joined to a cell of emf 6 V
through a resistance 10 Ω at time t=0. The emf across
(C) Half the ratio of the energy stored in the magnetic
field to the rate of dissipation of energy in the resistance.
the coil at time t=
(D) Square of the ratio of the energy stored in the
magnetic field to the rate of dissipation of energy in
the resistance.
Q.3 In the adjoining circuit, initially the switch S is open.
The switch‘s’ is closed at t=0. The difference between
(A) 3V
(B) 1.5V
2 s is:
(C) 0.75V
(D) 4.5V
Q.8 The network shown in the figure is part of a
complete circuit. If at a certain instant, the current I is
5A and it is decreasing at a rate of 103As -1 then VB-VA
equals.
P hysi cs | 23.35
(A) 20 V
+
1
A
5 mH
I
15 V
(B) 15 V
(C) 10
L
B
A
S
B
(D) 5 V
R1
12v
R2
Q.9 In the previous question, if I is reversed in direction,
then VB-VA equals
(A) 5 V
(B) 10 V
(C) 15 V
(D) 20 V
Q.10 Two resistors of 10 Ω and 20 Ω and an ideal
inductor of 10 H are connected to a 2 V battery as
shown in figure. The key K is inserted at time t=0. The
initial (t=0) and final (t>=00) current through battery
are
10 H
(A) VL=12 V; point A is at the higher potential
(B) VL=12 V; point B is at the higher potential
(C) VL=6 V; point A is at the higher potential
(D) VL=6 V; point B is at the higher potential
Q.13 The power factor of the circuit shown in figure is
1/ 2 . The capacitance of the circuit is equal to
20
10
V=2sin(100t)
K
10
2V
(A)
1
1
A, A
15 10
(B)
(C)
2
1
A, A
15 10
(D)
1
1
A,
A
10
15
1
2
A,
A
15
25
Q.11 In the circuit shown, the cell is ideal. The coil has
an inductance of 4H and zero resistance. F is a fuse zero
resistance and will blow when the current through it
reaches 5A. The switch is closed at t=0. The fuse will
blow
0.1H
C
(A) 400 µF
(B) 300 µF
(C) 500 µF
(D) 200 µF
Q.14 In the circuit, as shown in the figure, if the value
of R.M.S current is 2.2 ampere, the power factor of the
box is
100
C
1/ Henry
L
Box
fuse
Sw
Vrms=220 volt, -100 s
2V
(A) Just after t=0
(B) After 2
(C) After 5s
(D) After 10s
Q.12 The circuit shown has been operating for a long
time. The instant after the switch in the circuit labeled
S is opened, what is the voltage across the inductor VL
and which labeled point (A or B) of the inductor is at a
higher potential? Take R1=4.0 Ω , R2=8.0 Ω and L= 2.5
H.
(A)
1
2
(B) 1
(C)
3
2
-1
(D)
1
2
Q.15 When 100 V DC is applied across a solenoid, a
current of 1 A flows in it. When 100 V AC is applied
across the same coil, the current drops to 0.5 A. If the
frequency of the AC source is 50 Hz, the impedance
and inductance of the solenoid are:
(A) 100 Ω , 0.93 H
(B) 200 Ω , 1.0 H
(C) 10 Ω , 0.86 H
(D) 200 Ω , 0.55 H
2 3 . 3 6 | Alternating Current
I0 I1 sin ωt then its
Q.16 An ac current is given by I =+
rms value will be
(A) I02 + 0.5112
(B) I02 + 0.5102
(C) 0
(D) I0 / 2
(B) 15.7 ms
(C) 0.25 s
(D) 2.5 ms
(B) 5/6
(C)
4
3 3
(D) 3 3
4
Q.19 The effective value of current i=2 sin100 π t+2 sin
(100 π t+300) is:
(A)
2A
(C) 4
π
ahead, as shown in the figure. If the circuit consists
4
possibly only of R-C or R-L or L-C in series, find the
(2003)
relationship between the two elements:
e
(B) 2 2 + 3
(D) None of these
Q.20 In a series R-L-C circuit, the frequency of the
source is half of the resonance frequency. The nature of
the circuit will be
(A) R=1 K Ω ,C=10 µF
(C) R=1 K Ω ,L=10H
(B) R=1 K Ω ,C=1 µF
(D) R=1 K Ω ,L=1H
Q.2 The current I4 through the resistor and voltage vC
across the capacitor are compared in the two cases.
(2011)
Which of the following is/are true?
(a) IRA > IBR
(B) IRA < IRB
A
(D) IC < IC
B
A
(C) IC > IC
B
Q.3 The network shown in Figure is part of a complete
circuit. If at a certain instant the current (I) is 5A and is
decreasing at a rate of 103 A/s then VB − VA =
.........V
(1997)
(A) Capacitive
i
(B) Inductive
A
(C) Purely resistive
(D) Data insufficient
i
t
Q.18 Power factor an L-R series circuit is 0.6 and that of
a C-R series circuit is 0.5. If the element (L, C, and R) of
the two circuits are joined in series, the power factor of
this circuit is found to be 1. The ratio of the resistance
in the L-R circuit to the resistance in the C-R circuit is
(A) 6/5
Q.1 When an AC source of emf e=E0sin (100 t) is
connected across a circuit, the phase difference between
the emf and the current i in the circuit is observed to be
Q.17 The phase difference between current and voltage
in an AC circuit is π / 4 radians. If the frequency of AC
is 50 Hz, then the phase difference is equivalent to the
time difference:
(A) 0.78 s
Previous Years’ Questions
1
15 V 5 mH
B
Q.4 An arc lamp requires a direct current of 10 A and 80
V to function. If it is connected to a 220 V (rms), 50 Hz
AC supply, the series inductor needed for it to work is
(2016)
close to:
(A) 0.08 H
(B) 0.044 H
(C) 0.065 H
(D) 80 H
P hysi cs | 23.37
JEE Advanced/Boards
Exercise 1
R
L
Q.1 In the given circuit, find the ratio of i1 to i2 where i1
is the initial current (at t=0), i2is steady state (at t=∞)
current through the battery.
6
2 mH
4
10V
4
L
Q.2 Find the dimension of the quantity
, where
RCV
symbols have usual meaning.
Q.3 In the circuit shown, initially the switch is in position
1 for a long time. Then the switch is shifted to position
2 for long time. Find the total heat produced in R2.
R2
2
1
L
Q.7 Two coils, 1 & 2, have a mutual inductance = M and
resistance R each. A current flows in coils 1, which varies
with time as: I1 = kt2, where k is constant ‘t’ is time. Find
the total charge that has flown through coil 2, between
t = 0 and t = T.
Q.8 Find the value of an inductance which should be
connected in series with a capacitor of 5 F, resistance
of 10 Ω and an ac source of 50 Hz so that the power
factor of the circuit is unity.
Q.9 In an L-R series A.C circuit the potential difference
across an inductance and resistance joined in series
are respectively 12 V and 16 V. Find the total potential
difference across the circuit.
Q.10 A 50W, 100V lamp is to be connected to an ac
mains of 200V, 50Hz. What capacitance is essential to
be put in series with lamp.
S
E
S
E
R1
Q.4 Two resisters of 10 Ω and 20 Ω and an ideal
inductor of 10 H are connected to a 2V battery as
shown in figure. The key K is shorted at time t=0. Find
the initial (t=0) and final (t->∞) current through battery.
L = 10 H
Q.11 In the circuit shown in the figure, the switched
S1 and S2 are closed at time t=0. After time t = (0.1) In
2sec, switch S2 is opened. Find the current in the circuit
at time t = (0.2) ln 2sec.
S1
R = 10
K
Q.5 An emf of 15 V is applied in a circuit containing
5 H inductance and 10 Ω resistance. Find the ratio of
the current at time t=∞ and t=1 second.
Q.6 In the circuit in shown in figure, switch S is closed
at time t=0. Find the charge which passes through the
battery in one time constant.
S2
40
20
100 V
10
1H
Q.12 Find the value of i1and i2
S
100 V
i1
30
i2
20
2 3 . 3 8 | Alternating Current
(i) Immediately after the switch S is closed.
V
(ii) Long time later, with S closed.
(iii) Immediately after switch S is open
(iv) Long time after S is opened.
Q.13 Suppose the emf of the battery in the circuit
shown varies with time t so the current is given by
i(t) = 3+5t, where i is in amperes & t is in seconds. Take
R=4 Ω , L=6H & find an expression for the battery emf
as a function of time.
R
i(t)
L
O
T/4
T/2
3T/4
T
Find the amplitude of current in the steady state and
obtain the phase difference between the current and
the voltage. Also plot the variation of current for one
cycle on the given graph.
Exercise 2
Q.14 An LCR series circuit with 100 Ω resistance
is connected to an ac source of 200 V and angular
frequency 300rad/s. When only the capacitance is
removed, the current lags behind the voltage by 600.
When only the inductance is removed, the current
leads the voltage by 600. Calculate the current and the
power dissipated in the LCR circuit.
Single Correct Choice Type
Q.1 A square coil ABCD is placed in x-y plane with its
centre at origin. A long straight wire, passing through
origin, carries a current in negative Z-direction. Current
in this wire increases with time. The induced current in
the coil is
Q.15 A box P and a coil Q are connected is series with
an ac source of variable frequency. The emf source at
10V. Box P contains a capacitance of 1 µ F in series with
a resistance of 32 Ω . Coil Q has a self-inductance 4.9
mH and a resistance of 68 Ω series. The frequency
adjusted so that the maximum current flows in P and Q.
Find the impedance of P and Q atthis frequency. Also
find the voltage across P and Q respectively.
Q.16 A series LCR circuit containing a resister of 120
Ω has angularresonance frequency 4 × 105 rad s-1. At
resonance, the voltage across resistance and inductance
are 60V and 40V respectively. Find the values of L and C.
At what frequency current in the circuit lags the voltage
by 45o?
Q.17 In an LR series circuit, a sinusoidal voltage V=V0
sinωt is applied. It is given that
L=
35mH,R =
11Ω, Vrms =
220V,
And π =22 / 7 .
ω
=
50Hz
2π
y
B
C
x
A
D
(A) Clock wise
(B) Anti clockwise
(C) Zero
(D) Alternating
Q.2 An electric current i1 can flow in either direction
through loop (1) and induced current i2 in loop (2).
Positive i1 is when current is from ‘a’ to ‘b’ in loop (1)
and positive i2 is when the current is from ‘c’ to ‘d’ in
loop
Loop (1)
a
b
Loop (2)
c
d
P hysi cs | 23.39
(2) In an experiment, the graph of i2 against time ‘t’ is as
shown below by Figure which one (s) of the following
i2
above.
graphs couldi2have caused i2 to behave as give
i2
(A)
(A)
t
t
0
t
i2
i2
(B)
(B)
t
t
(C)
(D)
t
t
(C) Same
(D) Data is insufficient to decide
11
RR
and
and
and
RC
RC
LL
1
(C)
i2
(B)
(C)
(C)
(B) 2
(A)
t
i
i22
(A) 1
Q.5 L, C and R represents physical quantities inductance,
capacitance and resistance. The combination which has
the dimensions of frequency?
i2
(A)
(A)
(B)
Q.4 Two identical inductances carry currents that vary
with time according to linear laws (see in figure). In
which of the inductances is the self-inductance emf
greater?
t
t
i1
LC
(B)
11
RR
and
and
and
LL
RC
RC
(D)
C
L
Q.6 In the circuit shown, X is joined to Y for a long time,
and then X is joined to Z, the total heat produced in R2
is:
R2
i2
X
Q.3 In an L-R circuit connected to a battery of constant
e.m.f. E, switch S is closed at time t = 0. If e denotes the
(C)
magnitude of induced e.m.f.
across inductor and i the
t
current in the circuit at anytime t. Then which of the
following graphs shows the variation of e with i?
Y
L
E
R1
i1
I
i2
(A)
(B)
2
1
t
i3
t
e
e
(A)
Z
t
i4
(C)
(D)
(B)
t
t
e
t
(A)
e
(C)
(D)
t
t
t
LE2
2R12
(B)
LE2
2R 22
(C)
LE2
2R1R 2
(D)
LE2R 2
2R12
Q.7 An induction coil stores 32 joules of magnetic
energy and dissipates energy as heat at the rate of 320
watt when a current of 4 amperes is passed through it.
Find the time constant of the circuit when the coil is
joined across a battery.
(A) 0.2s
(B) 0.1s
(C) 0.3s
(D) 0.4s
2 3 . 4 0 | Alternating Current
Q.8 In an L-R decay circuit, the initial current at t=0 is 1.
The total charge that has inductor has reduced to onefourth of its initial value is
(B) LI / 2R
(A) LI / R
(C) LI / 2R (D) None
Q.9 An inductor coil stores U energy when i current
is passed through it and dissipates energy at the rate
of P. The time constant of the circuit, when the coil is
connected across a battery of zero internal resistance is
(A)
4U
P
(B)
U
P
(C)
2U
P
(D)
Q.14 The current I, potential difference VL across the
inductor and potential difference VC across the capacitor
in circuit as shown in the figure are best represented
vectorially as.
θ1 + θ2
(B) 1
(C)
1
2
(D)
(III)
i
I0
O
-I0
t
i
I0
O
-I0
2
t
i
I0
O
(IV)
t
(C) I3 > I4 > I2 = I1
(D) I3 > I2 > I1 > I4
Q.13 In series LR circuit XL=3R. Now a capacitor with XC=R
is added in series. Ratio of new to old power factor is
(B) 2
XC
I1
(C)
1
2i
R
(A)
X
π
− tan−1 L
2
R
(B) tan−1
(C)
X
π
+ tan−1 L
2
R
(D) tan−1
(D) 2
XL − X C
R
XL − X C
R
+
π
2
Multiple Correct Choice Type
Q.16 A circuit element is placed in a closed box. At time
t=0, constant current generator supplying a current of
1 amp, is connected across the box. Potential difference
across the box varies according to graph shown in
Figure. The element in the box is:
(C) Inductance of 2H
(B) I3 > I1 = I2 > I4
(A) 1
Q.15 In the shown AC circuit in figure, phase difference
between current I1 and I2 is
(A) Resistance of 2 Ω
(A) I1 = I2 = I3 = I4
VC
VC
1
i
I0
O
-I0
(II)
I
XL
Q.12 If I1, I2,I3 and I4 are the respective r.m.s values of the
time varying current as shown in figure the four cases
I.II,III and IV in. Then identify the correct relations.
(I)
I
VL
(D)
I2
Q.11 The power in ac circuit is given by P=ErmsIrms cos φ .
The value of cos φ in series LCR circuit at resonance is:
(A) Zero
VL
(D) None of these
2
I
VL
(B) tan=
θ tan θ2 − tan θ1
θ = θ1 − θ2
(B)
VL
2P
U
Q.10 When a resistance R is connected in series with
an element A, the electric current is found to be
lagging behind the voltage by angle θ1. When the
same resistance is connected in series with element B,
current leads voltage by θ2. When R, A, B, are connected
in series, the current now leads voltage by θ. Assume
same AC source in used in all cases. Then:
(C) θ =
I
(A)
(C)
(A)
VC
VC
(B) Battery of emf 6V
(D) Capacitance of 0.5F
8
2
3
t(s)
P hysi cs | 23.41
Q.17 For L-R circuit, the time constant is equal to
(A) The low resistance of P
(A) Twice the ratio of the energy stored in the magnetic
field to the rate of the dissipation of energy in the
resistance
(B) The induced-emf in L
(B) The ratio of the energy stored in the magnetic field
to the rate of the dissipation of energy in the resistance.
L
C
VC
VL
(C) Half of the ratio of the energy stored in the magnetic
field to the rate of the dissipation of energy in the
resistance.
(D) Square of the ratio of the energy stored in the
magnetic field to the rate of the dissipation of energy
in the resistance.
Q.18 An inductor L, a resistor R and two identical bulbs
B1 and B2 are connected to a battery through a switch S
as shown in the figure. The resistance of the coil having
inductance L is also R. Which of the following statement
gives the correct description of the happening when
the switch S is closed?
L
B1
R
B2
E
S
(A) The bulb B2 lights up earlier then B1 and finally both
the bulbs shine equally bright.
(B) B1 lights up earlier and finally both the bulbs acquire
brightness.
(C) B2 lights up earlier and finally B1 shines brighter
than B2.
(D) B1 and B2 lights up together with equal brightness
all the time.
Q.19 In figure, a lamp P is in series with an iron-core
inductor L. When the switch S is closed, the brightness
of the lamp rises relatively slowly to its full brightness
than it would to without the inductor. This is due to
P
L
S
B
(C) The low resistance of L
(D) The high voltage of the battery B
Q.20 Two different coils have a self-inductanceof
8mH and 2mH. The current in one coil is increased at
a constant rate. The current in the second coil is also
increased at the same instant of time. The power given
to the two coils is the same. At that time the current,
the induced voltage and the energy stored in the first
coil are I1 V1 and W1 respectively. Corresponding values
for the second coil at the same instant are I2, V2 and W2
respectively . Then:
(A)
(C)
I1
I2
1
3
(B)
=4
(D)
=
W1
W2
I1
I2
V2
V1
=4
=
1
4
Q.21 The symbol L, C, R represents inductance,
capacitance and resistance respectively. Dimension of
frequency is given by the combination.
(A) 1/RC
(B) R/L
(C)
1
LC
(D) C/L
Q.22 An LR circuit with a battery is connected at t=0.
Which of the following quantities is not zero just after
the circuit is closed?
(A) Current in the circuit
(B) Magnetic field
(C) Power delivered by the battery
(D) Emf induced in the inductor
Q.23 The switches in figure (a) and (b) are closed at t=0
)
C
R
E
R
L
E
(a)
(b)
(A) The charge on C just after t=0 is EC.
(B) The charge on C long after t=0 is EC.
(C) The charge on L just after t=0 is E/R.
(D) The charge on L long after t=0 is EC.
2 3 . 4 2 | Alternating Current
Q.24 Two coils A and B have coefficient of mutual
inductance M=2H. The Magnetic flux passing through
coil A changes by 4 Weber in 10 seconds due to the
change in current in B. Then
(A) Change in current in B in this time interval is 0.5 A
(B) The change in current in B in this time interval is 2A
Comprehension Type Question
Paragraph 1: A capacitor of capacitance C can be
charged (with the help of a resistance R) by a voltage
source V, by closing switch s1 while keeping switch s2
open. The capacitor can be connected in series with an
inductor ‘L’ by closing switch S2 and opening S1.
(C) The change in current in B in this time interval is 8A
(D) A change in current of 1A in coil A will produce a
change in flux passing through B by 4 Weber.
Assertion Reasoning Type
V
R
C
(A) Statement-I is true, statement-II is true and
statement-II is correct explaining for statement-I.
(B) Statement-I is true, statement-II is true and
statement-II is not correct explaining for statement-I
(C) Statement-I is true, statement-II is false.
(D) Statement-I is false, statement-II is true.
Q.25 Statement-I: when resistance of rheostat is
increased, clockwisecurrent is induced in the ring.
Statement-II: Magnetic flux through the ring is out of
the phase and decreasing.
L
S1
S2
Q.28 After the capacitor gets fully charged, s1 is opened
and S2 is closed so that the inductor is connected in
series with the capacitor. Then,
(A) At t=0, energy stored in the circuit is purely in the
form of magnetic energy.
(B) At any time t>0, current in the circuit is in the same
direction.
(C) At t>0, there is no exchange of energy between the
inductor and capacitor.
(D) At any time t>0, instantaneous current in the circuit
Q.26 Statement-I: Peak voltage across the resistance
can be greater than the peak voltage of the source in a
series LCR circuit.
Statement-II: Peak voltage across the inductor can be
greater than the peak voltage of the source in a series
LCR circuit.
Q.27 Statement-I: when a circuit having large inductance
is switched off, sparking occurs at the switch.
Statement-II: Emf induced in an inductor is given by
∈ =L
|e|
di
(A) Statement-I is true, statement-II is true
dt
and statement-II is correct explanation for statement-I.
(B) Statement-I is true, statement-II is true and statementII is not the correct explanation for statement-I.
(C) Statement-I is true, statement-II is false.
(D) Statement-I is false, statement-II is true.
is V C
L
Q.29 If the total charge stored in the LC circuit is Q0
then for t>=0
π
=
(A) The charge on the capacitor
is Q Q0 cos 
2
π
=
(B) The charge on the capacitor
is Q Q0 cos 
2
(C) The charge on the capacitor is Q = LC
(D) The charge on the capacitor is Q = −
+
t 

LC 
+
t 

LC 
d2Q
dt2
1 d2Q
LC dt2
P hysi cs | 23.43
Paragraph 2: In a series L-R circuit, connected with a
sinusoidal ac source, the maximum potential difference
across L and R are respectively 3 volts and 4 volts
Q.30 At an instant, the potential difference across
resistor is 2 V. The potential difference in volt, across
the inductor at the same instant will be:
(A) 3 cos30o
(B) 3 cos60o
(C) 3 cos45o
(D) None of these
ways as shown in column II. When a current I (steady
state for DC or rms for AC) flows through the circuit, the
corresponding voltage V1 and V2 (indicated in circuits)
(2010)
are related as shown in column I. Column I
Column II
(A) I ≠ 0, V1 is
(p)
V1
Proportional to I
V2
6 mH
Q.31 At the same instant, the magnitude of the potential
difference in volt, across the ac source may be
(B)
(A) 4 + 3 3
(C) 1 +
3
2
(D) 2 +
V
(B) I ≠ 0, V2 > V1
4+3 3
2
3 F
(q)
V1
3
2
V2
6 mH
3 F
V
Previous Years’ Questions
V1 0,=
V2 V
(C)=
(r)
V1
Q.1 A circuit containing a two position switch S is
shown in Figure.
R3
1
2
A
3V
V
R5
(D) I ≠ 0, V2 is
(t)
V1
Proportional to I
12 V
E2
L
R2
2
2 F
2 F 1
E1
2
S
6 mH
C
2
R1
V2
V2
3 F
1k
B
3
R4
10 mH
V
(s)
V1
(a) The switch S is in two position 1. Find the potential
difference VA − VB and the rate production of joule heat
in R1.
6 mH
Q.2 Match the Columns
You are given many resistances, capacitors and
inductors. They are connected to a variable DC voltage
source (the first two circuits) or in AC voltage source of
50 Hz frequency (the next three circuits) in difference
3 F
(b) If Now The switch S is put in position 2 at t=0. Find:
(i) Steady current in R4 and(ii) The time when current
in R4 is half the steady value. Also calculate the energy
(1991)
stored in the inductor L at that time.
V2
V
Paragraph 1 (Q.3 to Q.8)
The capacitor of capacitance C can be charged(with the
help of resistance R) by a voltage source V, by closing
switch S1 while keeping switch S2 open. The capacitor
can be connected in series with an inductor L by closing
switch S2 and opening S1.
2 3 . 4 4 | Alternating Current
Q.6 In the circuit shown, A and B are two cells of same
emf E but different internal resistance r1 and r2 ( r1 >r2)
respectively find the value of R such that the potential
difference across the terminals of cell A is zero a long
(2004)
time after the key K is closed
V
R
C
L
S1
R
S2
R
A
B
R
Q.3 Initially, the capacitor was uncharged. Now switch
s1 is closed and S2 is kept open. If time constant of this
(2006)
circuit is τthen (B) After time interval 2τ, charge on the capacitor is CV
(1-e-2)
(C) The work done by voltage source will be half of the
heat dissipated when the capacitor is fully charged
(D) After time interval 2τ, charge on the capacitor is CV
(1-e-1)
Q.4 After capacitor gets fully charged, S1 is opened and
S2 is closed so that the inductor isconnectedin series
(2006)
with the capacitor, then
S
Q.7 In an L-R series circuit, a sinusoidal voltage V = V0
sin ωt is applied. It is given that L=35 mH, R=11 Ω ,
=
Vrms 220V, ω=
/ 2π 50Hz and
=
π 22 / 7.
Find the amplitude of current in the steady state and
obtain the phase difference between the current and
the voltage. Also plot the variation of current for one
(2004)
cycle on the given graph. V
t
T/4
(B) At any time t>0, current in the circuit is in the same
direction.
(D) At any time t>0, instantaneous current in the circuit
C
may V
L
T/2
3T/4
T
Q.8 What is the maximum energy of the anti-neutrino ?
(2012)
(A) Zero
(B) Much less than 0.8 × 106 eV
Q.5 If the total charge stored in the LC circuit is Q0 then
(2006)
for t ≥ 0
π
t 
=
(A) The charge on the capacitor
is Q Q0 cos  +

LC 
2
π
t 
=
(B) The charge on the capacitor
is Q Q0 cos  −

LC 
2
(D) The charge on the capacitor is Q = −
C
R
(A) At t=0, energy stored in the circuit is purely in the
form of magnetic energy.
(C) At t>0, there is no exchange of energy between the
inductor and capacitor.
R
r1 r2
(A) After time interval τ, charge on the capacitor is CV/2
(C) The charge on the capacitor is Q = LC
L
R
d2Q
dt2
1 d2Q
LC dt2
(C) Nearly 0.8 × 106 eV
(D) Much larger than 0.8 × 106 eV
Q.9 At time t = 0 terminal A in the circuit shown in the
figure is connected to B by a key and an alternating
=
I(t) I0 cos (ωt) , with I0 = 1A and ω =500
current
rad/s starts flowing in it with the initial direction shown
7π
in the figure. At t =
, the key is switched from B to
6ω
D. Now onwards only A and D are connected. A total
charge Q flows from the battery to charge the capacitor
fully. If C = 20 µF, R =10Ω and the battery is ideal with
emf of 50 V, identify the correct statement(s). (2014)
P hysi cs | 23.45
B
(A) Magnitude of the maximum charge on the capacitor
7π
=
t
is 1 × 10−3 C
before
6ω
(B) The current in the left part of the circuit just before
7π
t=
is clockwise.
6ω
(C) Immediately after A is connected to D, the current
in R is 10 A.
D
A
50 V
C=20F
R=10
(D) Q= 2 × 10−3 C
MASTERJEE Essential Questions
JEE Main/Boards
JEE Advanced/Boards
Exercise 1
Exercise 1
Q. 15
Q.21
Q.23
Q.27
Q.22
Q.3
Q.4
Q.7
Q.14
Q.15
Q.16
Exercise 2
Exercise 2
Q. 1
Q. 3
Q. 11
Q.12
Q.2
Q.3
Q.12
Q.14
Q.22
Q.23
Q.28
Q.28
Q.29
Q.30
Q.31
Answer Key
JEE Main/Boards
Q.10 It is based up on the principle of electromagnetic
induction.
Exercise 1
Q.11 (i) By using laminated iron core, we minimize loss
of energy due to eddy current.
V 
Q.5. Vrms =  0 
 2
Q.6 No
(ii) By selecting a suitable materials for the core of a
transformer, the hysteresis loss can be minimized.
Q.7 The current lags behind the voltage by phase angle
π/2.
XC
Q.8 Capacitive reactance, =
1
1
=
ωC 2πfc
Q.13 (a) ≈ 311sin314t (b) 200V, 127.4V
Q.14 50Hz
Q.15 (a) 318.31 Ω (b) 0.527 A (c) 9 W
Q.16 6.124A
2 3 . 4 6 | Alternating Current
Q.23 (a) Yes. The same is not true for rms voltage,
because voltage across different element may not be
in phase.
Q.18 (a) 2.20A, (b) 484 W
Q.19 0.354A
Q.20 109.5 A
Q.21 (a) 50 rad s-1, (b) 40 Ω , 8.1A, (c) VLcms=1437.5

1 
=
V,Vvcrms=1437.5V,VRms=230 VLCrms= Irms  ω0L −
 0
ω0 C 

=
V V0 sin ωt
Q.22 (a) For
I
=
V0

π
sin  ω
t +  ; If R 0
=
2
1

ωL −
ωC
(b) The high induced voltage, when the circuit is broken,
is used to change the capacitor, thus avoiding sparks,
etc.
(c) For dc, impedance of L is negligible and C very
high (infinite), so the D.C. signal appears across C. For
frequency ac, impedance of L is high and that of C is
low. So, the A.C. signal appears across L.
(e) A choke coil reduces voltage across the tube without
wasting power. A resister would waste power as heat.
Where- sign appears if ωL >I/ ωC , and+sign appears
if ωL <I ωC .
=
I0 11.6A,I
=
8.24A
rms
Q.24 1.75 × 10−2 H; 5.5Ω
Q.25 0.04H
Q.26 Resonant frequency=39.79 Hz
(b) VLCrms =207V, VCrms =437 V
(i) 2000 Ω
(ii) 100 Ω
(c) Whatever be the current I in L, actual voltage leads
current by π / 2 . Therefore, average power consumed
by L is zero.
(iv) 900
(v)1800
Q.27 (i) 500;
(ii) 8.9kW
(iii) 2A
(d) For C, voltage lags by π / 2 . Again average power
consumed by C is zero.
(e) Total average power absorbed is zero.
Exercise 2
Q.1 B
Q.2 A
Q.3 C
Q.4 B
Q.5 C
Q.6 C
Q.7 A
Q.8 B
Q.9 C
Q.10 A
Q.11 D
Q.12 D
Q.13 C
Q.14 A
Q.15 D
Q.16 A
Q.17 D
Q18 D
Q.19 B
Q.20 A
Q.3 15V
Q.4 C
Previous Years’ Questions
Q.1 A
Q.2 B, C
JEE Advanced/Boards
Exercise 1
Q.1 0.8
Q.4
1
1
,
15A 10A
Q.7 q =
KLt2
C R
Q.2 [I]−1 Q.5
Q.8
e2 − 1
2
e
20
π2
≅ 2H Q.3
Q.6
LE2
2R12
EL
eR 2
Q.10 C = 9.2 .F
P hysi cs | 23.47
Q.11 6.94 A
Q.12 (i) i1 = i2 = 10/3A, (ii) i1 = 0, i2 = 30/11A, (iv) i1 = i2 = 0
Q.13 42+20t V
Q.14 2A, 400W
Q.15 Z = 100 Ω, VQ = 9.8 V
Q.16 0.2 mH,
Q.17 20A,
1
µF,8 × 105 rad / s
32

π
1
, ∴ Steady state current=20sin π  100t − 
4
4

Exercise 2
Single Correct Choice Type
Q.1 C
Q.2 D
Q.3 A
Q.4 A
Q.5 A
Q.6 A
Q.7 A
Q.8 B
Q.9 C
Q.10 B
Q.11 B
Q.12 B
Q.13 D
Q.14 D
Q.15 A
Q.19 B
Q.20 B, C, D
Q.21 A, B, C
Multiple Correct Choice Type
Q.16 D
Q.17 D
Q.18 A
Q.22 D
Q.23 B, D
Q.24 D
Assertion Reasoning Type
Q.25 C
Q.26 D
Q.27 A
Comprehension Type
Paragraph 1: Q.28 D
Q.29 C
Paragraph 2: Q.30 D
Q.31 B
Previous Years’ Questions
Q.1 (a) -5v, 24.5w (b) (i) 0.6A (ii) 1.386 × 10–3s, 4.5 × 10–4 J
Q.2 A → r, s, t; B → q, r, s, t; C →q, p; D → q, r, s, t
Q.3 B
Q.4 D
Q.5 C
Q.7 Amplitude = 20A, phase difference =
Q.9 C, D
π
4
Q.6 R =
Q.8 C
4
(r – r )
3 1 2
2 3 . 4 8 | Alternating Current
Solutions
JEE Main/Boards
Im
Exercise 1
z
Sol 1: In a resistance coil, when an alternating current
is flown, there will be a magnetic field generated across
the coil and so there will be an inductance induced into
the coil. Hence it will have more impedance compared
to the one withDC current.
R
ZR = R
ZL = iωL
R 
cos θ =   ⇒ power factor
Z
now for an ideal inductor, Z = ωL and R = 0
ZC = –i/ωC
∴ cos θ = 0
Hence power = VI (0) = 0
Sol 3: Impedance is the effective resistance of an
electric circuit or component to alternating current,
arising from the combined effect of ohmic resistance
and reactance.
R
C
L
znet=ZR + ZL + ZC (Since they all are in series)
Now we can write any quantity in phasor notation,
for V = V0 cos (ωt + θ)
we write this quantity in phasor notation as,
V = |V| ∠q
⇒ V = V0∠θ. [θ is the phase angle].
This is very helpful for us.
Now for the given potential, V = V0 sin ωt
V = V0 cos (ωt –
Now let ‘i’ (iota) be the complex number, square root
of –1.
Now, Impedance of resistance ‘R’ = R ≡ ZR
π
)
2
∴ V= V ∠ − π ........(1)
0
2

1 
Znet = R + i  ωL −

ω
C


1 
R 2 +  ωL −

ωC 

 −i 
Impedance of capacitor ‘C’ = 
 ≡ ZC
 ωC 
now |Znet| =
now net Impedance of the circuit (figure (i)) is

1 
 ωL −

ω
C
tan θ = 
R






=R–

1 
i
+ iωL = R + i  ωL −

ω
C
ωC

… (i)
We got Znet = ZR + ZL + ZC= R + i ωL –
Impedance of Inductor ‘L’ = i ωL ≡ ZL
Znet = ZR+ ZC + ZL
Re
Sol 4: As derived above,
Sol 2: We know that power dissipated = VI cosθ.
V = V0 cos t
z
x
2
i
ωL
P hysi cs | 23.49
Sol 6: No nothing is perfect. It is impossible to make a
perpetual machine.
Im
Sol 7: Using the notation used in Q.4 and Q.5;
z
R
L
Re
With this we can write
=
Znet | Znet | ∠θ V = V0 cos t
… (ii)
Now we known that
∴
V = I × Z [ V = I R]
In phasor notation: V0 = V0∠ 0
π

2
I =  V  ; I =

Z 0 ∠θ
Z
ZL = iωL ⇔ ZL = |ZL| ∠
v0∠ −
I  V0
=
Z
 0
∴
[
I = I ∠ –  π + θ  

0
2

use complex analysis in maths.]
⇔ ZL = ωL ∠

π
 ∠ − − θ
2

π
2
π
.
2
Now we know that V = I Z L
… (iii)
π

Phase of current = –  + θ 
2


π
Phase of voltage = –
2
∴ Depending upon the ‘θ’ we can speak more about
the relation between fV and φI.
V0 ∠0
π
ωL∠
2
= I
I V0 ∠ − π
=
ωL
2
π
⇒ I = I0 ∠ −
2
Phase of voltage = ∠ 0 = zero
Sol 5: Let V = V0 sin (wt + θ) be an ac voltage source.
Then
1/2
T 2 
 ∫ V dt 


Vrms =  0

T
 dt 
 ∫

 0

1/2
T 2

2
 ∫ V0 sin (ωt + θ) 


Vrms =  0

T






now for simplifying the calculation,
∴ We put θ = 0, and solve;
V 
we get Vrms =  0 
 2
V(0)
I
2
Phase of current = ∠ −
π
π
= −
2
2
Hence current lags behind the voltage by an angle of
π
 .
2
Sol 8: ω = 2pf
Now as derived in Q.4;
2 3 . 5 0 | Alternating Current
ZC =
−i
−i
=
ωC 2πfC
∴ vrms =
Sol 9: V= V0 ∠0 [In phasor]
…..(i)
V0
2
V0 = (vrms) 2
V0 = (220) ( 2)
C
V0 = 311 V.
V = V0 cos t
And given f = 50 HZ;
ω = 2pf= 2π(50) = 100 p
ω = 314
∴ v = 311 sin (314 t)
π
−i  1 
ZC =
= 
∠ −
2
ωC  ω C 
(b) Given Vrms = 100 2V;

π
Z C | Z C | ∠ θ ; for i → ∠ − 
=
2


π

−i → ∠− 

2
We know that Vrms =

V0 ∠0
I V=
Now =
Z  1 
π

∠ −
2
 ωC 
V = 200 sin (ωt)
2
Comparing both of them;
V0 = 200 V
V = 200 sin (314 t)
Now; ω =
I = V ωC∠ π
0
2
I= I ∠ π
0
2
…..(ii)
Now power dissipated P = V I standard notation get
familiar with this

π
P = (V0∠ 0)  I0 ∠ 
2

π
P = V 0 I0∠ 0 +
2
P = V 0 I0∠
And cos
π
2
π
= zero
2
Hence P = 0.
Sol 10: Refer to theory.
Sol 11: Refer to theory.
Sol 12: Refer to theory.
Sol 13: (a) Instantaneous voltage V = V0 sinωt now V0 is
the maximum possible voltage (or amplitude)
220 V given is the RMS value of voltage
V0
2π
T
 2π t 
⇒ V = 200 sin 

 T 
T/2
∫
Average =
0
 2πt 
200 sin 
 dt
 T 
T/2
∫
= 127 V.
dt
0
Sol 14: Let ‘f’ be the required frequency
ω = 2pf
now V = V0cos (2pft)
we are given Vrms = 200 V
∴ Vrms =
V0
2
V0 = 200 2 V
=
⇒ V 200 2∠0 ZL = iωL = i(ω) (i)
= iw≡ i2pf
π
Z L = 2pf ∠
2
V
now I =
Z
….(i)
P hysi cs | 23.51
∴ Imax = 5.0 2 A
I = 200 2∠0
π
2πf∠
2
π
I = I ∠ −
0
2
we want I0 =
∴f=
Let ∴ I = 5 2 sin(100πt)
when t =
then
200 2
= 0.9
2πf
I = 5 2 sin
200 2
H ≡ 50 Hz
2π(0.9) Z
Sol 15: V0 = Vrms .
1
sec
300
3
π
=
2.5 6 A
= 5 2×
2
3
Sol 17: Vrms = 220
V0 =
2
2 (Vrms)
V0 = 220 2
100
10f
ω = 2pf
V = 220 2 cos (2pft)
V = V0 cos t
V= 220 2 ∠ 0
V = V0 cos (t)
C
(a) V0 = 100 2
ω = 2π (50) = 100 p
∴ V = 100 2 cos (100 pt) = 100 2 ∠ 0
ZR= R = 100
 −i 
−i
ZC = 
= –i (318) W
=
 ωC  (100π)(10 × 10−6 )
∴ Resistance of capacitor is |ZC| ≈ 318 W
(b) now Znet = ZR + ZC
Znet = 100 – i (318)
Znet=
 −318 
(100)2 + (318)2 ∠ tan−1 

 100 
ZC =
−i
1
π
=
∠ − [In phasor notation]
ωC ωC
2
 1 
π
=
Z C 
∠ −
ω
C
2


V0 ∠0
V
Now I =
=
Z
 1 
π
C

∠ −
2
 ωC 
I = V ωC∠ π + 0
0
2
Znet = 334 ∠ –72.5°
I = V ωC∠ π
0
2

I V= 100 2∠0 = 0.42 ∠72.5=0.527 A
=
Z 334∠ − 72.5
∴ Phase of current =
(c) Pavg= Vrms Irms cos f
Phase of voltage = 0
 0.42 
= (100) 
 . cos(72.5) = 29.9 cos (72.5)
 2 
Pavg = 9 watt
Sol 16: f = 50 Hz ∴ ω = 2π × 50 = 100 p
Irms= 5.0 A
∴ φI– fv=
π
2
π
π
–0=
2
2
Sol 18: V = 220 2 cos (50 (2π) t)
V = 220 2 cos (100 pt)
=
V 220 2∠0
2 3 . 5 2 | Alternating Current
R = 100
Sol 20: V = 220 2 cos (2π (50) t)
10
220 V
50 H
C
A
220 V
50 H
(a) ZR = R = 100 ⇔ ZR = 100 ∠ 0
V = I Z
V = 220 2 cos (100pt)

I V= 220 2∠0
=
100∠0
Z
V = 220 2 ∠ 0 Now let ‘C’ be the capacitance of the circuit;
I 2.2 2∠0
=
=
2
 1 
π

∠ − 2
π
fc
2


ZR = R = 10Ω = 10 ∠ 0 now I0 = (2.2) 2
Irms =
−i
−i
=
=
ωC 2πfc
ZC =
⇒ I = (2.2) 2 cos (100pt)
I0
… (i)
(2.2)( 2)
2
=
= 2.2 Amp.
Znet= (10 + ZC) = 10 –
(220)2
= 484 watt
=
100
R
1
i
2πfc
 1 
(10)2 + 

 2πfc 
|=
Znet |
2

1 
−
 
1 
tan θ =  2πfc  =  −

 R   2πfcR 




Sol 19: V = 110 2 cos (2π (70) t)
1H
110 V
… (iii)
Now Znet = ZR + ZC
(b) Net power over a full cycle
(Vrms )2
 −1 
θ = tan–1  2πfRC 


 −1 
∴ Z = (10)2 + (X C )2 ∠ tan−1 
  2πfRC 
10HZ
Now V = 
IZ
V = 110 2 cos (140 pt)= 110 2 ∠ 0
…(i)
ZL = iωL = i (140 π) = i (140 π)
|ZL| = 440 W
ZL= 440 ∠
π
2

I V= 110 2∠0
=
π
Z
440∠
2
I
=
I0 =
 1 

π
π
∠− = 
 cos  140πt − 
2
2
2 2

2 2 
1
1
2 2
… (ii)
= 0.354 Amp.

I = V
Z
I
=
…(ii)
220 2
 −1 
∠0 − tan−1 

 2πfRC 
100 + X 2C
220 2
Now I0 =
Irms =
100 + X 2C
I0
2
=
220
100 + X 2C
Irms = 2A (Given)
⇒2=
220
100 + X 2C
… (iv)
P hysi cs | 23.53
100 + X 2C =
(110)2
V = – (813 ∠ 0) (40)
XC = 109.5 A
V =
−325∠0
⇔ V = – 325 cos (50 t)
Sol 21: Z = ZR + ZL + Z C = 40 + iωL –
1
ωC
= −230
2
(b) Inductance:

1 
Z= 40 + i  ωL −

ωC 

Now condition for resonance is Imaginary part of
Impedance is zero
R = 40
V
5H
−325
Vrms =
80F
( ) = – (8.13 ∠ 0)  50 × 5∠ 2π 
V = − 
IZL
(c) Capacitor:
(
V = − I Z c
V = 230 2 cos(ωt)
=
V 230 2 < 0 → (1)
1
=0
ωC
⇒ ω2 =
1
, ωC =
LC
ω=
=
1
5 × 80 × 10
−6



π 
− 2033cos  50t +  
V=
2 


∴ ωL –


π
V=
−  2033 ∠  → (x1)
2


… (i)
1


π 
1
∠ − 
VC= – ( 8.13∠0 ) 
2 
 50 × 80 × 10−6


π
8.13
∠− 
VC = − 
−6
2
 50 × 80 × 10

π
VC = − 2033∠ −  → (x2)
2


LC
=
)
1


π 
⇔ VC = – 2033cos  50t −  
2 


20 × 10−3
Now from equations (x1) and (x2)
1000
= 50 rad/s
20
we get VL + VC = 0.
w 50 25
~
f= = =
− 8HZ
2π 2π
π
Study more effectively on Resonance conditions.
Z =40 + i(0) =40
Sol 22: ZL = iωL = i (100 π) (80 × 10–3)
⇔ Z = 40 < 0 → … (ii)
80 mH
80 mH
V
Now from (i) and (ii), I =
Z
2∠0 23 2
I 230 =
=
∠0
40∠0
4
I = 23 2 cos (50t)
4
I = 8.13 cos (50 t)
230 V
230 V,50 ~
HZ
50 Hz
=
V 230 2 cos(2π(50)t)
V 230 2 cos(100πt)
=
=
V 230 2 < 0
Now potential drop across
(a) Resistance:
ZL = i (8π)
V = – IR
ZL ⇔ 8π <
( )
60F
60 µF
π
2
… (i)
2 3 . 5 4 | Alternating Current
500i
−i
−i
=
= −
−
6
3π
ωC 100π × 60 × 10
500
π
ZC ⇔
∠− 3π
2
500i
Znet
= ZL + Z C = 8pi –
3π


500
Znet=  8π −
i
3π  ⇒ Znet = – 28 i

π
⇔ Znet = 28 ∠ − 2
V
Now V = I Znet ⇒
= I
Znet
Power transferred to Inductor
ZC =
… (ii)
( ) ( I ) = (290 <π)
= VL

π
→ (1)
From(i)Xand
(ii)and (4)
 11.6∠  From
2


3π
2
 3π 
= 290 × 6 cos  
 2 
= (290 × 6) ∠
… (iii)
= Zero
Similarly zero for the capacitor to.
230 2∠0
⇒ I =
π
28∠ −
2
Total power absorbed by the circuit is
( ) ( I ) = ( 230
I 230 2 ∠ π
=
28
2
I = 6 ∠ π 2

π
⇒ I = 6 cos  100πt + 
2

11.6
I0 = 6 and Irms=
= 8.2 amp
2
Potential drop across;
(a) Inductor;

π 
π
VL= I . ZL =  11.6∠   8π∠ 
2
2 

… (iv)
(
P = ( 230
)
π
2 × 11.6 ) cos
2
P = zero
Sol 23: Explained in the key.
Sol 24: Initially
XL = 22 at f1= 200 HZ
[ω1 = 2π × 200]
(XL)A = ω 1L = 22
VL = (11 6 × 8π) <p
VL = 290 ∠ p→ (x1)
⇒ 2π × 200 L = 22 VL = 290 cos (100 pt + π)
L=
( )
VL = 290; VL =
O rms
O
290
= 205 V
2
(b) Capacitor
( ) ( Z C )
VC = I
VC =

π   500
π
∠− 
=  11.6∠  
π
2
2
2

 

11.6 × 500
∠0
3π
616
2
=4
22
= 1.75 × 10–2 H and finally;
400π
ω22 2π (50)
X2 = ω2L (1)
2π × 200 × L
(i) ⇒ x1 =
(2)
x
2π × 50 × L
(ii)
2
x2
⇔ VC = 616 cos (100 pt + 0)
… (i)
f2 = 50Hz
x1
VC = 616 ∠ 0 → (x2)
(VC)O = 616 (VC)rms=
)

π
2 ∠ 0  11.6 ∠ 
2


π
P = 230 2 × 11.6 ∠
2
P = V
x=
2
=4
x1 22
= 5.5 ohm.
=
4
4
… (ii)
P hysi cs | 23.55
Sol 25: Resistance of the lamp
90 8kW
8 × 100
⇒ Pi =
kW
=
100
Pi
90
220 V
= 22 ohm.
=
10
Let ‘L’ be the Inductance of the lamp;
⇒ Pi =
XL = ωL = (100 π) L
Znet = 22 + i (100 πL)

I = V =
Znet
I =
Sol 28: XL = ωL ; XC =
 100πL 
(22)2 + (100πL)2 ∠ tan−1 
 Now
 22 
Z=
net
250 2∠0
 100πL 
(22)2 + (100πL)2 ∠ tan−1 

 22 
250 2
2
484 + (100πL)
I0 =
 100πL 
∠ − tan−1 

 22 
250 2
484 + (100πL)2
and Irms=
I0
2
484 + (100πL)2
Put we are given that Irms = 10 A;
250
∴ 10 =
484 + (100πL)2
484 + (100 πL)2 = 625
141 ⇒ L =
100 πL =
⇒L=
141
100π
11.9
L = 0.04 H.
100π
Sol 26: Current drawn in circuit is maximum when the
circuit is in Resonance i.e. the Imaginary part of the
circuit is zero.
Now solve this question exactly as solved in Q. 21.
Sol 27:
VS
VP
=
NP
x1
x2
=
ω1
ω2
⇒ x2 =
ω2
ω1
. x1
⇒ x2= 2x
 −1 
xc = 

 ωC 
x1
=
w2
x 
⇔ x2 =  1 
w1
 2
Phasor method:Let V = V0 cos (ωt + q1) be the emf of an AC-source,
then can write this is phasor method as,
V = |V| ∠ q1⇔ V = V0 ∠ q1
Now for I = I0cos ( ω t + q2)
⇔ I = I0 ∠ q2
Now let Impedance (Z) ;
ZRe sis tance = R
Z capacitor =
−i
ωC
(i is iota; complex number)
Zinductor = iωL
Now in a circuit with series RCL;
i
+ i ωL
ωC

1 
Znet = R + i  ωL −
→ ωC 

NS
220
=
2200 5000
… (i)
Now let us write this in phasor notation,
NS = 500 turns.
8 kW
x=
Pi
Sol 29: XL = ωL
Znet
= Z=
Z C + ZL = =
R
R
NS
n (efficiency) =
−i
ωC
Now as ω is increased, both XL and XC increase.
x2
250
⇒ Irms =
80
kW ⇒ Pi= 8.9 kW.
9
Output power
Imput power
=
Znet | Znet | ∠ q
| Znet=
|

1 
R 2 +  ωL −

ωC 

2
2 3 . 5 6 | Alternating Current
E = (0.1) (0.1) 1 ⇒ E = 10–2 V

1 
 ωL −

ω
C
θ = tan–1 
R






now applying KVL in mesh (i)
E – i (i) – i1 (2) = 0

1 
2
ωL −




1
−1
ωC 
= R 2 +  ωL −
∴ Z net
 ∠ tan 
ωC 
R







Now for a source of emf V = V0 cos (ωt + q1)
Img
⇔ V = V0<q1
I =

1 
 ωL −

ωC 
| ∠ tan 
R






−1
⇒ i = i1 + i2
... (iii)
1
A.
220
L 
T (time constant) =  
R 
Sol 3: (C) At t = 0, inductor is open circuited
At t = ∞, inductor is short circuited
At t = 0,

1 
 ωL −

ωC 
–1 
∠ q1– tan
R






| Znet |
... (ii)
1
Now energy stored in magnetic field is LI2 and rate of
2
dissipation of energy is I2R.
V0 ∠θ1
V0
⇒ E = i + 3i2
Sol 2: (A) For an L–R circuit,
Re
R
| Znet
⇒ E – i (i) – 3i2 = 0
Znet
V
=
Z
net
In mesh-(ii);
From this we get i =
1
L C
I
=
... (i)
E = i + 2i1
10 V
I0
For resonance, imaginary part in eq. (i) is zero!
Exercise 2
10
Sol 1: (B) Emf induced in rod = BLv
At t = ∞
A
10
D
2
10
10
=i = 1 amp
10
3
10
R
B
C
10
i2
i1
i
1
2
(1)
=i
3
E (2)
10V 10V
= 2 amp.
=
Rnet
5
∴ Difference = (2 – 1) amp= 1 amp.
P hysi cs | 23.57
 L 
Sol 4: (B) T1 (time constant) during build up =  
 2R 
 L 
T2 during decay =  
 3R 
T
3
∴ 1 =
T2 2
Sol 5: (C) Energy stored per unit time = Li
= 2 (2) (4) = 16 J/s.
di
= 4 amp/s.
Sol 6: (C) i = 2 amp
dt
Applying KVL,
di
dt
1
⇒ i = 0.6   ⇒ i = 0.3 amp
2
Emf across coil = L
di
dt
di
di
= i0 (– (–2) e–2t) ⇒
= 2 i0 e–2t
dt
dt
Emf = 2L (0.6) e–2t
⇒ E = 6 e–2t ⇒E = 6 e−2ln
ln2
E= 6e
−1
⇒E=6×
2
1
E = 3V
2
Sol 8: (B) i = 5 amp
di
= –103 A/S
dt
4V
[Since decreasing; –ve sign]
1
1
1H
⇒ 4 – i (1) – L
3F
di Q
− =
0
dt C
VA – i(1) + 15 – L
Q
=0
C
⇒ 4 – 2 (1) – 1 (4) –
B
15V
di
= VB
dt
VA – VB = i – 15 + L
⇒Q=–2×3
di
dt
VA– VB = 5 – 15 + 5 × 10–3 (–10+3)
⇒ Q = 6C.
Sol 7: (A)
=i i0
A
5mH
−Rt

1 − e L


6V
VA – VB = 5 – 15 – 5 ⇒ VA – VB = –15 V




Sol 9: (C) When ‘i’ is reversed,
R
1
L
A i
5mH
15V
B
 di 
VA + i (1) + 15 – L   = VB
 dt 
=
i0
6v
= 0.6
10
VA – VB = –i – 15 + L
−10t 

–2t
=i 0.6  1 − e 5  ⇒ i = 0.6 (1 – e )




Put t = ln 2
(
2 n
=
⇒ i 0.6 1 − e

⇒ i 0.6  1 − e
=

2
n 2−1
)
=i 0.6  1 − 1 


 ⇒
2


di
dt
= –5 – 15 + 5 (+10-3) × 103
di
[i is decreasing against the direction of KVL. Hence
dt
= 103].
VA – VB = – 5 – 15 + 5
VA – VB = –15 V
2 3 . 5 8 | Alternating Current
Sol 12: (D) Just before the switch is opened, let us find
the currents,
Sol 10: (A) At t = 0, inductor is open circuited,
at t = ∞, it is short circuited
at t = 0,
L
A
i2
B
i1
R1
12 V
10
R2
20
i=
2V
l2 V
Rnet
 RR
Rnet=  1 2
 R1 + RL
2V
2
⇒ i1 =
i=
Rnet
10 + 20
 4×8 8
=
Ω
=
12
3

9
12
⇒ i = amp.
8
2
3
Now just at the instant switch is opened, i would remain
same
L B
A
i=
2
1
⇒=
amp.
i1 =
30 15
Finally; at t = ∞
10
12 V
20
R2
R1
2V
∴ VR = i R1 =
2V
2
⇒ i2 =
amp
i2 =
Rnet
20
i2 =
1
9
× 4 VR = 18V
1
2
Now applying KVL;
12 + (VB – VA) – 18 = 0 ⇒ VB – VA = 6 V.
1
amp.
10
Sol 13: (C) Power factor,
Sol 11: (D) At t = 0, no current flows in the circuit.
Fuse

R
cos φ = 

2
2
 (X C − XL ) + R
L = 4H
~
S




V=2sin (100t)
2V
10Ω
As time starts, current starts flowing and at t = ∞,
current in the circuit is infinity.
Hence at t = 10, i → ∞ so the fuse will get blown
∴
[ Infinity is just an unknown number !]
0.1H
C
10
0.1H
C
P hysi cs | 23.59

I V=
=
Z
Im
z
220 2

1 
(100) +  100 −

100
πC 

(XC-XL)
R
cos φ =
1
2
∴ tan φ =
⇒ θ=
R
π
4
| X C − XL |
R
⇒ |XC – XL| = R

1 
 100 −

100
πC 
–tan–1 
100






irms =
XL = ωL = (0.1) (100) ⇒ XL = 10 Ω.
⇒ |XC – XL| = R
220
2

1  ≡ 2.2
(100)2 +  100 −

100πC 

220
=
2.2
1
= R + XL
ωC
2
2

1 
(100)2 +  100 −

100πC 

2
C=
1
1
⇒C=
100(20)
ω(R + XL )

1 
∴ (100) = 100 +  100 −

100πC 

C=
1
× 10−3 ⇒ C = 500 µF.
2
1
=0
100πC

−1 
∴ XC = – 100 ∴ X C = 
ωC 

Sol 14: (A) ZL = iωL =
1
H
2
⇒ 100 −
1
× 100 π = i 100 W
π
100 C
ZR = 100 W
R
Hence,=
Znet = ZR + ZL + Z C
Znet=

1 
 100 −

−1
πC 
100
tan 
100






100
= 100Ω
1
Now for AC source of 100 V
i
= 100 + i (100) –
100πC

1 
(100) +  100 −

100πC 




Sol 15: (D) For 100 V D.C. source, i =1 amp.
−i
−i
=
=
ZC
ωC 100πC
2
 XC
Now power factor; φ = tan–1  R

 −100 
π
φ = tan–1  100  ⇒ φ = −


4
π 1
Power factor; cos φ = cos   =
2
4
Znet
2
2
i=
2
100
1 100
Znet ⇒ 2 = Z
net
⇒ Znet = 200
Z=
net
R 2 + XL2
∴ R2 + XL2 = (200)2 ⇒ XL2 = (200)2 – (100)2
XL = 174 W
2 3 . 6 0 | Alternating Current
ω L = 174
L=
174
⇒ L = 0.55 H.
100π
I2rms
=
∫I
2
dt
T
∫ dt
=
∫
O
(I20
+
I12
It is similar to superimposition of two vectors with an
angle of 30° in between them
2
T
inet = i0 sin (100 pt + θ)
O
i0 =
22 + 22 + 2(2)(2)cos(30°)
i0 =
8 + 8 3 ⇒=
i0 2 2 + 3
∫ dt
2
Irms
=
I12 T
+0
2
⇒ Irms
=
T
Sol 17: (D)
=
Sol 19: (B) i = 2 sin 100 pt + 2 sin (100 pt + 30°)
sin ωt + 2I0 I1 sin ωt)dt
I20 T +
3 3
4
R2
Use tan θ and simplify!
O
T
R1
(*) Don’t run to catch cos θ.
2π
T=
ω
O
3
∴
Sol 16: (A) I = I0 + I1 sin ω t
T
tan q2=
I20
+
I12
2
Phase diagram will be shown as
π
π
1
= ωt ;= 100πt ; t =
s.
4
4
400
o
i2 = 2 sin (100t + 30 )
Sol 18: (D) for LR circuit;

R1
cos q1= 
 2
2
 R 1 + XL
ieffective

 = 0.6 

... (i)
i1 = 2 sin 100t
for CR circuit;

R2
cos q2= 
 2
2
 R2 + XC
/3

 = 0.5 

... (ii)
Sol 20: (A) We can speak on nature by observing the
phase of final Impedance. If the phase of Impedance is
negative then it is capacitive, else it is inductive.
ω
1
=
2 2 LC
∴ ZR = R
Now when L, C, R of two circuits are joined;
∴ ω' =

R1 + R 2
cos θ = 

2
2
 (R1 + R 2 ) + (X C − XL )
ZL = i ω L = i .




Given that cos θ = 1
∴ XC = XL = X
X
tan q1=  L
 R1



X
tan q2=  C
 R2



tan θ1 XL R 2  R 2 
=
.
≡ 
tan θ2 R1 X C  R1 
4
tan q1=
3
ZC =
−i
=
ωC
1
2 LC
−i
1
2 LC
.L=i
= −2i
.C
1 L
2 C
L
C
3i L
3i L
−
∴ ZL + Z C =
; Znet= R −
2 C
2 C
 −3i L 
Znet= Z 0 ∠ tan−1 

 2R R 


∴ -ve phase
Hence capacitive.
P hysi cs | 23.61
Previous Years’ Questions
Sol 4: (C) For the lamp with direct current,
V = IR
π
⇒ R =8Ω and P =80 × 10 =800 W
Sol 1: (A) As the current i leads the emf e by , it is an
4
R–C circuit.
For ac supply
1
2
Erms
XC
2
π
ω
C
=
P
I
=
R
R
or tan =
\ωCR = 1
tan φ =
rms
Z2
4
R
R
As ω = 100 rad/s
The product of C–R should be
1 –1
s
100
⇒
(220)2 × 8
Z2 =
800
⇒
Z=
22Ω
⇒ R 2 + ω2L2 =(22)2
Sol 2: (B, C) Z =
R 2 + X 2C =
 1 
R2 + 

 ωC 
2
⇒ ωL = 420
In case (b) capacitance C will be more. Therefore,
impedance Z will be less. Hence, current will be more.
⇒ L=
0.065 H
∴Option (b) is correct.
JEE Advanced/Boards
Further,
Exercise 1
V 2 – VR2
Vc =
=
Sol 1: At t = 0, we can replace the inductor by open
circuit and at t = ∞, the inductor can be short circuited
V 2 – (IR)2
at t = 0,
In case (b), since current I is more.
Therefore, Vc will be less.
Sol 3:
6
dI
= 103 A/s
dt
A
1Ω
10
I
15 V
5 mH
B
di
∴Induced emf across inductance, |e| = L |e| =
dt
(5 × 10–3) (103) V = 5 V
Since, the current is decreasing, the polarity of this emf
would be so as to increase the existing current. The
circuit can be redrawn as
A
1Ω
15 V
Now VA – 5 + 15 + 5 = VB
\VA – VB = – 15 V
or VB – VA = 15 V
5 mH
4
4
4
10
i1 =
10
= 1 amp.
10
At t = ∞,
6
10
I = 5A
4
B
=
i2
10
10
=
amp
R eff
8
i1
1
8
=
=
= 0.8 amp
i2 10 10
8
2 3 . 6 2 | Alternating Current
Sol 2:
L
RCV
V = IR ⇒
L
L
⇒
RC(IR)
R(RC)I
10
Now {RC} = time constant in RC circuit
∴
20
L 
[RC] = [T] and   = time constant in LR circuit
R 
2V
I=
∴ L 
  = [T]
R 
 L 
[T]
−1
∴ =
 = [I] .
RCV
[T]
[I]


Sol 3: Let us calculate the total energy stored in the
inductor before switch is shifted.
2
Rnet
I1 =
2
10 + 20
I1 =
2
amp. 30
… (i)
at t = ∞,
R2
10
20
L
2V
Here the resistor 10 Ω is shorted.
E
R1
1 2
Energy stored in inductor =
LI
2
1
= L [Iat t = ∞]2
2
It =∞
2
=
2
1
amp.
=
20 10
Rnet
Sol 5: Let us now derive the current in the circuit as a
function of time
R=10
 E 
=  
 R1 
1
∴E= L
2
E=
=
I2
 E 
 
 R1 
L=5
i
V
2
LE2
2R12
Now this is the total heat produced in R2.
Sol 4: This is similar to the Questions 1 (Ex. I).
At t = 0; Inductor is open circuited,
at time t = t; current = i amp;
using KVL;
V – iR – L
di
di
= 0 ⇒ V – iR = L
dt
dt
1
di
dt =
V − iR
L
⇒
At t = ∞, Inductor is short circuited.
Integrating;
At t = 0;
1
di
⇒ i = i0 1 − e−Rt/L
dt = ∫
∫
L0
V − iR
i
t
i
0
(
)
P hysi cs | 23.63
At t = 0, i = zero
E = 2kLt
At t = ∞, i = i0 = constant
Current in the coil II is
Now R = 10Ω, L = 5
(
i = i0 1 − e−2t
i=
)
At t = 1 sec
q=
dq
⇒ q = ∫ i dt
dt
t
−RT
L
t0
q=
q=
i0L
R
−
( )
R

L 
−  0 + 
R 


L i0L
1 − e−1
−
R R
R
Given that cos θ = 1 ⇒ θ = 0
0
∴ |XL + XC| = 0 ⇒ XL = – XC
XL = ω L
−1
1
⇒ ωL =
ωC
ωC
XL =
ω =
i0L 
−Rt 
1 − e

R 
L 
(
0
2KL 2
t
2R
t
1
LC
⇒L=
1
ω2C
ω = 2π (50) = 100 p
L 
One time constant ⇒ t =  
R 
⇒ q = i0 .
⇒ q=
XL+XC
−Rt  


⇒ q = i0 t − L  1 − e L  

 R



⇒ q=
i 0t −
t
KLt2
C
R

 dt


−Rt
eL
 2KL 
 t dt
R 
∫ 
Img
−Rt 

⇒ q = i0 t −  − L  e L 


  R 


⇒ q = i0 t + L
 R
⇒q=
Sol 8: Power factor is cos (θ)
−Rt 

=
q ∫ i0  1 − e L  dt




=
q i0 ∫ 1 − e
∫ i dt
 2KL  t2
q=
.
 R  2
Sol 6: i = i0 (1 – e–Rt/L)
i=
t
t =0
 e2 − 1
 =
e2

 i

 i0
dq
dt
q=
i
−2
i = i0 (1 – e–2) ⇒ = (1 − e )
i0
E  2kL 
=
t
R  R 
L=
)
1
20
= = 2H.
(100π) C π2
2
Sol 9: We know that VR and VL will have a phase
π
difference of .
2
i0L
i0L 
1
1 −  ⇒ q =
Re
R 
e
Vnet =
EL
VR2 + VC2 =
162 + 122 = 20V.
R 2e
Sol 7: Given mutual inductance between coils = M
And I1 = kt2
∴ EMF induced in second coil= L
dI
= L [2kt]
dt
Sol 10: Resistance of Lamp = R
 V 2   100 × 100 
=
=
R  =

 200Ω
 P  
50



Maximum current the lamp can sustain,
2 3 . 6 4 | Alternating Current
imax =
i=
max
now at t = 0.1 ln 2, S2 is open;
Pmax
V
50
1
amp.
=
100 2
40
100 V
Now in the given conditions;
10
(200 V, 50 HZ )
200V
= 1 amp which is greater than 0.5 amp.
200Ω
Hence we need to increase the Impedance by using a
capacitor of capacitance ‘C’. Such that ‘ I’ will be equal
1
to
amp.
2
i=
 1 
R +

 ωC 
2
Z
∴=
I=
∴ t’ = t – 0.1 ln 2
200
 1 
R2 + 

 ωC 
∴ inew = i0 1 − e−50(t −0.1 ln2)  ; t ≥ 0.1 ln 2 

2
=
i0
... (iii)
100
= 2 amp.
50
using equation (iii) at time t = 0.1 ln 2, i = 0
But this is not true, since there is a current flowing in
the circuit guided by the equation,
2
 1 
2
(200)2 + 
 = (400)
ω
C


i = 10 (1 – e–10t) [from eq.(i)]
ω = 2π (50) = 100 p
now at t = 0.1 ln 2
Solving this will give the value of ‘C’.
i = 10 (1 – e–10t(0.1)ln2)

1
=i 10  1 −  ⇒ i = 5 amp.
2

−10t 
100 
1 − e e 
Sol=
11: i

10 


∴ inew = 5 + 2 1 − e−50(t −0.1 ln2) 


S1
At time t = 0.2 ln2
inew = 5 + 2 1 − e−50(t −0.1 ln2)  = 5 + 2


S2
40
100 V

1
=5 + 2 1 − 
25 

10
 31 
inew = 5 + 2  
 32 
1H
)
… (ii)
Also t’ = 0 ⇒ t = 0.1 ln 2 sec.
 1 
2
R2 + 
 = (400)
ω
C


(
)
But this is not true; Since there is a current flowing in
the circuit at that instant.
2
=i 10 1 − e−10t (
)
But this equation; at t’ = 0, we get inew = 0
2
1
1
I = amp ⇒ =
2
2
(
100
1 − e−50t
50
∴ inew =
inew
= 2 1 − e−50t 2
200
 1 
R2 + 

 100 
1H
… (i)

31 
inew =  5 +
 amp. = 6.94 amp.
16



ln2−5 
1 − e

P hysi cs | 23.65
Sol 12: After switch is closed;
10
(i) t = 0; open circuiting the inductor;
i3 =
30
A
B
i2
100 10
amp.
i1= i2=
=
30
3
20
10
30
i1
i2
100 V
D
C
∴ Hence the current in loop ABCD will be
20
20
amp.
11
And this current will start decaying to zero
∴ At t = ∞, i = zero.
(ii) now at t = ∞ ;
inductor is short circuited,
Sol 13: Applying KVL;
10
30
i1
R
i(t)
i2
100
20
100
Rnet
E – i (t) R – L
i1 =
100 50
=
amp
22 11
i(t) = 3 + 5t ⇒
10
i1
i3
di
=0
dt
E = 42 + 20t
20
Sol 14: Now when capacitance is removed;
R
2i2 = 3i3 ⇒ i3=
i 1 = i2 +
2i2
3
200 V, ~
300 rad/s.
2i2
3
⇒ i1 =
di
=5
dt
E = R i(t) + L(5) ⇒ E = 4(3 + 5t) + 5(6)
30
i2
and i1= i2 + i3
L
E
i1 =
100
20
11
L
CR
200 V
300 rad/s
5i2
3
3  50  30
3
i2 =
i=  =
amp.
5 1 5  11  11
20
i3 =
amp.
11
V = 200 2 cos(300t)
=
V 200 2∠0
Znet
= ZR + ZL
(iii) Now when switch is open
Znet = R + iωL
(a) Immediately after that, current through inductor will
be same as just before
Znet =

I = V
Z net
 ωL 
R 2 ω2L2 ∠ tan−1 

 R 
L
C
2 3 . 6 6 | Alternating Current
I =
Sol 15: Maximum current flows when the circuit is in
resonance
P
Q
200 2∠0
 ωL 
R 2 + ω2L2 ∠ tan−1 

 R 
P
 ωL 
∠ − tan−1 

 R 
R 2 + ω2L2
Now given that current lags behind voltage by 60°,
 ωL 
∴ tan–1  R  = 60


∴
R 3
100 3
L=
⇒L=
ω
300
L=
3
i.e. ω =
 π
 =
 3
1
1 × 10
=
∴ ZP
1
1
=R 3 ⇒ C =
ωC
R 3ω
C
⇒=
× 4.9 × 10
−3
⇒ ω=
1
49 × 10−10
(32)2 + (X C )2
−1
−1
= –70W
=
ωC 1
5
−6
× 10 × 10
7
(32)2 + (70)2
| ZP | = 77 ohm,
100
3 3
µF
And impedance of coil Q is
Now when all together are present
XL= ω L =
Znet = ZR + ZL + Z C = 100 + iR 3 − iR 3
XL = 70 W
[From X1and X2]
(68)2 + (70)2
| Z Q | = 98 W
Znet
= 100∠0
V
200 2∠0
I 2 2∠0
I =
⇒=
=
Z
100∠0
net
  = (200 2) (2 2)cos(0)
power = VI
Znet = 32 – 70 i + 68 + 70 i
Znet = 100 W
2
I 10 2 ∠0 ⇒=
I
=
∠0
100
10
Voltage across P; VP = (Irms) ( | ZP | )
P = 800 W
 200 2 
Pavg = Vrms . Irms = 


2 

(68)2 + (XL )2
1
× 105 × 4.9 × 10–3
7
∴ Impedance =
Znet = 100
Pavg = 400 W.
−6
1
× 105 rad/s.
7
=
XC
⇒ X C = 3 R → (x2)
100. 3 (300)
ω=
LC
Impedance of Box P is
By intuition we can say that
1
4.9 H 68
=
V 10 2∠0
1
ω=
Now when the inductance is removed;
C=
4.9 H 68
1F 32
H.
 XC
tan–1  R

XC
= 3
R
R
R
=
V 10 2 cos(ωt)
ωL
= 3 ⇔ XL = R 3 → (x1)
R
1
~
Q
L
1µF 32
200 2
I
=
L
2 2 


 2 


 2


 10 

 . (77)
=
2
VP = 7.7 V
P hysi cs | 23.67
Voltage across Q; VQ = (Irms) ( | ZP | ) =
1
(98)
10
VQ = 9.8 V.
Sol 16: ωr = 4 × 10 rad/s.
5
ωL −
ω2L − 1
=R;
ωC
ω 2L – ωCR – 1 = 0
Solving this would give us
Given Va – Vb = 60 V
ω = 8 × 105 rad/s.
and Vb – VC = 40 V
120
a
1
=R
ωC
|XL – XC| = R;
b
c
d
Sol 17: V = 220 2 sin (100 pt)
=
V 220 2∠0
Znet
= ZL + ZR = i (100 π × 35 × 10–3) + 11
Znet = 11 i + 11
We know that during resonance,
VL + VC = 0
VC = – 40 V
∴
Vc – Vd = – 40V
(Va – Vb) = irms R

I V= 220 2∠0 ; I= 20∠ − π
=
π
4
Z
11 2∠
4
1
60 = irms . 120 ⇒ irms =
amp.
2
Now, Vb – VC = (irms) . ZL
40 40
40 = (irms) . ( ZL ) ⇒ ZL =
=
= 80Ω
1
irms
2
ω L = 80
(4 × 105) L = 80; L = 0.2 mH
−1
= −80
ωC
Sol 2: (D)
1
1
;C=
80ω
80 × 4 × 105
1
C=
µF.
32
C=
t
i1
Current i2 is constant and positive i.e. from ‘c’ to ‘d’ have
i1 has to be from ‘b‘ to ‘a’. Hence negative
di
L
dt
Also i2 =
RL
|


XL+XC
∴
di
= constant
cons tant
dt
Hence i1 versus t is as shown.
R
π
Now at θ =
4
Exercise 2
Sol 1: (C) Current is induced by varying magnetic flux.
Here there is no such phenomena as flux linked with
the coil is zero. Hence induced current is zero.
i.e. irms . Z c = − 40
 | X − XC
tan θ =  L
R


π
⇔ I = 20 sin  100πt − 
4

Single Correct Choice Type
Now Vc – Vd = – 40
Z c = −80 ;
 11 
112 + 112 ∠ tan−1  
 11 
π
Z
=
11 2∠
net
4
Znet=
Sol 3: (A) Emf induced across inductor = L
di
dt
2 3 . 6 8 | Alternating Current
Sol 10: (B) Let Z A be the Impedance of element A, and
ZB be that of element B.
Rt 

i = i0  1 − e − L 




Initially; when R is connected to A;
Rt
  R  − Rt 
di
di i0R  − L 
e 
= i0  −  −  e L  ⇒
=
  L


dt
dt
L 




e = i 0R . e
−
Rt
L
Znet = R + Z A .
… (i)

i = V
Z
Rt
i = i 0 – i0 e − L
e
R
=i i0 −
i
=
e
= −i + i0
R
Hence graph A.
Sol 4: (A) Self-induction Emf = –L
dt
<
di2
dt
⇒–
di1
dt
>–
di
dt
Z=
di2
dt
i
=
E1> E2.
L
will have dimensions
Sol 5: (A) We know that RC and
R
R
1
of time. Hence
and
will have dimensions of
L
RC
frequency.
Sol 6: (A) Refer to Questions – 3 (Ex –I JEE Advanced)
Sol 7: (A)
1 2
LI = 32J 2
I2R = 320
… (i)
… (ii)
(1) L 2 × 32
= =
(2) R
320
L
τ= = 0.2s.
R
Sol 8: (B) In an L-R decay circuit, the initial current at
t=0 is 1. The total charge that has inductor has reduced
to onefourth of its initial value is LI/2R
Sol 9: (C)
1 2
LI = U
2
I2R = P
T=
L 2U
=
R
P
Z
∠ − tan−1  A
 R
Z 2A + R 2



V
Given that current is lagging behind voltage by angle
‘q1’
 ZA 
… (i)
∴ tan–1  R  = q1 

When R is connected to B
e = R (–i + i0) [y = –mx + c]
di1
Z 
Z 2A + R 2 ∠ tan−1  A 
 R 
⇔ Znet=
Z
ZB2 + R 2 ∠ tan−1  B
 R
Z
∠ − tan−1  B
 R
+R
V
ZB2



2



Given that current leads voltage by ‘q2’
∴
 ZB
q2= –tan–1  R


 
… (ii)
Using same method, when R, A, B are connected,
 Z A + ZB
θ = –tan–1 
R


 
… (iii)
tan (– θ) = tan (–q2) + tan q1
tan θ = tan q2– tan q1
Sol 11: (B) Resonance is a condition of maximum power
Hence cos φ = 1.
Sol 12: (B) In calculating the rms value, we square each
value.
A
Hence both A and B have same square value at every
point.
P hysi cs | 23.69
2
Znet =

1 
π
 ωL −
 ∠
ωC 
2


i = V ⇒
Z
net
B
Hence irmsA = irmsB
Here we have every value greater than that of Irmsin
graph A or graph B.
VL = i Z L
∴ (irms)C> IA = IB.
VL = V1 ∠
Sol 13: (D) Initially in LR circuit;
VC = i Z C
V0

1 
 ωL −

ωC 

2
∠−
π
2





V0
π 
π
 ωL ∠
=
VL 
∠−
2
2 
2
 
  ωL − 1 

 

ωC 



R
cos θ1 = 
 2
2
 R + 9R

 ⇒cos q1=


 R 


 R 10 
10
Now finally

R

cos θ2 =
 2
2
 R + 4R
1
P2 =
5
P1
P2
=
π π
−
2 2
Hence phase difference between VL and VC will be π and
π
between VL and I will be ± . Graph D satisfies all the
2
conditions.
XL – XC = 3R – R = 2R
⇒
VC= V1 ∠ −
π π 
VC= V1 ∠ −  +

2 2 
1
P1 =
π π
−
2 2
1
10
. 5 ⇒




Sol 15: (A) Let us consider mesh (1);
XC
P2
P1
i1
= 2
(2)
i2
Sol 14: (D) Znet
= ZL + Z C
C
L
(1)
XL
R
=
V V0 < 0
Z=
Z=
1
C
V = V0 cos t
Znet
 −i 
= i ωL + 

 ωC 

1 
⇒ Znet = i  ωL −

ωC 

i
=
1
V
=
Z 1
π
1
∠−
ωC
2
V0 ∠0
1
π
∠−
ωC
2
i1 = V0 ωC∠
π
2
Now in mesh (2)
Z 2 = ZR + ZL = R + i ω L
... (i)
2 3 . 7 0 | Alternating Current
 ωL 
R 2 + (ωL)2 ∠ tan−1 

 R 
Z=
2
V
=
Z 2
i
=
2
B2
V∠0
B2
 ωL 
R 2 + (ωL)2 ∠ tan−1 

 R 
i = i1 ∠ − tan−1  ωL 


2
0
 R 
 ωL 
π
− tan−1 
Phase difference between i1 and i2=

2
 R 
X 
π
= − tan−1  L 
2
R 
At t = ∞;
i
R
i
R
B2
B2
2i
Multiple Correct Choice Type
Sol 16: (D) Using intuition;
Let us go for capacitance in the box
∴ Q = CV
Hence B2 lights up early; but finally both B1 and B2 shine
with equal brightness.
dq
dv
=C
dt
dt
Sol 19: (B)
Given i =
dq
= constant
dt
L
i1
i
dv
= constant
∴
dt
∴ Graph looks like a straight line.
(1)
dv
i=C
dt
Slope of the graph =
∴
1
2
8−2
=2
3
This opposition is due to the induced EMF in ‘L’.
C = 0.5 C.
Sol 20: (B, C, D) Emf induced in coil 1 = L1 di1
dt
di
E2 = L2 2
dt
L 
Sol 17: (D) Time constant τ = 
R 
Energy stored in magnetic field =
Power dissipated in resistor = I R
2
1 2 
 LI 
∴ 22
= τ
2
 IR 


Sol 18: (A) At t = 0;
(2)
Just after switch is closed, Inductor tries to oppose the
current ‘i1’. Hence i1< i2. As time goes on, the opposition
given by inductor reduces.
i = 2C = 1 amp
C=
i2
1 2
LI
2
Given that
di1
dt
=
di2
dt
E1 L1
= 4
∴ =
E2 L2
∴
V2
V1
=
1
4
And also given that power given to the two coils is
same,
∴ Vi i1 = V2 i2
P hysi cs | 23.71
i1
i2
=
V2
V1
⇒
i1
i2
=
Assertion Reasoning Type
1
4
Sol 25: (C) Magnetic field is into the page
1
1
W1 = L1 I12 and W2 = L2 I22
2
2
2
L   I 
W
8 1
=  1   1  ⇒ 1 =    
W2  2   4 
W2  L2   I2 
W1
∴
W1
W2
2
1
.
4
=
L
will have the dimensions
R
R
1
of time and hence
and
will have dimensions of
L
RC
frequency.
Sol 21: (A, B, C) RC and
Sol 22: (D) When just after battery is connected, current
is zero in the circuit, and hence will follow magnetic
1 
field energy  LI2  and power delivered (I2R) is also
2 
zero.
 di 
EMF induced is  L
 . Hence there is a finite value.
 dt 
Sol 23: (B, D) At time t = 0, capacitor is short circuited,
Inductor is open circuited.
At t = ∞, capacitor is open circuited,
Inductor is short circuited.
Hence both the options follow from this.
Sol 24: (D) M.
M
∆iB
∆t
=
diB
dt
=
dφ A
dt
∆φA
∴ ∆ iB =
∆t
∆φA
M
4
∆IB =
2
∆t
=M
∴ Magnetic field decreases.
Hence there will b e a clockwise current in the ring.
Sol 26: (D) In an LCR circuit,
|Z|=
imax =
R 2 + (XL − X C )2
Vmax
R 2 + (XL − X C )2
(VR)max =
(VL )max =
R . Vmax
R 2 + (XL − X C )2
;
ωL . Vmax
2
R + (XL − X C )2
Now (VR)max = Vmax; at resonance condition, (XL – XC = 0),
now for (VL)max; we can set conditions,
(a) R ?0 and (b) XL = XC;
This will lead to (VL)max> Vmax.
Sol 27: (A) When circuit is suddenly switched off, there
will be a change in current, and it will lead to induced
EMF.
|E| = L
∆IB = 2A
∆φB
As resistance is increasing, current decreases
di
dt
Now for large ‘L’, E is also high.
∆iA
∆t
∆φB = 2(1) = 2
But given the values of 4 weber.
Hence options D isn’t true.
2 3 . 7 2 | Alternating Current
Comprehension Type
Paragraph 2
Paragraph 1
Sol 30: (D)
R
L
Sol 28: (D) Now when S1 is opened and S2 is closed
V = V0 cos t
CV -
+ CV
L
At t = 0; energy stored is purely in capacitor.In this type
of circuits, charge and current will be in the form of sin
or cos. Thus oscillatory.
 1 
q = Q0 cos 
t  ; Q0 = CV
 LC 
−1
i=
=i
Q0 sin ω t
LC
Q0
=
LC
CV
C
= V
L
LC
Hence option D.
 1 
t
Sol 29: (C) q = Q0 cos 
 LC 
Znet = R + i ω L
 ωL 
Z net | Z net | ∠ tan−1 
R 2 + ω2L2 ;=

 R 
V0 ∠0
| Znet
=
|
V
=
Z
I
=
 ωL 
| Znet | ∠ tan−1 

 R 
 ωL 
V
∠ − tan−1 

| Znet |
 R 
I
=
Now potential difference across resistance,
VR = i × Z R
 V
 ωL  
=  0 ∠ − tan−1 
  R∠0 
 R  
 | Znet |
VR
=
V0 R
| Znet
(VR)max =
 ωL 
∠ − tan−1 

|
 R 
V0R
R 2 + XL2
≡ 4 volts (given)
 1 
dq −Q0
=
sin 
t
dt
LC
 LC 
( VL ) = ( i ) ( Z L )
d2q
 V
X
=  0 ∠ − tan−1  L
R
 | Znet |
dt2
d2q
dt2
=
−Q0
 1 
cos 
t
LC
 LC 
= −
1
q,
LC
Hence option ‘C’.

 
 

π
ωL∠ 
2


 V0 XL ∠ π − tan−1  XL
=
VL
R
 2
2
2

 R + XL
( )
(VL )max =
V0 XL
R 2 + XL2
… (i)
≡3V 
 
 
… (ii)
(i)
(1)
R
4
= =
(ii) XL 3
(2)
∴
R
4
3R
= ⇒ XL =
XL 3
4
… (iii)
P hysi cs | 23.73
R 2 + XL2 = R 2 +
Previous Years’ Questions
9R 2 25R 2
=
16
16
5R
R 2 + XL2 = 4
In equation (i)
V0R
2
R +
XL2
= 4;
... (iv)
5R
4
=4
you can just start from here if you understand how I
wrote them



X
π
VL = 3 ∠ − tan−1  L
2
R
12 V
i1 - i2
i1
i1
i1
2
A

X
⇔ VR = 4 cos  ωt − tan−1  L

R

1
i2
1
2
X
VR = 4∠ − tan−1  L
R
2F
i2
2
V0R
V0 = 1 V
Sol 1: In steady state no current will flow through
capacitor. Applying Kirchhoff’s second law in loop 1:
2
3V

 

B
3
10



– 2i2 + 2(i1 – i2) + 12 = 0

X
π
⇔ VL = 3 cos  ωt + − tan−1  L

2
R

\2i1 – 4i2 = – 12

 

or i1 – 2i2 = – 6
…(i)

X 
VL≡ 3 sin  ωt − tan−1 L 
R 

Applying Kirchhoff’s second law in loop 2:
Given VR = 2
4i1 – 2i2 = – 9 – 12 – 2(i1 – i2) + 3 – 2i1 = 0

X
∴ 2 = 4 cos  ωt − tan−1  L

R


X
1
= cos  ωt − tan−1  L

2
R

 XL
∴ ω t – tan  R

–1
4+3 3
2
i2 = 2.5 A and i1 = – 1A
Now, VA + 3 – 2i1 = VB

 

or VA – VB = 2i1 – 3
= 2 (–1) – 3 = – 5V
PR = (i1 – i2)2 R1 = (– 1 – 2.5)2 (2) = 24.5 W
1
VL = 3 sin 60° =
Sol 31: (B) Vsource = VL + VR =
Vsource =
Solving Equations (i) and (ii), we get

 

 π
 = → (X )
1
 3
π
Now VL = 3 sin   ;
3
…(ii)
3 3
2
(b) In position 2: Circuit is as under
3V
2
3
3 3
+2
2
10
Steady current in R4:
i0 =
3
= 0.6 A
3+2
2 3 . 7 4 | Alternating Current
Time when current in R4 is half the steady value
t1/2 = τL (In 2) =
–3
L
(10 × 10 )
 n (2) =
 n (2)
R
5
= 1.386 × 10–4 s
U=
1 2 1
Li = (10 × 10–3) (0.3)2 = 4.5 × 104J
2
2
Sol 6: After a long time, resistance across an inductor
becomes zero while resistance across capacitor
becomes infinite. Hence, net external resistance,
R
+R
3R
Rnet = 2
=
4
2
2E
Current through the batteries, i =
In circuit (q): V1 = 0 and V2 = 2I = V (also)
3R
+r +r
4 1 2
Given that potential across the terminals of cell A is
zero.
In circuit (r): V1 = XLI = (2pfL)I
∴E – iri = 0
= (2π × 50 × 6 × 10–3)I = 1.88I
V2 = 2I


2E
or E – 
 r1 = 0
 3R / 4 + r1 + r2 
In circuit (s): V1 = XLI = 1.88 I
Solving this equation, we get, R =
 1 
V2 = XCI = 

 2πfC 
Sol 7: Inductive reactance XL = ωL
Sol 2: In circuit (p): I can’t be non-zero in steady state.


1
= 
= I = (1061) I
–6 
 2π × 50 × 3 × 10 
= (50) (2π) (35 × 10–3) = 11W
= 11 2 Ω
V2 = XCI = (1061)I
Given Vrms = 220 V
(A) → r, s, t
(B) → q, r, s, t
(C) → q or p, q
(D) → q, r, s, t
Sol 3: (B) Charge on capacitor at time t is
q = q0 (1 – e )
–t/τ
Here q0 = CV & t = 2t
Here q0 = CV(1 – e–2τ/τ) = CV (1 – e–2)
Sol 4: (B) From conservation of energy,
C
1 2
1
LImax = CV 2 ∴ lmax = V
L
2
2
Sol 5: (C) Comparing the LC oscillations with normal
SHM, we get
2
dQ
dt2
=–wQ
2
1
Here, w =
LC
2
\Q = – LC
d2Q
dt2
(11)2 + 11)2
R 2 + XL2 =
Impedance Z =
In circuit (t): V1 = IR = (1000) I
Therefore the correct options are as under
4
(r – r )
3 1 2
Hence, amplitude of valtage V0 =
2 Vrms
= 220 2 V
∴ Amplitude of current i0 =
V0
Z
=
220 2
11 2
or i0 = 20 A
X
Phase difference φ = tan–1  L
R
π
φ=
4

–1  11 
 = tan  
 11 

In L-R circuit voltage leads the current. Hence,
instantaneous current in the circuit is,
i = (20 A) sin (wt – π/4)
Corresponding i-t graph is shown in figure.
V,I
20
V=
i=20 sin (t-/4)
9T/8
O
-10 2 T/8
T/2 5T/8
t
P hysi cs | 23.75
Sol 8: (C) When e– has zero kinetic energy total energy
is shared by antineutrino and proton. This time energy
of antineutrino is its maximum possible kinetic energy.
As antineutrino is very light mass in comparison to
proton so it will have almost contribution in total
energy.
∴ Its energy is almost 0.8 × 106 eV
Sol 9: (C, D) As current leads voltage by π / 2 in the
given circuit initially, then ac voltage can be represented
as
=
V V0 sin ωt
∴=
q CV0 sin=
ωt Q sin ωt
Where, Q= 2 × 10−3 C
••
••
=i
3
7 π / 6ω ; I =
−
I and hence current is
At t =
2 0
anticlockwise
Current ‘i’ immediately after t =
7π
is
6ω
Vc + 50
= 10 A
R
Q final − Q(7 π /6ω) =
2 × 10−6 C
Charge flow =