Alternating Current | Physics JEE Main questions with Solutions

Q 1 :
In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to

a) pure inductor

b) pure capacitor

c) pure resistor

d) combination of an inductor and capacitor

Choose the correct answer from the options given below [2024]

b, c and d only
a, b and d only
a and b only
a, b and c only

1

2

3

4

(B) This is possible when phase difference is $\frac{π}{2}$ between current and voltage.

Q 2 :
A series LCR circuit is subjected to an ac signal of 200 V, 50 Hz. If the voltage across the inductor (L = 10 mH) is 31.4 V, then the current in this circuit is [2024]

68 A
63 A
10 A
10 mA

1

2

3

4

(C) Voltage across inductor $V_{L} = I X_{L}$

$31.4 = I [L ω]$

$31.4 = I [L (2 π f)]$

$31.4 = I [10 \times 10^{- 3} (2 \times 3.14) \times 50] \Rightarrow I = 10 A$

Q 3 :
Given below are two statements:

Statement I: In an LCR series circuit, current is maximum at resonance.

Statement II: Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to same voltage source.

In the light of the above statements, choose the correct from the options given below: [2024]

Statement I is true but Statement II is false
Statement I is false but Statement II is true
Both Statement I and Statement II are false
Both Statement I and Statement II are true

1

2

3

4

(D) Statement-I

$I_{m} = \frac{V_{m}}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}, at resonance X_{L} = X_{C}$

$Thus, I_{m} = \frac{V_{m}}{R}$

$∴$ Impendence is minimum therefore $I$ is maximum at resonance.

Statement-II

$I = (\frac{V}{R}) in purely resistive circuit .$

Q 4 :
A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters the same. The current amplitude at resonance will be now [2024]

halved
Zero
same
double

1

2

3

4

(D) At resonance: Z = R

So $I \propto \frac{1}{R}$

$R \to halved \Rightarrow I = 2 I, I becomes doubled .$

Q 5 :
In series LCR circuit, the capacitance is changed from C to 4C. To keep the resonance frequency unchanged, the new inductance should be [2024]

reduced to L/4
increased by 2L
reduced by L/3
increased to 4L

1

2

3

4

(A) $ω^{'} = ω$

$\frac{1}{\sqrt{L^{'} C^{'}}} = \frac{1}{\sqrt{L C}} ∴ L^{'} C^{'} = L C$

$L^{'} (4 C) = L C$

$L^{'} = \frac{L}{4}$

Q 6 :
For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF. If resistance of 200 $Ω$ and 100 mH inductor are being used in the given circuit.

The frequency of ac source is _________ $\times 10^{3}$ Hz. (given $π^{2} = 10$ ) [2024]

0%

(10) For maximum current, circuit must be in resonance.

At resonance, $X_{C} = X_{L}$

$ω L = \frac{1}{ω C} \Rightarrow ω^{2} = \frac{1}{100 \times 10^{- 3} \times 2.5 \times 10^{- 9}} = 4 \times 10^{9}$

${(2 π f)}^{2} = 4 \times 10^{9}$

$f^{2} = 10^{8} \Rightarrow f = 10^{4} Hz \Rightarrow f = 10 \times 10^{3} Hz$

Q 7 :
A series LCR circuit with $L = \frac{100}{π} m H$ , $C = \frac{10^{- 3}}{π} F$ and $R = 10 Ω$ , is connected across an ac source of 220V, 50 Hz supply. The power factor of the circuit would be _________ . [2024]

0%

(1) $X_{c} = \frac{1}{ω C} = \frac{π}{2 π \times 50 \times 10^{- 3}} = 10 Ω$

$X_{L} = ω L = 2 π \times 50 \times \frac{100}{π} \times 10^{- 3} = 10 Ω$

$∵ X_{C} = X_{L}, Hence, circuit is in resonance$

$∴ power factor = \frac{R}{Z} = \frac{R}{R} = 1$

Topic Question Set

Q 2 :
A series LCR circuit is subjected to an ac signal of 200 V, 50 Hz. If the voltage across the inductor (L = 10 mH) is 31.4 V, then the current in this circuit is [2024]

Q 4 :
A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters the same. The current amplitude at resonance will be now [2024]

Q 5 :
In series LCR circuit, the capacitance is changed from C to 4C. To keep the resonance frequency unchanged, the new inductance should be [2024]

Q 6 :
For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF. If resistance of 200 $Ω$ and 100 mH inductor are being used in the given circuit.

The frequency of ac source is _________ $\times 10^{3}$ Hz. (given $π^{2} = 10$ ) [2024]

0%

Q 7 :
A series LCR circuit with $L = \frac{100}{π} m H$ , $C = \frac{10^{- 3}}{π} F$ and $R = 10 Ω$ , is connected across an ac source of 220V, 50 Hz supply. The power factor of the circuit would be _________ . [2024]

0%

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 2 : A series LCR circuit is subjected to an ac signal of 200 V, 50 Hz. If the voltage across the inductor (L = 10 mH) is 31.4 V, then the current in this circuit is [2024]

Q 4 : A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters the same. The current amplitude at resonance will be now [2024]

Q 5 : In series LCR circuit, the capacitance is changed from C to 4C. To keep the resonance frequency unchanged, the new inductance should be [2024]

Q 6 : For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF. If resistance of 200 Ω and 100 mH inductor are being used in the given circuit. The frequency of ac source is _________ ×103 Hz. (given π2=10) [2024]

0%

Q 7 : A series LCR circuit with L=100πmH, C=10-3πF and R=10Ω, is connected across an ac source of 220V, 50 Hz supply. The power factor of the circuit would be _________ . [2024]

0%

Want Us to Email you About Special Offers & Updates?

Q 2 :
A series LCR circuit is subjected to an ac signal of 200 V, 50 Hz. If the voltage across the inductor (L = 10 mH) is 31.4 V, then the current in this circuit is [2024]

Q 4 :
A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters the same. The current amplitude at resonance will be now [2024]

Q 5 :
In series LCR circuit, the capacitance is changed from C to 4C. To keep the resonance frequency unchanged, the new inductance should be [2024]

Q 6 :
For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF. If resistance of 200 $Ω$ and 100 mH inductor are being used in the given circuit.

The frequency of ac source is _________ $\times 10^{3}$ Hz. (given $π^{2} = 10$ ) [2024]

Q 7 :
A series LCR circuit with $L = \frac{100}{π} m H$ , $C = \frac{10^{- 3}}{π} F$ and $R = 10 Ω$ , is connected across an ac source of 220V, 50 Hz supply. The power factor of the circuit would be _________ . [2024]