Chapter-wise DPP of Selected Questions for NEET

Alternating Current

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1 The impedance of a circuit consists of 3 𝑜ℎ𝑚 resistance and 4 𝑜ℎ𝑚 reactance. The power factor of the
circuit is
a) 0.4 b) 0.6 c) 0.8 d) 1.0
2. An alternating 𝑒mf is applied across a parallel combination of a resistance 𝑅, capacitance 𝐶 and an
inductance 𝐿. If 𝐼𝑅 , 𝐼𝐿 , 𝐼𝐶 are the current through 𝑅, 𝐿 and 𝐶 respectively, then the diagram which correctly
represents the phase relationship among 𝐼𝑅 , 𝐼𝐿 , 𝐼𝐶 and source 𝑒mf 𝐸, is given by
a) IL b) IR c) IC d) IR
E E E E
IR IL IR IC
IC IC IL IL

3. In 𝐴𝐶 series circuit, the resistance, inductive reactance and capacitive reactance are 3Ω, 10Ω and 14Ω
respectively. The impedance of the circuit is
a) 5Ω b) 4Ω c) 7Ω d) 10Ω
4. The values of L, C and R for a circuit are 1H, 9F and 3Ω. What is the quality factor for the circuit at
resonance?
a) 1 b) 9 1 1
c) d)
9 3
5. The value of alternating emf 𝐸 in the given circuit will be

a) 220 V b) 140 V c) 100 V d) 20 V

6. At high frequency, the capacitor offers
a) More reactance b) Less reactance c) Zero reactance d) Infinite reactance
7. An 𝐿𝐶𝑅 series circuit with 𝑅 = 100Ω is connected to a 200 𝑉, 50 𝐻𝑧 a.c. source when only the capacitance
is removed, the current leads the voltage by 60°. When only the inductance is removed, the current leads
the voltage by 60°. The current in the circuit is
√3 2
a) 2𝐴 b) 1𝐴 c) 𝐴 d) 𝐴
2 √3
8. An ac source of variable frequency 𝑓 is connected to an 𝐿𝐶𝑅 series circuit. Which of the graphs in figure
represents the variation of current 𝐼 in the circuit with frequency 𝑓
a) I b) I c) I d) I

0 f 0 f 0 f 0 f

9. The current in series 𝐿𝐶𝑅 circuit will be maximum when 𝜔 is

a) As large as possible b) Equal o natural frequency of 𝐿𝐶𝑅 system
c) √𝐿𝐶 d) √1/𝐿𝐶
10. An alternating voltage (in volt) given by 𝑉 = 200√2 sin(100𝑡) is connected to1𝜇𝐹 capacitor through an AC
ammeter. The reading of the ammeter will be
a) 10 mA b) 20 m A c) 40 mA d) 80 mA
11. The instantaneous values of current and emf in an ac circuit are 𝐼 = 1/√2 sin 314 𝑡 𝑎𝑚𝑝 and 𝐸 =
√2 sin(314 𝑡 − 𝜋/6) 𝑉 respectively. The phase difference between 𝐸 and 𝐼 will be
a) −𝜋/6 𝑟𝑎𝑑 b) −𝜋/3 𝑟𝑎𝑑 c) 𝜋/6 𝑟𝑎𝑑 d) 𝜋/3 𝑟𝑎𝑑
12. The time taken by an alternating current of 50 Hz in reaching from zero to its maximum value will be
a) 0.5 s b) 0.005 s c) 0.05 s d) 5 s
13. The voltage of domestic ac is 220 𝑣𝑜𝑙𝑡. What does the represent
a) Mean voltage b) Peak voltage
c) Root mean voltage d) Root mean square voltage
14. The initial phase angle for 𝑖 = 10 sin 𝜔𝑡 + 8 cos 𝜔𝑡 is
4 5 4
a) tan−1 ( ) b) tan−1 ( ) c) sin−1 ( ) d) 90°
5 4 5
15. If 𝐴 and 𝐵 are identical bulbs, which bulb glows brighter
100 mH A

10 pF B

a) 𝐴 b) 𝐵 c) Both equally bright d) Cannot say

16. In an AC circuit the voltage applied is 𝐸 = 𝐸0 sin 𝜔𝑡. The resulting current in the circuit is 𝐼 =
𝜋
𝐼0 sin (𝜔𝑡 − 2 ). The power consumption in the circuit is given by
𝐸0 𝐼0 b) P = zero 𝐸0 𝐼0
a) 𝑃 = c) 𝑃 = d) 𝑃 = √2𝐸0 𝐼0
√2 2
17. The phase difference between the voltage and the current in an ac circuit is 𝜋/4. If the frequency is 50 𝐻𝑧
then this phase difference will be equivalent to a time of
a) 0.02 𝑠 b) 0.25 𝑠 c) 2.5 𝑚𝑠 d) 25 𝑚𝑠
18. In the non-resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant
frequency
a) Resistive b) Capacitive c) Inductive d) None of the above
19. The graphs given below depict the dependence of two reactive impedances 𝑋1 and 𝑋2 on the frequency of
the alternating e.m.f. applied individually to them. We can then say that
Impedance

Impedance

X1 X2

Frequency Frequency
a) 𝑋1 is an inductor and 𝑋2 is a capacitor b) 𝑋1 is a resistor and 𝑋2 is a capacitor
c) 𝑋1 is a capacitor and 𝑋2 is an inductor d) 𝑋1 is an inductor and 𝑋2 is a resistor
20. In an electrical circuit 𝑅, 𝐿, 𝐶 and an a.c. 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
a) 1/2 b) 1/√2 c) 1 d) √3/2

21. The phase difference between the alternating current and emf is 𝜋/2. Which of the following cannot be the
constituent of the circuit?
a) C alone b) R, L c) L, C d) L alone
22. In a series L – C – R circuit, resistance 𝑅 = 10 Ω and the impedance 𝑍 = 10 Ω. The phase difference
between the current and the voltage is
a) 0° b) 30° c) 45° d) 60°
23. The output current versus time curve of a rectifier is shown in the figure. The average value of output
current in this case is
Current

I0

Time

a) 0 𝐼0 2𝐼
b) c) 0 d) 𝐼0
2 𝜋
24. A transistor-oscillator using a resonant circuit with an inductor 𝐿 (of negligible resistance) and a capacitor
𝐶 in series produce oscillation of frequency 𝑓. If 𝐿 is doubled and 𝐶 is changed to 4𝐶, the frequency will be
a) 𝑓/2√2 b) 𝑓/2 c) 𝑓/4 d) 8𝑓
25. In a series 𝐿𝐶𝑅 circuit, operated with an ac of angular frequency 𝜔, the total impedance is
1/2
1 2
a) [𝑅2 + (𝐿𝜔 − 𝐶𝜔)2 ]1/2 b) [𝑅2 + (𝐿𝜔 − ) ]
𝐶𝜔
−1/2 1/2
1 2 1 2
c) [𝑅2 + (𝐿𝜔 − ) ] d) [(𝑅𝜔)2 + (𝐿𝜔 − ) ]
𝐶𝜔 𝐶𝜔
26. In pure inductive circuit, the curves between frequency 𝑓 and reciprocal of inductive reactance 1/𝑋𝐿 is

a) b) c) d)

f f f f

27. The maximum voltage in DC circuit is 282V. The effective voltage in AC circuit will be
a) 200 V b) 300 V c) 400 V d) 564 V
28. A resistor 30 Ω, inductor of reactance 10 Ω and capacitor of reactance 10 Ω are connected in series to an
AC voltage source 𝑒 = 300√2 sin(𝜔𝑡). The current in the circuit is
a) 10√2 A b) 10 A c) 30√11 A d) 30/√11 A
29. Q-factor can be increased by having a coil of
a) Large inductance, small ohmic resistance
b) Large inductance, large ohmic resistance
c) Small inductance, large ohmic resistance
d) Small inductance, small ohmic resistance
30. An alternating voltage is represented as 𝐸 = 20 sin 300𝑡. The average value of voltage over one cycle will
be
a) Zero 20
b) 10 𝑣𝑜𝑙𝑡 c) 20√2 𝑣𝑜𝑙𝑡 d) 𝑣𝑜𝑙𝑡
√2

31. The voltage across a pure inductor is represented by the following diagram. Which of the following
diagrams will represent the current
 V

a) i b) i c) i d) i

t t t t

32. The current which does not contribute to the power consumed in an AC circuit is called
a) non-ideal current b) wattles current
c) convectional current d) inductance current
33. In a purely resistive ac circuit, the current
a) Lags behind the 𝑒.m.f. in phase
b) Is in phase with the 𝑒.m.f.
c) Leads the 𝑒.m.f. in phase
d) Leads the 𝑒.m.f. in half the cycle and lags behind it in the other half
34. An alternating voltage is connected in series with a resistance 𝑅 and an inductance 𝐿. If the potential drop
across the resistance is 200 𝑉 and across the inductance is 150 𝑉, then the applied voltage is
a) 350 𝑉 b) 250 𝑉 c) 500 𝑉 d) 300 𝑉
35. An 𝐿𝐶𝑅 series ac circuit is at resonance with 10 𝑉 each across 𝐿, 𝐶 and 𝑅. If the resistance is halved, the
respective voltage across 𝐿, 𝐶 and 𝑅 are
a) 10 V, 10 V and 5 V b) 10 V, 10 V and 10 V c) 20 V, 20 V and 5 V d) 20 V, 20 V and 10 V
36. If 𝐿 and 𝑅 represent inductance and resistance respectively, then dimension of 𝐿/𝑅 will be
a) [𝑀𝐿0 𝑇 0 ] b) [𝑀0 𝐿0 𝑇 −1 ] c) [𝑀0 𝐿0 𝑇 −2 ] d) [𝑀0 𝐿𝑇 −2 ]
37. In the circuit shown in figure neglecting source resistance, the voltmeter and ammeter readings will be
respectively

a) 0 V, 3 A b) 150 V, 3 A c) 150 V, 6 A d) 0 V, 8 A
38. The resistance of a coil for dc is in ohms. In ac, the resistance
a) Will remain same b) Will increase c) Will decrease d) Will be zero
39. An 𝐿𝐶𝑅 series circuit is at resonance. Then
a) The phase difference between current and voltage is 90°
b) The phase difference between current and voltage is 45°
c) Its impedance is purely resistive
d) Its impedance is zero

40. In the adjoining ac circuit the voltmeter whose reading will be zero at resonance is
 a) 𝑉1 b) 𝑉2 c) 𝑉3 d) 𝑉4
41. In an ac circuit the reactance of a coil is √3 times its resistance, the phase difference between the, voltage
across the coil to the current through the coil will be
a) 𝜋/3 b) 𝜋/2 c) 𝜋/4 d) 𝜋/6
42. In a pure inductive circuit or In an ac circuit containing inductance only, the current
a) Leads the 𝑒.m.f. by 90° b) Lags behind the 𝑒.m.f. by 90°
Sometimes leads and sometimes lags behind the
c) d) Is in phase with the 𝑒.m.f.
𝑒.m.f.
43. In L – R circuit, resistance is 8 Ω and inductive reactance is 6 Ω , then impedance is
a) 2 Ω b) 14 Ω c) 4 Ω d) 10 Ω
44. From figure shown below a series L – C – R circuit connected to a variable frequency 200 V source. 𝐶 =
80 𝜇𝐹 and 𝑅 = 40 Ω. Then the source frequency which drive the circuit at resonance is

a) 25 Hz b)
25
Hz c) 50 Hz d)
50
Hz
𝜋 𝜋
45. An AC voltage source of variable angular frequency 𝜔 and fixed amplitude 𝑉0 is connected in series with a
capacitance C and an electric bulb of resistance R (inductance zero). When 𝜔 is increased
a) The bulb glows dimmer b) The bulb glows brighter
c) Total impedance of the circuit is unchanged d) Total impedance of the circuit increases
 : HINTS AND SOLUTIONS :
1 (b) 1
∴ 𝐿𝜔 = 𝐶𝜔 This is a resonance circuit
𝑍 = √𝑅 + 𝑋 = √4 + 3 = 5
2 2 2 2 𝐸𝑟𝑚𝑠
𝑅 3 𝑍 = 𝑅; 𝐼𝑟𝑚𝑠 = , 𝐸𝑟𝑚𝑠 = 200 𝑉
∴ cos 𝜙 = = = 0.6 𝑅
𝑍 5 200𝑉
∴ 𝐼𝑟𝑚𝑠 = = 2𝐴
2 (c) 100Ω
𝜋
𝐼𝐿 lags behind 𝐼𝑅 by a phase of 2 , while 𝐼𝐶 leads by 8 (d)
𝜋 As explained in solution (1) for frequency 0 −
a phase of 2
𝑓𝑟 , 𝑍 decreases hence (𝑖 = 𝑉/𝑍) increases and for
3 (a) frequency 𝑓𝑟 − ∞, 𝑍 increases hence 𝑖 decreases
Here, Resistance, 𝑅 = 3Ω 9 (d)
Inductive reactance, 𝑋𝐿 = 10Ω At resonant frequency current in series 𝐿𝐶𝑅
Capacitive reactance, 𝑋𝐶 = 14Ω circuit is maximum
The impedance of the series 𝐿𝐶𝑅 circuit is 10 (b)
𝑍 = √𝑅2 + (𝑋𝐶 + 𝑋𝐿 )2 = √(3)2 + (14 − 10)2
𝑍 = 5Ω 11 (a)
4 (c) Phase difference relative to the current
𝜔𝐿 1 1 𝜋 𝜋
𝑄= = × ×𝐿 𝜙 = (314𝑡 − ) − (314 𝑡) = −
𝑅 𝑅 √𝐿𝐶 6 6
1 𝐿
12 (b)
= 𝑅 √𝐶 An alternating current is one whose magnitude
1 1 1 changes continuously with time between zero and
= 3 × √9 = 9 a maximum value and whose direction reverses
5 (c) periodically. The relation between frequency (𝑓)
For series L – C – R circuit and time (𝑇) is.
𝑉 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2
= √(80)2 + (40 − 100)2
= 100 V
6 (b)
Capacitive reactance is given by
1
𝑋𝐶 = 𝜔 𝐶
Where C is capacitance and 𝜔 the angular
frequency (𝜔 = 2𝜋𝑓).
1
∴ 𝑋𝐶 =
2𝜋𝑓𝐶
1
⇒ 𝑋𝐶 ∝ 𝑓
1 1
Hence, when frequency 𝑓 increases capacitive 𝑇= = = 0.02 𝑠
𝑓 50
reactance decreases. As is clear from the figure time taken to reach the
7 (a) maximum value is
If the capacitance is removed, it is an 𝐿 − 𝑅 circuit 𝑇 0.02
= = 0.005 s
𝜙 = 60° 4 4
𝑋𝐿 13 (d)
tan 𝜙 = = tan 60° = √3
𝑅
If inductance is removed, it is a capacitative
circuit or 𝑅 − 𝐶 circuit. |𝜙| is the same
14 (a) 21 (c)
Current 𝑖 = 𝑖0 sin(𝜔𝑡 + 𝜙) (i) In a circuit having C alone, the voltage lags the
𝜋
𝑖𝑝 = 𝑖0 sin 𝜔𝑡 cos 𝜙 + 𝑖0 cos 𝜔𝑡 sin 𝜙 current by 2 .
Thus, 𝑖0 cos 𝜙 = 10 (ii) In a circuit containing R and L, the voltage
𝑖0 sin 𝜙 = 8 leads the current by .
𝜋
4 2
Hence, tan 𝜙 = 5 (iii) In L – C circuit, the phase difference between
15 (a) current and voltage can have any
∵ (𝑋𝐶 ) >> (𝑋𝐿 ) 𝜋
value between 0 to 2 depending on the
16 (b)
values of L and C.
For given circuit current is lagging the voltage by
(iv) In a circuit containing L alone, the voltage
𝜋/2, so circuit is purely inductive and there is no 𝜋
leads the current by 2 .
power consumption in the circuit. The work done
by battery is stored as magnetic energy in the 22 (a)
inductor. Impedance,
17 I 𝑍 = √𝑅2 + (𝑋𝐿 − 𝑋𝐶 )2
𝑇 (1/50) 𝜋 1
Time difference = × 𝜙 = × = 𝑠= ∴ 10 = √(102 + (𝑋𝐿 − 𝑋𝐶 )2
2𝜋 2𝜋 4 400
2.5𝑚-𝑠 ⇒ 100 = 100 + (𝑋𝐿 − 𝑋𝐶 )2
18 (b) ⇒ 𝑋𝐿 − 𝑋𝐶 = 0
In non resonant circuits …(i)
1 Let 𝜙 is the phase difference between current and
Impedance 𝑍 = 2
, with rise in
√ 12 +(𝜔𝐶− 1 ) voltage
𝑅 𝜔𝐿
𝑋𝐿 −𝑋𝐶
frequency 𝑍 decreases, 𝑖. 𝑒., current increases so tan 𝜙 = 𝑅
circuit behaves as capacitive circuit 0
∴ tan 𝜙 =
19 I 𝑅
1
We have 𝑋𝐶 = 𝐶×2𝜋𝑓 and 𝑋𝐿 = 𝐿 × 2𝜋𝑓 ⇒ 𝜙=0 [From Eq.(i)]
23 (c)
20 I 𝑇/2 𝑇/2
∫0 𝑖 𝑑𝑡 ∫0 𝐼0 sin(𝜔𝑡)𝑑𝑡
𝐼𝑎𝑣 = 𝑇/2
=
𝑋𝐶 𝜋 ∫0 𝑑𝑡 𝑇/2
= tan
𝑅 3 cos ( 2 ) cos 0°
𝜔𝑇
2𝐼0 − cos 𝜔𝑡 𝑇/2 2𝐼0
= [ ] = [− + ]
𝑇 𝜔 0 𝑇 𝜔 𝜔
2𝐼0 2𝐼 2𝐼
= [− cos 𝜋 + cos 0°] = 0 [1 + 1] = 0
𝜔𝑇 2𝜋 𝜋
24 (a)
𝜋 1
𝑋𝐶 = 𝑅 tan … (i) Frequency of 𝐿𝐶 oscillation = 2𝜋√𝐿𝐶
3
𝑋𝐿 𝜋 𝑓1 1 𝐿2 𝐶2 1/2
= tan ⇒ = √𝐿 𝐶 = ( )
𝑅 3 𝑓2 √𝐿1 𝐶1 2 2 𝐿1 𝐶1
2𝐿 × 4𝐶 1/2
=( ) = (8)1/2
𝐿×𝐶
𝑓 𝑓1 𝑓
∴ 𝑓1 = 2√2 ⇒ 𝑓2 = 2√2 or, 𝑓2 = 2√2 [∵ 𝑓1 = 𝑓]
2
25 (b)
𝜋
𝑋𝐿 = 𝑅 tan … (ii) 26 (c)
3 𝑋𝐿 = 2𝜋𝑓
Net impedance 𝑍 = √𝑅2 + (𝑋𝐿 − 𝑋𝐶 )2 = 𝑅 ⇒ 𝑋𝐿 ∝ 𝑓
𝑅
Power factor cos 𝜙 = 𝑍 = 1 1 1
⇒ ∝
𝑋𝐿 𝑓
1
𝑖. 𝑒., graph between 𝑋 and 𝑓 will be a hyperbola
𝐿

Page|7
27 (a) 33 (b)
Maximum voltage is AC circuit
𝑉0 = 282 𝑉 34 (b)
𝑉0 282 The applied voltage is given by 𝑉 = √𝑉𝑅2 + 𝑉𝐿2
𝑉= =
√2 √2 𝑉 = √(200) 2 + (150)2 = 250 𝑣𝑜𝑙𝑡
282 28200
𝑉= = 35 (d)
1.41 141
𝑉 = 200 𝑉
28 (b) 36 (b)
𝐿/𝑅 represents time constant of R-L circuit.
𝑒 = 300√2 sin 𝜔𝑡
Therefore, its dimensions are [𝑀0 𝐿0 𝑇1 ].
𝑒0 300√2
𝐼0 = = 37 (d)
𝑍 √(30) + (10 − 10)2
2
The voltage 𝑉𝐿 and 𝑉𝐶 are equal and opposite so,
{∵ 𝑍 = √𝑅2 + (𝑋𝐿 − 𝑋𝐶 )2 } voltmeter reading will be zero.
=
300√2
= 10√2 A Also, 𝑅 = 30 Ω, 𝑋𝐿 = 𝑋𝐶 = 25 Ω
30 𝑉
𝐼0 So, 𝑖=
∴ Current 𝐼 = = 10 A √𝑅 2+(𝑋𝐿 −𝑋𝐶 )2
√2 𝑉 240
29 (a) =𝑅= = 8A
30
1 𝐿 38 (b)
𝑄 factor is given by 𝑅 √𝐶
The coil has inductance 𝐿 besides the resistance 𝑅.
So, for large quality factor the inductance should Hence for ac it’s effective resistance √𝑅2 + 𝑋𝐿2 will
be large and resistance and capacitance must be be larger than it’s resistance 𝑅 for dc
small 39 (c)
30 (a) In series LCR, the impedance of the circuit is given
by
31 (d)
𝑍 = √𝑅2 + (𝑋𝐿 − 𝑋𝐶 )2
In purely inductive circuit voltage leads the
At resonance, 𝑋𝐿 = 𝑋𝐶
current by 90°
∴𝑍=𝑅
32 (b)
true power
At resonance, the phase difference between the
As, power factor = apparent power current and voltage is 0°. Current is maximum at
= cos 𝜙 resonance
=
𝑅 40 (d)
√𝑅 2 +(𝑋𝐿 −𝑋𝐶 )2
At resonance net voltage across 𝐿 and 𝐶 is zero
𝑅
∴ power factor= cos 𝜙 = 𝑍 41 (a)
In a non-inductive circuit, 𝑋𝐿 = 𝑋𝐶 𝑋𝐿 √3 𝑅
𝑅 𝑅 tan 𝜙 = = = √3 ⇒ 𝜙 = 60° = 𝜋/3
∴ Power factor = cos 𝜙 = = =1 𝑅 𝑅
√𝑅2 𝑅 42 (b)
∴ 𝜙 = 0°
This is the maximum value of power factor. In a 43 (d)
pure inductor or an ideal capacitor In series L – R circuit, impedance is given by
𝜙 = 90° 𝑍 = √𝑅2 + 𝑋𝐿2
∴ Power factor= cos 𝜙 = cos 90° = 0 Where R is the resistance and 𝑋𝐿 the inductive
Average power consumed in a pure inductor or reactance.
ideal capacitor Given, 𝑅 = 8Ω, 𝑋𝐿 = 6Ω
𝑃 = 𝐸𝑣 . 𝐼𝑣 cos 90° = zero
∴ 𝑍 = √(8)2 + (6)2
Therefore, current through pure L or pure C,
= √64 + 36
which consumes no power for its maintenance in
= √100 = 10 Ω
the circuit is called ideal current or wattles
current.

Page|8
44 (b) 45 (b)
𝑉𝑟𝑚𝑠
𝑍 = √𝑅2 + 𝑋𝐶2 ∶ 𝐼𝑟𝑚𝑠 = 2 𝑅
: 𝑃 = 𝐼𝑟𝑚𝑠
𝑍
1
Where 𝑋𝐶 = 𝜔𝐶
As 𝜔 is increased, 𝑋𝐶 will decrease or 𝑍 will
decrease. Hence 𝐼𝑟𝑚𝑠 𝑜𝑟 𝑃 will increase.
Therefore, bulb glows brighter.
Hence the correct option is (b).

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