AC Circuits
Ranker
PHYSICS TEST
AC CIRCUITS
1.
The voltage of an ac supply varies with time
voltage and frequency respectively are
A) 120 volts, 100 Hz
2.
3.
4.
5.
6.
7.
B)
as
120
2
V  120 sin 100  t cos 100 t.
the maximum
volts, 100 Hz
C) 60 volts, 200 Hz
D) 60 volts, 100 Hz
In the circuit shown in the figure, the ac source gives a voltage V  20 cos( 2000 t).
6
Neglecting source resistance, the voltmeter and ammeter reading
A
will be
5mH 4 50 F
A) 0V, 0.47A
B) 1.68V, 0.47A
C) 0V, 1.4 A
D) 5.6V, 1.4 A
V
A telephone wire of length 200 km has a capacitance of 0.014 F
per km. If it carries an ac of frequency 5 kHz, what should be the value of an inductor
required to be connected in series so that the impedance of the circuit is minimum
A) 0.35 mH
B) 35 mH
C) 3.5 mH
D) Zero
Instantaneous current in branch having capacitor C will be:
A) 20 2 sin  t  3 / 4 
B) 40 2 sin  t   / 4 
C) 60 2 sin  t   / 4
D) None of above
In the above problem, angle between current through inductor and capacitor will be
A) 143°
B) 90°
C) 53°
D) None
In the above problem, potential drop across XL by a.c. voltmeter will be
A) 160 volt
B) 120 volt
C) 200 volt
D) 160 2 volt
The r.m.s. current in an ac circuit is 2 A. If the wattless current be 3 A , what is the
power factor?
A)
8.
(t )
1
3
B)
1
2
C)
1
2
D)
1
3
A bulb and a capacitor are in series with an ac source. On increasing frequency how
will glow of the bulb change
A) The glow decreases
B) The glow increases
C) The glow remain the same
D) The bulb quenches
Page 1
AC Circuits
9.
Ranker
A virtual current of 4A and 50 Hz flows in an ac circuit containing a coil. The power
consumed in the coil is 240 W. If the virtual voltage across the coil is 100 V its
inductance will be
A) 1 H
B) 1 H
C) 1 H
D) 1 H
3
10.
7
9
For a series RLC circuit R = XL = 2XC. The impedance of the circuit and phase
difference (between) V and i will be
A)
11.
5
5R
, tan 1 (2)
2
B)
5R
1
, tan 1  
2
2
C)
5 X C , tan
1
D)
(2 )
1 
5 R, tan 1  
2
In the adjoining ac circuit the voltmeter whose reading will be zero at resonance is
A) V1
B) V2
V
C) V3
D) V4
4
V1
L
12.
The reading of ammeter in the circuit shown will be
A) 2A
X = 5
B) 2.4 A
C) Zero
D) 1.7 A
C
XL = 5
13.
15.
16.
V5
V3
C
R
A
V
110 V
R = 55
An ac source of angular frequency  is fed across a resistor r and a capacitor C in
series. The current registered is I. If now the frequency of source is changed to  /3 (but
maintaining the same voltage), the current in then circuit is found to be halved.
Calculate the ratio of reactance to resistance at the original frequency 
A)
14.
V2
3
5
B)
2
5
C)
1
5
D)
4
5
An LCR series circuit with a resistance of 100 ohm is connected to an ac source of 200
V (r.m.s.) and angular frequency 300 rad/s. When only the capacitor is removed, the
current lags behind the voltage by 60º. When only the inductor is removed the current
leads the voltage by 60º. The average power dissipated is
A) 50 W
B) 100 W
C) 200 W
D) 400 W
When 100 volts dc is supplied across a solenoid, a current of 1.0 amperes flows in it.
When 100 volts ac is applied across the same coil, the current drops to 0.5 ampere. If
the frequency of ac source is 50 Hz, then the impedance and inductance of the solenoid
are
A) 200  and 0.55 Henry
B) 100  and 0.86 Henry
C) 200  and 1.0 Henry
D) 100  and 0.93 Henry
In an LR-circuit, the inductive reactance is equal to the resistance R of the circuit.
An e.m.f. E  E0 cos(t) applied to the circuit. The power consumed in the circuit is
A)
E02
R
B)
E02
2R
C)
E 02
4R
D)
E02
8R
Page 2
AC Circuits
17.
Ranker
In figure below if ZL  Zc and reading of ammeter is 1 A. Find value of source voltage
V.
A) 80 Volt
C) 100 Volt
18.
B) 60 Volt
D) None of these
In the circuit given below, what will be the reading of the voltmeter?
V
A) 300 V
B) 900 V
100V
100V
C) 200 V
D) 400 V
200V, 100 Hz
19.
20.
21.
22.
In the circuit shown below, what will be
100 
the readings of the voltmeter and ammeter?
A) 800 V, 2A
B) 300 V, 2A
V
A
300 V
300 V
C) 220 V, 2.2 A
220 V, 50 Hz
D) 100 V, 2A
A bulb and a capacitor are connected in series to a source of alternating current. If its
frequency is increased, while keeping the voltage of the source constant, then
A) Bulb will give more intense light
B) Bulb will give less intense light
C) Bulb will give light of same intensity as before
D) Bulb will stop radiating light
An alternating e.m.f. of angular frequency  is applied across an inductance. The
instantaneous power developed in the circuit has an angular frequency


A)
B)
C) 
D) 2
4
2
The graph of alternating current versus time is shown in figure. The average value for
the positive half cycle, is
I
A) 0
B) 2I 0
3

C) 5I 0
8
D) I 0
2
23.
As shown in figure value of inductive reactance XL will be if source voltage is 100 volt
A) 40 
B) 30 
C) 50 
D) Can have any value
24.
In the series LCR circuit, the voltmeter and ammeter readings are
respectively:
A) V = 250V, I = 4 A
B) V = 150V, I = 2 A
C) V = 1000V, I = 5 A
D) V = 100V, I = 2 A
Page 3
AC Circuits
25.
26.
27.
28.
29.
30.
Ranker
Following figure shows an ac generator connected to a "black box" through a pair of
terminals. The box contains possible R, L, C or their combination, whose elements and
arrangements are not known to us. Measurements outside the box reveals that e = 75
sin  t volt, i = 1.5 sin ( t + 45o) amp then, the wrong statement is
A) There must be a capacitor in the box
B) There must be an inductor in the box
?
C) There must be a resistance in the box
D) The power factor is 0.707
A given LCR series circuit satisfies the condition for resonance with a given AC source.
If the angular frequency of the AC source is increased by 100% then in order to
establish resonance, without changing the value of inductance, the capacitance must be
A) Increased by 100%
B) Reduced by 50%
C) Increased by 75%
D) reduced by 75%
2 .5
 F Capacitor and 3000-ohm resistance are joined in series to an ac source of 200 volt

and 50 sec 1 frequency. The power factor of the circuit and the power dissipated in it
will respectively
A) 0.6, 0.06 W
B) 0.06, 0.6 W
C) 0.6, 4.8 W
D) 4.8, 0.6 W
The self inductance of a choke coil is 10 mH. When it is connected with a 10V dc
source, then the loss of power is 20 watt. When it is connected with 10 volt ac source
loss of power is 10 watt. The frequency of ac source will be
A) 50 Hz
B) 60 Hz
C) 80 Hz
D) 100 Hz
In an LCR circuit R  100 ohm. When capacitance C is removed, the current lags behind
the voltage by  / 3 . When inductance L is removed, the current leads the voltage by  / 3 .
The impedance of the circuit is
A) 50 ohm
B) 100 ohm
C) 200 ohm
D) 400 ohm
In the circuit shown in figure neglecting source resistance the voltmeter and ammeter
reading will respectively be
V
A) 0V, 3A
B) 150V, 3A
C) 150V, 6A
A
R = 30
XL = 25
XC = 25
240 V
D) 0V, 8A
Page 4
AC Circuits
Ranker
KEY
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
D
D
A
A
A
D
C
B
B
B
D
C
A
D
A
C
C
C
C
A
D
D
C
D
B
D
C
C
B
D
Page 5
AC Circuits
Ranker
SOLUTIONS
1.
D
V  120 sin 100 t cos 100 t  V  60 sin 200t
And   100Hz
Vmax  60 V
2.
D
Z  (R)2  (X L  X C )2 ;
R  10  , X L   L  2000  5  10
1
1

C
2000  50  10
XC 
Maximum current
Hence
3.
irms 
2
6
i0 
 1. 4 A
2
3
 10 
 10  i.e . Z  10 
V0
20

 2A
Z
10
and
Vrms  4  1 . 41 
5.64 V
A
Capacitance of wire
C  0 . 014  10 6  200  2 . 8  10 6 F  2 . 8  F
For impedance of the circuit to be minimum
L
X L  XC

2  L 
1
2  C
1
1

2
4  C 4(3.14)  (5  10 3 )2  2.8  10 6
2 2
 0 . 35  10 3 H  0 . 35 mH
4.
5.
6.
7.
A
Current leads in capacitor by 900
A
90+Tan-1(4/3).
D
C
iWL  irms sin 

8.
9.
  60
o

so p.f.

3
2
sin 
 cos   cos 60 o 
1
2
.
B
This is because, when frequency  is increased, the capacitive reactance
XC 
1
2  C
decreases and hence the current through the bulb increases
B
R
Z 
P
2
irms

240
 15 
16
V
100

 25 
i
4
Now
X L  Z 2  R 2  (25)2  (15)2  20
 2  L  20  L 
10.
3  2 sin 
20
1

Hz
2   50
5
B
X L  R, X C  R / 2
X  XC
 tan   L

R
   tan
1
R
2 1
R
2
R
(1 / 2 )
Page 6
AC Circuits
Ranker
Also Z 
11.
12.
13.
R 2  (X L  X C )2 
R2 
R2
5

R
4
2
D
At resonance net voltage across L and C is zero
C
Given X L  X C  5, this is the condition of resonance. So VL  VC , so net voltage across L
and C combination will be zero.
A
At angular frequency, the current in RC circuit is given by
Vrms
...... (i)
irms 
2
 1 
R2  

 C 
irms
Vrms


2
2


 1 

R2  
 
C


 3 
Also
Vrms
...... (ii)
9
R  2 2
 C
2
From equation (i) and (ii) we get
1
5

3R  2 2  C 
R
 C
3
5
2
14.
 X
i.e.
XL
XC

 tan 60
R
R
 X
L
R 
V
100

 100 
i
1
For ac,
Z 
V
100

 200 
i
0 .5

XL
XC

R
R

= 400 W
200  (100)2  4 2 (50)2 L2
L  0 . 55 H
P  Erms irms cos  


3 R
V 2 200  200

R
100
P
A
For dc,
 Z  R 2  (L)2
E0

2
E0

Z 2
Given
17.

C
o
Z  R 2  (X L  X C )2  R
So average power
16.
3
5
D
tan  
15.
XC

R

XL  R
E0

i0
2

2
R
Z
E2R
R
P 02
Z
2Z
so,
Z
2 R P 
E02
4R
C
If ZL  ZC current will be same,
So, VZ  VZ ;
L
C
Page 7
AC Circuits
Ranker
 VL  1 2 30 
1
 60 Volt

VR  80 1  80volt ; V  VL2  VR2   80 2   60  2  100 Volt
18.
V 2  VR2  (VL  VC )2
Since
60W, 10V
hence
V  VR  200 V
L
i
10 V
i
V L  VC
VL
100V, 50Hz
19.
V 2  VR2  (VL  VC )2
Also
20.
21.
i

VR  V  220 V
220
 2 .2 A
100
When a bulb and a capacitor are connected in series to an ac source, then on increasing
the frequency the current in the circuit is increased, because the impedance of the
circuit is decreased. So the bulb will give more intense light.
The instantaneous values of emf and current in inductive circuit are given by
E  E 0 sin  t
So,
and


i  i0 sin t   respectively.
2



Pinst  Ei  E0 sin t  i0 sint  
2




 E0 i0 sin t sin t cos  cos t sin 
2
2

 E 0 i0 sin t cos t

22.
1
E 0 i0 sin 2  t
2
(sin 2t  2 sint cos t)
Hence, angular frequency of instantaneous power is 2 .
D
T/2
 I.dt
Iavg 
0
T/2
 dt
I 
 0 
2
0
23.
C
VI
 20  10
2

 X L  X L  X C2
2
100  2  30   30  XL  40
2
2

2
2
2
 50    30    X L  10 
2
 X L  10   80  20 ; XL 10  40 ; XL  50
24.
D
V = VR + (VL – VC)
100 = VR + (200 – 200)
VR = 100V
Page 8
AC Circuits
i
25.
Ranker
100
 2
50
B
1
(for resonance)
c
1
. . . (i)
2 
LC
26. L =
for case (ii)
 2 2 
1
LC1
. . . (ii)
(ii) divide (i)  4 =
C
C
 C1  .
C1
4
The capacitance is to be reduced by 75%.
27.
2
Z
1
 1 
R2  
  (3000 )2 
2
 2C 
2
.5


 10  6 
 2  50 



R 3000
 0.6
Z 5 103
So power factor cos  
28.
With dc:
P
With ac:
P
V2
R

2
Vrms
R
Z2
R

and power
Z  (3000)2  (4000)2  5 103 
P  Vrmsirms cos 
Z2 

P
(200)2  0.6
 4.8W
5 103
(10 ) 2  5
 50  2
10
  80 Hz.
When C is removed circuit becomes RL circuit hence
tan
When L is removed circuit becomes RC circuit hence
30.
2
Vrms
cos
Z
(10 ) 2
 5 ;
20
Also Z 2  R 2  4 2 2 L2
 50  (5)2  4 (3 .14 )2  2 (10  10 3 )2 
29.

X

 L
3
R
tan
X

 C
3
R
.....(i)
.....(ii)
From equation (i) and (ii) we obtain XL = XC. This is the condition of resonance and in
resonance Z = R = 100
D
The voltage V L and VC are equal and opposite so voltmeter reading will be zero.
Also R  30 , X L  X C  25 
V
V
240
So i 


8A
R 2  (X L  X C )2
R
30
Page 9