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- Alternating Current (AC) is the current which varies in both magnitude as well as direction alternatively and periodically. i = i (^) 0 sinω t or i = i (^) 0 cos ω t where, i 0 = peak value or maximum value of AC.
- RMS Value of AC is defined as the value of steady current that would generate the same amount of heat in a given resistor as would be generated by the given AC current over a complete cycle.
- Average or Mean Value of AC is defined as the value of steady current which would send same amount of charge through a circuit that is sent by the AC in the in half-cycle. i i av =^ = i
π
- The instantaneous alternating emf is given by V = V (^) 0 sinω t or V = V (^) 0 cos ω t Also , V
V
rms =^ = 0 2
- or V rms = 707. % of V 0
and V
V
av =^ =
π
. or V rms = 63 7. % of V 0
- Power In a AC circuit, both emf and current change continuously w.r.t. time, so in circuit, we have to calculate average power in complete cycle ( 0 → T ). P (^) av = V (^) rms i rmscosφ where, cos φ = Power factor.
- In an AC Circuit Containing Resistance Only Instantaneous value E is given by E = E (^) 0 sinω t Then, voltage and current are in same phase i = i (^) 0 sinω t
- In an AC Circuit Containing Inductor Only Instantaneous value E is given by E = E (^) 0 sinω t
Then, (i) Inductive reactance, X (^) L = ω L = 2 π fL (ii) Voltage leads the current by phase π 2
If V = V (^) 0 sin ω t ,then i = i t −
sin ω π
(iii) Power factor, cos φ cos π = = 2
Thus, average power consumption, P (^) av = V (^) rms rms i cos φ = 0
- In an L - R Series AC Circuit
Impedance, Z R X
V
i
= + L =
2 2 rms rms For the phase angle, tan φ ω = =
X
R
L
R
L (^) , voltage leads
current by phase φ.
- In an AC Circuit Containing Capacitor Only Instantaneous value E is given by E = E (^) 0 sinω t Then, (i) Capacitive reactance, X C (^) C fC
ω 2 π (ii) Capacitor offers infinite reactance in DC circuit as f = 0. (iii) Voltage lags behind the current by phase π 2
If V = V (^) 0 sin ω t ,then i = i t +
sin ω π
(iv) Power factor (cos φ) is minimum and equal to zero. ∴ Average power consumption (during a complete cycle), P (^) av = V (^) rms rms i cos φ = 0
Alternating Current
C H A P T E R
A Quick Recapitulation of the Chapter
1. Which current do not change direction with time?
(a) DC current (b) AC current (c) Both (a) and (b) (d) Neither (a) nor (b)
2. The electric mains supply in our homes and offices is
a voltage that varies like a sine function with time.
Such a voltage is called ..... and the current driven by
it in a circuit is called the ......
(a) DC voltage, AC current (b) AC voltage, DC current (c) AC voltage, DC voltage (d) AC voltage, AC current
3. Potential difference between two points is called
(a) AC current (b) voltage (c) DC current (d) resistor
4. When the current changes continuously in magnitude
and periodically in direction, several times per
second, the current is known as the
(a) direct current (b) induced current (c) displacement current (d) alternating current
5. Consider a source which
produces sinusoidally varying
potential difference across its
terminals, this potential difference
called AC voltage, be given by the
expression
(a) Vm sin ω t (b) Vm cos ω t (c) 2 V (^) m cos ω t (d) 2 V (^) m sin ω t
- In an C - R Series AC Circuit
Impedance, Z
V
i
= rms= R + XC rms
2 2
For the phase angle, tan φ ω
X
R CR
C^1
- In an L - C Series AC Circuit Impedance, Z
V
i
= rms= X (^) L − XC rms Phase difference between voltage and current is π /2. Thus, power factor, cos φ = 0
- In an L - C - R Series AC Circuit (i) Impedance, Z R X X
V
L C i = 2 + ( − ) 2 = rms rms (ii) If X (^) L > XC , then V leads i by φ and if X (^) L < XC , then V lags behind i by φ. where, tan φ =
X X −
R
V V
V
L C L C R
- In Resonant L - C - R Series AC Circuit (i) X (^) L = XC (ii) Impedance, Z = Z (^) min= R (iii) The phase difference between V and i is 0°. (iv) Resonant angular frequency, ω (^0)
LC
(v) Average power consumption P av becomes maximum. (vi) Current becomes maximum and i
V
max (^) R = rms
- L - C Oscillations When the charged capacitor is connected with the inductor, current flows through the inductor and energy stored in the inductor in the form of magnetic field and capacitor discharges and vice-versa. In this way, energy oscillates between capacitor and inductor. The frequency of oscillation is ω (^0)
LC
- Quality Factor It indicates the sharpness of resonance in an L - C - R series AC circuit.
Quality factor = = = = =
V
V
V
V
L
R CR R
L
C
L R
C R
ω ω
0 0
Quality factor is also defined as
Q =
2 π Maximum nergy stored Energy dissipated / cycle
e
- A transformer is device used either to obtain a high AC voltage from a low voltage AC source or vice-versa. For an ideal transformer, e e
V
V
N
N
i i
s k p
s p
s p
p s
where, k is known as transformation ratio. For a step-up transformer, k > 1 but for a step-down transformer k < 1. The efficiency of a transformer is given by η = =
Output power Input power
V i V i
s s p p For an ideal transformer, η = 100% or 1. However, for practical transformer, η ≈ 85-90%.
Objective Questions Based on NCERT Text
AC Voltage Applied to a Resistor
Topic 1
ε (^) R
23. Alternating current cannot be measured by DC
ammeter, because
(a) AC cannot pass through DC ammeter (b) average value of current in complete cycle is zero (c) AC is virtual (d) AC changes its direction
24. In an AC circuit, I = 100 sin 200 π t. The time required
for the current to achieve its peak value will be
(a)
s (b)
s
(c)
s (d)
s
25. A generator produces a voltage that is given by
V = 240 sin 120 t , where t is in seconds. The frequency
and rms voltage are
(a) 60 Hz and 240 V (b) 19 Hz and 120 V (c) 19 Hz and 170 V (d) 754 Hz and 70 V
26. An alternating current is given by the equation
i = i 1 cos ω + t i 2 sin ω t .The rms current is given by
(a)
( i + i ) (b)
( i + i )^2
(c)
2 2 ( i + i^2 )^1 /^2 (d) 1 2 1
2 2 ( i + i^2 )1 2 /
27. In a circuit, the value of the alternating current is
measured by hot wire ammeter as 10 A. Its peak value
will be
(a) 10 A (b) 20 A (c) 14.14 A (d) 7.07 A
28. A resistance of 20 Ω is connected to a source of an
alternating potential, V = 220 sin ( 100 π t ). The time
taken by current to change from its peak value to rms
value is
(a) 0.2 s (b) 0.25 s (c) 25 × 10 −^3 s (d) 2.5 × 10 −^3 s
29. If an AC main supply is given to be 220 V. What
would be the average emf during a positive half-cycle?
(a) 198 V (b) 386 V (c) 256 V (d) None of these
30. If an alternating voltage is represented as
E = 141 sin ( 628 t ), then the rms value of the voltage
and the frequency are respectively
(a) 141 V, 628 Hz (b) 100 V, 50 Hz (c) 100 V, 100 Hz (d) 141 V, 100 Hz
31. The voltage of an AC source varies with time
according to equation V =100 sinπ t cos 100 π t. where
t is in seconds and V is in volts. Then
(a) the peak voltage of the source is 100 V (b) the peak voltage of the source is 50 V (c) the peak voltage of the source is 100/ 2 V (d) the frequency of the source is 100 Hz.
From the above figure, which one of the following
option is correct?
(a) V L di dt
− = 0 (b) L V di dt
(c) L V di dt
33. Equation di dt / = =
V
L
( V m / L ) sin ω t implies that the
equation for i t ( ), the current as a function of time,
must be such that
(a) its slope di dt / is a sinusoidally varying quantity with the same phase as the source voltage (b) an amplitude given by V (^) m / L (c) Both (a) and (b) (d) Neither (a) nor (b)
34. i
V
L
= − m t +
cos (ω ) constant, in the given equation,
the integration constant has the dimension of
(a) resistor (b) current (c) voltage (d) inductor
35. The integration constant in above question, is
(a) time-independent (b) time-dependent (c) may be time-independent (d) never time dependent
36. The inductive reactance is directly proportional to the
(a) inductance (b) frequency of the current (c) Both (a) and (b) (d) amplitude of current
37. Which of the following figure shows that the current
phasor I is π /2 behind the voltage phasor V?
ε L
AC Voltage Applied to an Inductor
Topic^2
im sin ω t 1
I
V
ω t 1
(a) V (^) m sin ω t (^1)
I
V
ω t 1
V (^) m sin ω t (^1)
im sin ( ω t 1 – /2)π
(b)
(c) Both (a) and (b) (d)Neither (a) nor (b)
38. In a purely inductive AC circuit, the current reaches
its maximum value later than the voltage by
(a) one-fourth of a period (b) half of a period (c) three by fourth of a period (d) complete a period
39. A pure inductor of 25.0 mH is connected to a source
of 220 V. Find the inductive reactance if the
frequency of the source is 50 Hz.
(a) 785 Ω (b) 6.50 Ω (c) 7.85 Ω (d) 8.75 Ω
40. Refer the above question, the rms current in the
circuit is
(a) 25 A (b) 16 A (c) 11 A (d) 28 A
41. Which of the following graphs represents the correct
variation of inductive reactance X L with angular
frequency ω?
42. In a purely inductive AC circuit, L = 30.0 mH and the
rms voltage is 150 V, frequency ν = 50 Hz. The
inductive reactance is
(a) 15.9 Ω (b) 9.42 Ω (c) 10 Ω (d) 8.85 Ω
43. An inductance of negligible resistance whose
reactance is 120 Ω at 200 Hz is connected to a 240 V,
60 Hz, power line. The current in the inductor is
(a) 6.66 A (b) 6.60 A (c) 5.45 A (d) 54.5 A
44. In a circuit containing an inductance of zero
resistance, the emf of the applied AC voltage leads
the current by
(a) 90° (b) 45° (c) 30° (d) 0°
45. In an AC circuit, the current lags behind the voltage
by π /2. The components of the circuit are
(a) R and L (b) L and C (c) R and C (d) only R
46. The current ( ) I in the inductance is varying with time
according to the plot shown in figure. Which one of
the following is the correct variation of voltage with
time in the coil?
47. A resistance of 300 Ω and an inductance of 1 / π henry
are connected in a series to an AC voltage of 20 V
and 200 Hz frequency. The phase angle between the
voltage and current is
(a) tan −^1 4 3/ (b) tan −^1 3 4/ (c) tan −^1 3 2/ (d) tan −^1 2 5/
48. Two inductors L 1 (inductance 1mH, internal
resistance 3 Ω) and L 2 (inductance 2 mH, internal
resistance 4 Ω), and a resistor R (resistance 12 Ω) are
all connected in parallel across a 5 V battery. The
circuit is switched on at time t = 0. The ratio of the
maximum to the minimum current ( I max / I min)
drawn from the battery is [JEE Advanced 2016]
(a) 2 (b) 4 (c) 6 (d) 8
49. In an L-R circuit connected to a battery, the rate at
which energy is stored in the inductor is plotted
against time during the growth of current in the
circuit. Which of the following figure best represents
the resulting curve?
XL
ω
XL
ω
XL
ω
XL
ω
(a)
(c)
(b)
(d)
t
V
T /2 T
(a) t
V
T /2 T
(b)
t
V
T /2 T
(c) X
Y
T /2 (^) T
(d)
t
V 0
(a)
0
dU dt
Time
(b)
0
dU dt
Time
(c)
0
dU dt
Time
(d)
0
dU dt
Time
t
T /
I
65. A resistor and a capacitor are connected in series
with an AC source. If the potential drop across the
capacitor is 5V and that across the resistor is 12 V,
then applied voltage is
(a) 13 V (b) 17 V (c) 5 V (d) 12 V
66. A resistor of 200 Ω and a capacitor of 15 μF are
connected in series to a 220 V, 50 Hz AC source. The
current in the circuit is
(a) 755 A (b) 7.55 mA (c) 0.755A (d) 0.755 mA
67. Consider the figure, the
resistor, inductor and capacitor
are in series, therefore
(a) the AC current in each element is same at any time (b) amplitude and phase are same in each element (c) Both (a) and (b) (d) Neither (a) nor (b)
68. Which one of the following phasor diagrams
correctly represents the relation between the
phasors V R , VL and VC of a series L-C-R circuit?
69. In the given circuit, the AC source has ω =100 rad/s.
Considering the inductor and capacitor to be ideal,
the correct choice(s) is (are) [IIT JEE 2012]
(a) The current through the circuit, I is 0.3 A (b) The current through the circuit, I is 0.3 2 A (c) The voltage across 100 Ω resistor = 10 2 V (d) The voltage across 50 Ω resistor = 10 V
70. Which of the following graph, is correct for a series
L-C-R circuit, where X C > XL?
71. The current in the series L-C-R circuit is
(a) i = i (^) m sin ( ω t +φ) (b) i
V
R X X
m t c L
sin ( ω φ)
(c) i = 2 i (^) m cos ( ω t +φ) (d) Both (a) and (b)
72. In an L-C-R series AC circuit, then voltage across each
of the components. L C , and R is 50 V. The voltage
across the C-R combination will be
(a) 50 V (b) 50 2 V (c) 100 V (d) zero
73. In a series L-C-R circuit, the frequency of 10 V AC
voltage source is adjusted in such a fashion that the
reactance of the inductor measures 15 Ω and that of the
capacitor 11 Ω. If R = 3 Ω, the potential difference
across the series combination of L and C will be
(a) 8 V (b) 10 V (c) 22 V (d) 52 V
74. In a circuit, L C , and R are connected in series with an
alternating voltage source of frequency f. The current
leads the voltage by 45°. The value of C is
(a)
2 π f ( 2 π fL + R )
(b)
π f ( 2 π fL + R )
(c)
2 π f ( 2 π fL − R )
(d)
π f ( 2 π fL − R )
AC Voltage Applied to a Series L-C-R Circuit
Topic 4
ε C L
R
VRm
VR
φ Vm ω t V (^) C +VL
V V
Cm
(a) (b)
(c) (d) VRm
V (^) C + VL
ω t Vm φ V
V V
Cm Lm
VR
VRm
VR
φ Vm ω t V (^) C + VL
VCm Lm
–V
VR
All of these
V
(a) (b) O^ ω t
V
i ω t 1 π 2 π
φ
O ω t
V i ω t 1
φ
2 π 3 π
(c) (d) O ω t
V i
φ
O ω t π
V i 4 π ω t 1
φ
ω t 1 2 π^2 π
100 μF (^100) Ω
0.5H (^50) Ω
20 V
I
75. In an L-C-R series AC circuit, the voltage across each
of the components, L C , and R is 50 V. The voltage
across the L-C combination will be
(a) 50 V (b) 50 2 V (c) 100 V (d) 0 V
76. In the given circuit, the readings of voltmeters V 1 and
V 2 are 300 V each. The readings of the voltmeter V 3
and ammeter A are respectively
(a) 100 V, 2.0 A (b) 150 V, 2.2 A (c) 220 V, 2.2 A (d) 220 V, 2.0 A
77. A sinusoidal voltage of peak value 300 V and an
angular frequency ω = 400 rads −^1 is applied to series
L - C - R circuit, in which R = 3 Ω, L = 20 mH and
C = 625 μF. The peak current in the circuit is
(a) 30 2 A (b) 60 A (c) 100 A (d) 60 2 A
78. For series L-C-R circuit, right statement is
(a) applied emf and potential difference across resistance are is same phase (b) applied emf and potential difference at inductor coil have phase difference of π / 2 (c) potential difference at capacitor and inductor have phase difference of π / 2 (d) Potential difference across resistance and capacitor have phase difference of π / 2.
79. In an L - C - R series circuit, the potential difference
between the terminals of the inductance is 60 V,
between the terminals of the capacitor is 30 V and
that across the resistance is 40 V. Then, supply
voltage will be equal to
(a) 50 V (b) 70 V (c) 130 V (d) 10 V
80. An AC source of angular frequency ω 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 ω /3 (but maintaining the same voltage),
the current in the circuit is found to be halved.
Calculate the ratio of reactance to resistance at the
original frequency ω.
(a)
(b)
(c)
(d)
81. The phenomenon of resonance is common among
systems that have a tendency
(a) to oscillate at a particular frequency (b) to get maximum amplitude (c) Both (a) and (b) (d) Neither (a) nor (b)
82. At resonant frequency, the current amplitude of an
R-L-C circuit is
(a) minimum (b) maximum (c) may be minimum (d) never maximum
Figure shows the variation of im with ω in a
(a) R-L-C circuit (b) R-L circuit (c) R-C circuit (d) None of these
84. In R-L-C series circuit with L =1.00 mH,
C = 1.00 nF two values of R are (i) R = 100 Ω and
(ii) R = 200 Ω. For the source applied with Vm =100 V.
Resonant frequency is
(a) 1 × 103 rad/s (b) 1 × 106 rad/s (c) 1.56 × 106 rad/s (d) 1.75 × 103 rad/s
85. Resonant circuits are used in
(a) the tuning mechanism of a radio (b) TV set (c) Both (a) and (b) (d) Neither (a) nor (b)
86. In tuning, we vary the capacitance of a capacitor in
the tuning circuit such that the resonant frequency of
the circuit becomes nearly equal to the frequency of
the radio signal received. When this happens,
the... A ... with the frequency of the signal of the
A V 1 V 2 V 3
L C R =100 Ω
220 V
Resonance
Topic 5
im
(A)
ω (rad/s)
0.5 1.0 1.5 2.
ω 0
(i) (ii)
100. Calculate the wavelength of the radiowaves radiated
out by a circuit containing 0.0 2 μF capacitor and 8 μH
inductance in series.
(a) 703.8 m (b) 460 m (c) 398 m (d) 753.8 m
101. In a series resonant L - C - R circuit, the voltage across
R is 100 V and R =1 kΩ with C = 2 μF. The resonant
freuency ω is 200 rads −^1. At resonance the voltage
across L is
(a) 2 5. × 10 − 2 V (b) 40 V (c) 250 V (d) 4 × 10 −^3 V
102. In an AC circuit, the average power dissipated depends
(a) on the voltage (b) current (c) cosine of the phase angle φ between them (d) All of the above
103. In an AC circuit, the instantaneous values of emf and
current are e = 200 sin ( 314 ) t V and
I = sin ( 314 t +π / ) 3 A.The average power consumed is
(a) 200 W (b) 100 W (c) 50 W (d) 25 W
104. The potential difference V and the current i flowing
through an instruments in an AC circuit of frequency
f are given by^ V = 5 cos ω t volts and^ i = 2 sin ω t
amperes (where, ω = 2 π f ).
The power dissipated in the instrument is
(a) zero (b) 10 W (c) 5 W (d) 2.5 W
105. In an AC circuit, V and I are given by
V = 100 sin ( 100 t ) V, i = t +
sin.
mA The
power dissipated in circuit is
(a) 10^4 W (b) 10 W (c) 2.5 W (d) 5 W
106. In an AC circuit, the current is given by
i = t −
sin
and the AC potential is
V = 200 sin ( 100 ) V. Then, the power consumption is
(a) 20 W (b) 40 W (c) 1000 W (d) 0 W
107. The average power supplied to an inductor over one
complete cycle is
(a) i Vm m / 2 (b) i Vm m (c) 3 i Vm m / 4 (d) zero
108. If a current I is given by I 0 sin ( ω t − π/ ) 2 flows in an
AC circuit across which an AC potential of
E = E 0 sin ω t has been applied, then the power
consumption P in the circuit will be
(a) P
E
I
(b) P = 2 E 0 I 0
(c) P
E
I
(d) P = 0
109. As in the case of inductor, the average power in
capacitor
(a) i V m m t 2
< sin ( 2 ω )> (b) i Vm m < sin ( ω t )>
(c) i Vm m < sin ( 2 ω t )> (d) 0
110. Power dissipated in an L - C - R series circuit connected
to an AC source of emf ε is
(a) ε
ω ω
2
2
R
R L
C
(b)
ε ω ω
2 2
2 1 R L C R
(c)
ε ω ω
2 2
2 1 R L C R
^
(d) ε
ω ω
2
2
2 1
R
R L
C
111. Which of the following components of a L-C-R
circuit, with AC supply, do not dissipatesel energy?
(a) L C , (b) R C , (c) L R , (d) L C R , ,
112. Which of the following components of a L-C-R
circuit with AC supply, dissipates energy?
(a) L (b) R (c) C (d) All of these
113. A coil of self-inductance L is connected in series with
a bulb B and an AC source. Brightness of the bulb
decreases when [NEET 2013]
(a) frequency of the AC source is decreased (b) number of turns in the coil is reduced (c) a capacitance of reactance X (^) C − XL is included in the same circuit (d) an iron rod is inserted in the coil
114. A lamp consumes only 50% of peak power in an AC
circuit. What is the phase difference between the
applied voltage and the circuit current?
(a) π 6
(b) π 3
(c) π 4
(d) π 2
Power in AC Circuit
Topic 6
115. A choke is preferred to a resistance for limiting
current in AC circuit, because
(a) choke is cheap (b) there is no wastage of power (c) choke is compact in size (d) choke is a good absorber of heat
116. A value of ω for which the current amplitude is 1 / 2
times its maximum value. At this value, the power
dissipated by the circuit becomes
(a) double (b) one-fourth (c) one-third (d) half
117. In an electrical circuit R L C , , and an AC voltage
source are all connected in series. When L is removed
from the circuit, the phase difference between the
voltage and the current in the circuit is π / 3. If
instead, C is removed from the circuit, the phase
difference is again π /3. The power factor of the
circuit is
[CBSE AIPMT 2012]
(a) 1/2 (b) 1/ 2 (c) 1 (d) 3/
118. In an AC circuit the power factor
(a) is zero when the circuit contains an ideal resistance only (b) is unity when the circuit contains an ideal resistance only (c) is unity when the circuit contains a capacitance only (d) is unity when the circuit contains an ideal inductance only
119. Power factor is maximum in a L-C-R circuit when
(a) X (^) L = XC (b) R = 0 (c) X (^) L = 0 (d) XC = 0
120. A coil of inductive reactance 31 Ω has a resistance of
8 Ω. It is placed in series with a condenser of
capacitive reactance 25 Ω. The combination is
connected to an AC source of 110 V. The power
factor of the circuit is
(a) 0.56 (b) 0.64 (c) 0.80 (d) 0.
121. A voltage of peak value 283 V and varying frequency
is applied to a series L-C-R combination in which
R = 3 Ω, L = 25 mH and^ C = 400 μF. The frequency
(in Hz) of the source at which maximum power is
dissipated in the above circuit is
(a) 51.5 Hz (b) 50.7 Hz (c) 51.1 Hz (d) 50.3 Hz
122. When a capacitor (initially charged) is connected
to a inductor, the change on the capacitor and the
current in the circuit exhibit the phenomenon of
(a) electrical oscillations (b) induction (c) power factor (d) All of these
123. d x dt^2 /^2 + ω 02 x = 0 , in the equation of SHM, ω 0 refers
to
(a) k / m (b) k m / (c) 2 k m / (d) 2 k m /
124. ω 0 = k m / , angular frequency in SHM, k refers to
(a) power constant (b) spring constant (c) quality factor (d) None of these
125. An inductor 20 mH, a capacitor 50 μF and a resistor
40 Ω are connected in series across a source of emf
V = 10 sin 340 t. The power loss in AC circuit is
[NEET 2016]
(a) 0.67 W (b) 0.76 W (c) 0.89 W (d) 0.51 W
126. A charged 30 μF capacitor is connected to a 27 mH
inductor. What is the angular frequency of free
oscillations of the circuit?
(a) 1.1 s (b) 1.1 × 10 3 s−^1 (c) 2 × 10 3 s−^1 (d) 2.5 × 103 s−^1
127. Suppose the initial charge on the capacitor in above
question is 6 mC. What is the total energy stored in the
circuit initially? What is the total energy at later time?
(a) 0.6 J, 0.6 J (b) 66.7 J, 67 J (c) 5.75 J, 0.92 J (d) 14.4 J, 10.5 J
128. A 10 μF capacitor is charged to 25 V of potential. The
battery is then disconnected and a pure 10 mH coil is
connected across the capacitor so that L-C oscillation
are set up. The maximum current in the coil is
(a) 0.25 A (b) 0.01A (c) 2.5 A (d) 1.6 A
129. A resonant AC circuit contains a capacitor of
capacitance 10 −^6 F and an inductor of 10 −^4 H. The
frequency of electrical oscillations will be
(a) 10^5 Hz (b) 10 Hz
(c)
5 π
Hz (d)
2 π
Hz
130. A charged 60 μF capacitor is connected to a 54 mH
inductor. What is the angular frequency of free
oscillations of the circuit?
(a) 5.5 s −^1 (b) 5.5 × 10 2 s−^1 (c) 1.2 s −^1 (d) 1.1 × 10 −^3 s−^1
145. Assertion When the capacitor is connected to an
AC source, it limits or regulates the current, but does
not completely prevent the flow of charge.
Reason The capacitor is alternately charged and
discharged as the current reverses each half-cycle.
146. Assertion Capacitor serves as a barrier for DC and
offers an easy path to AC.
Reason Capacitor reactance is inversely
proportional to frequency.
147. Assertion If X C > XL , φ is positive and the circuit
is predominantly capacitive. The current in the
circuit leads the source voltage.
Reason If X C < XL , φ is negative and the circuit is
predominantly inductive, the current in the circuit
lags the source voltage.
148. Assertion In a series R-L-C circuit, the voltages
across resistor, inductor and capacitor are 8V, 16V
and 10V, respectively. The resultant emf in the
circuit is 10 V.
Reason Resultant emf of the circuit is given by the
relation.
E = V R^2^ + ( V L − VC )^2
149. Assertion Resonance phenomenon is exhibited by a
circuit only if both L and C are present in the circuit.
Reason Voltage across L and C cancel each other
and the current amplitude is V m / R , the total source
voltage appearing across R causes resonance.
150. Assertion In series L-C-R circuit resonance can take
place.
Reason Resonance takes place if inductance and
capacitive reactances are equal and opposite.
151. Assertion The wire used for the windings of
transformer has some resistance.
Reason Energy is lost due to heat produced in the
wire ( I^2 R^ ).
II. Statement Based Questions Type I
n Directions (Q. Nos. 152-155) In the following
questions, a statement I is followed by a
corresponding statement II. Of the following
statements, choose the correct one.
(a) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I. (b) Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I. (c) Statement I is correct but Statement II is incorrect. (d) Statement I is incorrect but Statement II is incorrect.
152. Statement I The alternating current lags behind the
emf by a phase angle of π /2, when AC flows through
an inductor.
Statement II The inductive reactance increases as
the frequency of AC source decreases.
153. Statement I The opposition offered by AC circuits to
the flow of AC through it is defined as impedance. It’s
unit is ohm.
Statement II The opposition offered by inductor or
capacitor or both to the flow of AC through it is defind
as reactance.
154. Statement I A capacitor of suitable capacitance can
be used in an AC circuit in place of the choke coil.
Statement II A capacitor blocks DC and allows AC
only.
155. Statement I There is always some flux leakage;
i e. ., not all of the flux due to primary passes through
the secondary due to poor design of the core or the air
gaps in the core.
Statement II Flux leakage can be reduced by winding
the primary and secondary coils one over the other.
Statement Based Questions Type II
156. Consider the statements.
I. Most of the electrical devices we use require AC voltage.
II. Most of the electrical energy sold by power
companies is transmitted and distributed as alternating
current.
III. AC voltages can be easily and efficiently converted
from one voltage to the other by means of transformers.
(a) I is correct, II and III are incorrect (b) I, III are correct, II is incorrect (c) I, II are correct, III is incorrect (d) I,II and III are correct
Consider the figure,
I. The vertical components of phasors V and I represent the
sinusoidally varying quantities V and i.
II. The magnitudes of phasors V and I represent the
amplitudes or the peak values Vm and im of these
oscillating quantities.
im sin ω t 1
I
V
ω t 1
V I
(^0) ω t 1 π (^2) π ω t
V (^) m sin ω t (^1)
(a) (b)
III. The projection of voltage and current phasors on
vertical axis, i e.. , V m sin ω t and i m sin ω t , respectively
represent the value of voltage and current at that instant.
Which of the above statements is/are correct? Choose
the correct option.
(a) I and II (b) I and III (c) II and III (d) All of these
158. I. When a capacitor is connected to a voltage source in a
DC circuit, current will flow for the short time
required to charge the capacitor.
II. As charge accumulates on the capacitor plates, the
voltage across them increases, opposing the current.
III. A capacitor in a DC circuit will limit or oppose the
current as it charges.
IV. When the capacitor is fully charged, the current in the
circuit falls to zero.
Which of the above statements are incorrect? Choose
the correct option.
(a) I, II and III (b) II, III and IV (c) I and IV (d) None of these
III. Matching Types
159. Match the following.
Column I Column II A. VR 1. π/ 2 ahead of I B. VC 2. Parallel to I C. VL 3. π/ 2 behind I A B C A B C (a) 1 2 3 (b) 2 3 1 (c) 3 2 1 (d) 1 3 2
160. Match the following.
Column I Column II A. VRm 1. i (^) m XL B. VCm 2. i (^) mR C. VLm 3. i (^) m XC
A B C A B C (a) 1 2 3 (b) 3 2 1 (c) 1 3 2 (d) 2 3 1
161. Match the following.
Column I Column II A. Resistive circuit 1. No power is dissipation B. Purely inductive or capacitive circuit
- Maximum power dissipation because of X (^) C = XL
C. L-C-R series circuit 3. Power dissipated only in
the resistor D. Power dissipated at resonance in L-C-R circuit
- Maximum power dissipation
A B C D A B C D
(a) 1 2 4 3 (b) 4 1 3 2 (c) 3 1 4 2 (d) 2 1 3 4
162. Match the following Column I and Column II.
When oscillations on spring are compared with L-C
oscillations.
Column I Column II
A. Mass m 1. Reciprocal of capacitance i.e., 1/ C
B. Force constant k 2. Current, i = dq dt / C. Displacement x 3. Inductance L D. Velocity, v = d x / dt 4. Electromagnetic energy
E. Mechanical energy 5.^ U^ =^ q^ C^ + Li 1 2
1 2
(^2) / 2
F. E^ =^ k^ + mv
1 2
1 2
x^2^2
- Charge q
A B C D E F (a) 1 5 4 3 2 6 (b) 6 4 2 3 1 5 (c) 3 1 6 2 4 5 (d) 2 4 5 6 3 1
IV. Passage Based Questions
n Directions (Q. Nos. 163-166) Answer the following
questions based on given passage.
A sinusoidal voltage of peak value 283 V and
frequency 50 Hz is applied to a series L-C-R circuit in
which R = 3 Ω, L =25.48 mH and C = 796 μF.
163. The impedance of the circuit and the phase difference
between the voltage across the source and the current
will be
(a) 5 Ω, 53.1° (b) 3 Ω, 50.3° (c) 4 Ω, − 50.3° (d) 5Ω, − 53.1°
164. The power dissipated in the circuit and the power
factor will be
(a) 480 W, 6.7 (b) 13.35 W, 66. (c) 4800 W, 0.6 (d) 11.09 W, 0.
165. Let the frequency of the source can be varied. What is
the frequency of the source at which resonance
occurs?
(a) 13.35 Hz (b) 66.7 Hz (c) 35.4 Hz (d) 25.5 Hz
166. Calculate the impedance, the current and the power
dissipated at the resonant condition.
(a) 4 Ω, 13.35 A, 60 W (b) 2Ω, 65 A, 13 kW (c) 8 Ω, 66.7 A, 13.35 kW (d) 3 Ω, 66.7 A, 13.35 kW
NCERT
176. A 100 Ω resistor is connected to a 220 V, 50 Hz AC
supply, then the rms value of current in the circuit is
(a) 2.2 A (b) 4.2 A (c) 3.2 A (d) 2.4 A
177. The peak voltage of an AC supply is 300 V, then the
rms voltage will be
(a) 212.1 V (b) 312.1 V (c) 84.2 V (d) 85.2 V
178. A 44 mH inductor is connected to 220 V, 50 Hz AC
supply. Determine the rms value of the current in the
circuit.
(a) 20.4 A (b) 15.9 A (c) 21.4 A (d) 22.4 A
179. A 60 μF capacitor is connected to a 110 V, 60 Hz AC
supply. The rms value of the current in the circuit will
be
(a) 4.49 A (b) 2.29 A (c) 2.49 A (d) 3.49 A
180. Obtain the resonant frequency ω of a series L-C-R
circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What
is the Q -value of this circuit?
(a) 36 (b) 27 (c) 24 (d) 25
181. A charged 30 μF capacitor is connected to a 27 mH
inductor. What is the angular frequency of free
oscillations of the circuit?
(a) 1.1 × 104 rads−^1 (b) 1.1 × 103 rads−^1 (c) 1.1 × 102 rads−^1 (d) 1.1 × 10 rads−^1
182. A series L-C-R circuit with R = 20 Ω, L = 1.5 H and
C = 35 μF is connected to a variable frequency 200 V
AC supply. When the frequency of the supply equals
the natural frequency of the circuit, what is the
average power transferred to the circuit in one
complete cycle?
(a) 2 kW (b) 3 kW (c) 4 kW (d) 5 kW
183. A radio can tune over the frequency range of a portion
of MW broadcast band: (800 kHz to 1200 kHz). If its
L-C circuit has an effective inductance of 200 μH,
what must be the range of its variable capacitor?
(a) 49 to 79 (b) 88 to 198 (c) 100 to 200 (d) 110 to 200
184. A coil of inductance 0.50 H and resistance 100 Ω is
connected to a 240 V, 50 Hz AC supply. What is the
maximum current in the coil?
(a) 1.824 A (b) 2.824 A (c) 3.824 A (d) 4.824 A
185. A 100 μF capacitor in series with a 40 Ω resistance is
connected to a 110 V, 60 Hz supply. What is the
maximum current in the circuit?
(a) 3.00 A (b) 3.24 A (c) 4.24 A (d) 2.24 A
186. A power transmission line feeds input power at
2300 V to a step-down transformer with its primary
windings having 4000 turns. What should be the
number of turns in the secondary in order to get
output power at 230 V?
(a) 400 (b) 450 (c) 800 (d) 230
187. At a hydroelectric power plant, the water pressure
head is at a height of 300 m and the water flow
available is 100 m 3 /s. If the turbine generator
efficiency is 60%, the electric power available from
the plant will be
(a) 184.4 MW (b) 176.4 MW (c) 190.4 MW (d) 290.4 MW
NCERT Exemplar
188. If the rms current in a 50 Hz AC circuit is 5 A, the
value of the current 1/300 s after its value becomes
zero is
(a) 5 2 A (b) 5 3 / 2 A (c) 5 / 6 A (d) 5 / 2 A
189. An alternating current generator has an internal
resistance Rg and an internal reactance X g. It is used
to supply power to a passive load consisting of a
resistance Rg and a reactance X L. For maximum
power to be delivered from the generator to the load,
the value of X L is equal to
(a) zero (b) X (^) g (c) − X (^) g (d) Rg
190. When a voltage measuring device is connected to AC
mains, the meter shows the steady input voltage of
220 V. This means
(a) input voltage cannot be AC voltage, but a DC voltage (b) maximum input voltage is 220 V (c) the meter reads not v but < v^2 > and is calibrated to read < v^2 > (d) the pointer of the meter is stuck by some mechanical defect
191. To reduce the resonant frequency in an L-C-R series
circuit with a generator.
(a) The generator frequency should be reduced (b) Another capacitor should be added in parallel to the first (c) The iron core of the inductor should be removed (d) Dielectric in the capacitor should be removed
NCERT & NCERT Exemplar Questions
192. Which of the following combinations should be
selected for better tuning of an L-C-R circuit used
for communication?
(a) R = 20 Ω, L = 15. H, C = 35 μ F (b) R = 25 Ω, L = 2 5. H, C = 45 μF (c) R = 15 Ω, L = 3 5. H, C = 30 μF (d) R = 25 Ω, L = 15. H, C = 45 μF
193. An inductor of reactance 1Ω and a resistor of 2Ω are
connected in series to the terminals of a 6V (rms)
AC source. The power dissipated in the circuit is
(a) 8 W (b) 12 W (c) 14.4 W (d) 18 W
194. The output of a step-down transformer is measured
to be 24 V when connected to a 12 W light bulb.
The value of the peak current is
(a) 1 / 2 A (b) 2 A (c) 2 A (d) 2 2 A
195. As the frequency of an AC circuit increases, the
current first increases and then decreases. What
combination of circuit elements is most likely to
comprise the circuit?
(a) Inductor and capacitor (b) Resistor and inductor (c) Resistor and capacitor (d) Resistor, inductor and capacitor
196. In an alternating current circuit consisting of
elements in series, the current increases on increasing
the frequency of supply. Which of the following
elements are likely to constitute the circuit?
(a) Only resistor (b) Resistor and an inductor (c) Resistor and a capacitor (d) Only a capacitor
197. Electrical energy is transmitted over large distances at
high alternating voltages. Which of the following
statements is (are) correct?
(a) For a given power level, there is a lower current (b) Lower current implies less power loss (c) Transmission lines can be made thinner (d) It is easy to reduce the voltage at the receiving end using step-down transformers
198. For a L-C-R circuit, the power transferred from the
driving source to the driven oscillator is P = I^2 Z cos φ.
(a) Here, the power factor cos φ ≥ 0 , P ≥ 0 (b) The driving force can give no energy to the oscillator ( P = 0 in some cases) (c) The driving force cannot syphon out ( P < 0 the energy) out of oscillator (d) The driving force can take away energy out of the oscillator
199. When an AC voltage of 220 V is applied to the
capacitor C
(a) the maximum voltage between plates is 220 V (b) the current is in phase with the applied voltage (c) the charge on the plates is in phase with the applied voltage (d) power delivered to the capacitor is zero
200. The line that draws power supply to your house from
street has
(a) zero average current (b) 220 V average voltage (c) voltage and current out of phase by 90° (d) voltage and current possibly differing in phase φ such that φ < π/ 2
Answers
1. (a) 2. (d) 3. (b) 4. (d) 5. (a) 6. (b) 7. (c) 8. (d) 9. (d) 10. (c) 11. (b) 12. (b) 13. (c) 14. (c) 15. (a) 16. (b) 17. (b) 18. (a) 19. (c) 20. (a) 21. (d) 22. (c) 23. (b) 24 (d) 25. (c) 26. (c) 27. (c) 28. (d) 29. (a) 30. (c) 31. (a) 32. (a) 33. (c) 34. (b) 35. (a) 36. (c) 37. (b) 38. (a) 39. (c) 40. (d) 41. (b) 42. (b) 43. (a) 44. (a) 45. (a) 46. (d) 47. (a) 48. (c) 49. (c) 50. (b) 51. (a) 52. (b) 53. (d) 54. (d) 55. (a) 56. (c) 57. (d) 58. (c) 59. (c) 60. (d) 61. (b) 62. (a) 63. (c) 64. (b) 65. (a) 66. (c) 67. (c) 68. (a) 69. (a,c) 70. (d) 71. (d) 72. (b) 73. (a) 74. (a) 75. (d) 76. (c) 77. (b) 78. (d,a) 79. (a) 80. (a) 81. (a) 82. (b) 83. (a) 84. (a) 85. (c) 86. (c) 87. (b) 88. (b) 89. (a) 90. (b) 91. (a) 92. (c) 93. (c) 94. (b) 95. (a) 96. (c) 97. (c) 98. (a) 99. (b) 100. (d) 101. (c) 102. (d) 103. (c) 104. (a) 105. (c) 106. (d) 107. (d) 108. (d) 109. (d) 110. (a) 111. (b) 112. (b) 113. (d) 114. (b) 115. (b) 116. (d) 117. (c) 118. (b) 119. (a) 120. (c) 121. (d) 122. (a) 123. (b) 124. (b) 125. (d) 126. (b) 127. (a) 128. (a) 129. (c) 130. (b) 131. (b) 132. (a) 133. (a) 134. (b) 135. (b) 136. (c) 137. (c) 138. (b) 139. (b) 140. (a) 141. (d) 142. (a) 143. (a) 144. (a) 145. (a) 146. (a) 147. (b) 148. (a) 149. (a) 150. (a) 151. (b) 152. (c) 153. (b) 154. (b) 155. (b) 156. (d) 157. (d) 158. (d) 159. (b) 160. (a) 161. (b) 162. (c) 163. (d) 164. (c) 165. (c) 166. (d) 167. (c) 168. (b) 169. (d) 170. (a) 171. (b) 172. (a,b, c) 173. (a,b,c ,d) 174. (a,b, c) 175. (a,b, c) 176. (a) 177. (a) 178. (b) 179. (c) 180. (d)
Angular displacement i.e., (^) ω = 628 and (^2) π f = 628 ∴ f = ×
Hz
32. ( a ) Using the Kirchhoff’s loop rule, Σε ( ) t = 0 and since there is no resistor in the circuit. An AC source connected to an inductor V − L ( di dt / ) = 0 36. ( c ) Inductive reactance X (^) L = ω L = 2 π fL 39. ( c ) The inductive reactance, X (^) L = 2 πν L = 2 × 3.14 × 50 × 25 × 10 − 3 = 7.85 Ω 40. ( d ) The rms current in the circuit is
I = V X (^) L
= 220 V=
28 A
41. ( b ) Inductive reactance, X (^) L = ω L ⇒ X (^) L ∝ ω
Hence, inductive reactance increases linearly with angular frequency.
42. ( b ) The inductive reactance, X (^) L = 2 πν L = 2 × 314. × ( 50 s− 1 )× (30.0 × 10 − 3 H)= 9.42 Ω 43. ( a ) The reactance ( X (^) L )of the inductance at 200 Hz is 120 Ω.
As, X (^) L = ω L = 2 πν× L L
X L
2 × −
πν (^2) π 200
s 1
10 π
H
If X ′ L denotes the reactance of the same inductance at 60 Hz, X (^) ′ = L ω′ L = 2 πν′ L or X ′ = L (2 π × 60 − = 36 π
s )
3H
If I rms is the current that flows through the inductance when connected to 240 V and 60 Hz power line, then I rms = rms^ 6.66 A ′
ε = = X
V
L
46. ( d ) V = − L ( di dt / ), V is proportional to the slope of the i - t graph, which is constant and positive for the first half (0 to T /2) and negative and constant for the second half ( T / 2 to T ). Note : | V | = L ( di dt / )in this case. For first half V is –ve and for the second half it is + ve. 47. ( a ) Phase angle,
tan φ ω^ πν^ π π
=^ L = = ×^ × =
R
L
R
⇒ φ = tan −^1 4 3/
48. ( c ) When t = 0 due to large impedance of two inductor current will flow only in 12Ω. ∴ I min = 5 12./ After sometime current become is steady then R = 12 Ω will go out of circuit only r 1 and r 2 will be effective route of current flow. r eff = 2 Ω ⇒ I max = 5 2
⇒ I
I
max min
49. ( c ) Energy stored in an inductor L carrying current i is U = ( / )1 2 Li^2
Rate at which energy is stored = = ^
^
dU dt
L i di dt
Li di dt
At (^) t = 0 , i = 0 ⇒ dU dt / = 0 At t = ∞ , i = i 0 (constant) ∴ di dt
dU dt
50. ( b ) Current I across the capacitor is i (^) m sin (ω t + π/ ). 51. ( a ) The amplitude of the oscillating current is I (^) m = V (^) m / XC =ω CVm. 57. ( d ) The current reaches its maximum value earlier than the volage by one-fourth of a period. 58. ( c ) The capacitive reactance is X C (^) C
× −
πν 2 π (50 Hz) (15.0 10 6 F)
59. ( c ) The rms current is i = = =
V
XC
220 V
1.04 A
The peak current is i (^) m = 2 i = (1.41) (1.04 A) =1.47 A
60. ( d ) For the first circuit, i
V
Z
V
L
ω So, increase in ω will cause a decrease in i. For the second circuit, i V C
1/ ω Hence, increase in ω will cause an increase in i.
61. ( b ) Reading of ammeter = i = =
V
X
V C
C
rms rms 0 2
ω Q X C (^) C
ω
=
200 2 × 100 × 1 × 10 −
= 2 × 10 − 2 A = 20 mA
62. ( a ) If the frequency is doubled, the capacitive reactance is halved and the current is doubled. 63. ( c ) Capacitive reactance, X C C =^ = fC
ω 2
π ⇒ X C (^) f
With increase in frequency, XC decreases. Hence, option (c) represents the correct graph.
64. ( b ) X C (^) C
2 πν
×
2 (3.14) (60 s ) (60 10 F)
I rms rms^
110 V
ε XC Ω
= 2.49 A
65. ( a ) Let the applied voltage be V volt.
Here, VR = 12 V, VC = 5 V V = V (^) R^2^ + VC^2 = ( 12 )^2 +( ) 52 = 144 + 25 = 169 =13 V
66. ( c ) Impedance of the circuit Z = R^2 + XC^2 = R^2 + ( 2 πν C )−^2 = ( 200 Ω ) 2 + ( 2 × 3.14 × 50 × 15 × 10 −^6 F)− 2 = ( 200 Ω) 2 +( 212 Ω )^2 = 291.5 Ω Therefore, the current in the circuit is i = V / Z = =
220 V
0.755 A
69. ( a,c )
Circuit 1 X C (^) C
ω
Ω ⇒ Z 1 = ( 100 )^2 + ( 100 )^2 = 100 2 Ω
φ 1 1 1 1
cos− (^) = °
R
Z
In this circuit current leads the voltage. i V (^1) Z 1
= = = A ⇒ V (^) 100 100 i 1 100 1 Ω (^5 )
V
= 10 2 V
Circuit 2 X (^) L = ω L = ( 100 )(. ) 0 5 = 50 Ω Z 2 = ( 50 )^2 + ( 50 )^2 = 50 2 Ω
φ 2 1 2 2
cos− (^) = °
R
Z
In this circuit voltage leads the current.
i
V
2 Z 2
= = = A
V (^) 50 50 i 2 50 2 5
Ω =^ =^10
( ) = V
Further, i 1 and i 2 have a mutual phase difference of 90°.
∴ i = i 1^2 + i 22 = 1 + 50
A ≈ 0 3. A
70. ( d )
For (^) X (^) C > XL , peak of (^) i comes before peak of (^) V.
72. ( b ) V (^) CR = VC^2^ + VR^2 = ( 50 ) 2 +( 50 )^2
= 2500 + 2500 = 5000 = 10 50 = 50 2 V
73. ( a ) Given, R = 3 Ω , X (^) L = 15 Ω , XC = 11 Ω ⇒ V rms = 10 V
∴Current through the circuit
i V R X (^) L XC
rms (^2) ( ) 2
2 A
Since (^) L C , and (^) R are connected in series combination, then potential difference across R is V (^) R = i XR = 2 × 3 =6V Across L V , (^) L = i XL = 2 × 15 =30 V Across (^) C V , (^) C = i XC = 2 × 11 =22 V So, potential difference across series combination of L and C = V (^) L − VC = 30 − 22 =8 V
74. ( a ) tan φ = X^ − X R
C L
⇒ tan 45
π
π fC
fL
R
⇒ C
f f L R
2 π ( 2 π )
75. ( d ) Net voltage across L - C combination = V (^) L − VC =0 V. 76. ( c ) As V = ( V (^) L − VC ) 2 + VR^2 , 220 = ( 300 − 300 ) 2 + VR^2
or V i
V
R R
= 220 V, = R = 220 V=
2.2 A
77. ( b ) The impedance of the circuit is Z = R^2 + ( X (^) L − XC )^2 X (^) L = ω L = 400 × 20 × 10 − 3 = 8 H
⇒ X C (^) C
× ×
ω
F
⇒ Z = ( ) 3 2 + ( 8 − 4 )^2 = 5 ⇒ i E Z
60 A
79. ( a ) In L-C-R series circuit V = V (^) R^2^ + ( V (^) L − VC )^2 = ( 40 ) 2 + ( 60 − 30 )^2
= 1600 + 900 = 2500 =50 V
80. ( a ) At angular frequency ω, the current in R - C circuit is given by I
V
R
C
rms = rms (^2) +
ω
…(i)
Also,
I V
R
C
V
R
C
rms rms rms (^2 ) 3
2
(^2 ) 2 2
ω / ω
…(ii)
From Eqs. (i) and (ii), we get
3
R = ω (^2) C 2 ⇒
ω C 5 R
X
R
C = 3
84. ( a ) ω 0
= = ×
LC
1.00 10 3 rad / s.
86. ( c ) At resonance, current in the circuit is maximum. 87. ( b ) Bandwidth of the resonant R-L-C circuit is ∆ω = R 2 L
88. ( b ) The quantity (ω 0 / 2 ∆ ω)is regarded as measure of the sharpness of resonance. The smaller the ∆ω, the sharper is the resonance.
V I (^) φ ω t 1 ω t 1 π^2 π ω t
φ
O
V^ I
(a) (b)
100 μF (^100) Ω
0.5H (^50) Ω
I
I 1
I 1
Z 1 Z 2
