Alternating Current

Practice Questions:

Question 41: difficult

A series LCR circuit containing \(5.0\text{ H}\) inductor, \(80 \mu\text{F}\) capacitor and \(40 \Omega\) resistor is connected to \(230\text{ V}\) variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be

1. 42 rad/s and 58 rad/s

2. 25 rad/s and 75 rad/s

3. 50 rad/s and 25 rad/s

4. 46 rad/s and 54 rad/s

View Answer

The resonant angular frequency is \(\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} = 50\text{ rad/s}\) and the bandwidth is \(Delta \omega = \frac{R}{L} = \frac{40}{5} = 8\text{ rad/s}\). The half-power frequencies are \(omega_0 \pm \frac{\Delta \omega}{2} = 50 \pm 4\text{ rad/s}\), which gives \(46\text{ rad/s}\) and \(54\text{ rad/s}\).

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Question 42: easy

The value of capacitance \(C\) in an LCR series circuit with inductance \(L = 50\) mH, connected with a voltage source of angular frequency \(100\) rad/s so that current in the circuit becomes maximum, is:

1. 1 mF

2. 2 mF

3. 1 \(\mu\)F

4. 4 \(\mu\)F

View Answer

Current is maximum at resonance, where \(\omega = \frac{1}{\sqrt{LC}}\). Solving for \(C\): \(C = \frac{1}{\omega^2 L} = \frac{1}{(100)^2 \times 50 \times 10^{-3}} = 2 \times 10^{-3}\)⇒ C = 2 mF.

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Question 43: easy

Consider the following statements:

A. Average power supplied to an inductor by an ac source over one complete cycle is zero.

B. Capacitive reactance is inversely proportional to both frequency of ac source and capacitance.

C. Power factor of AC series LCR circuit is equal to the product of resistance and impedance of the circuit.

The correct statement(s) is/are

1. Both (B) and (C)

2. Both (A) and (B)

3. Both (A) and (C)

4. All (A), (B) and (C)

View Answer

Statement A is correct as average power of an inductor over a cycle is zero. Statement B is correct because \(X_C = \frac{1}{2pi f C}\). Statement C is incorrect because power factor is the ratio of resistance to impedance, \(cos \phi = R/Z\).

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Question 44: easy

The peak voltage of the AC source is equal to

1. The value of voltage applied to the circuit

2. The rms value of the source

3. \(\sqrt{2}\) times the rms value of the AC source

4. \(\frac{1}{\sqrt{2}}\) times the average value of the AC source

View Answer

For a sinusoidal AC voltage source, the relation between peak voltage \(V_0\) and root-mean-square voltage \(V_{\text{rms}}\) is \(V_0 = \sqrt{2}V_{\text{rms}}\).

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Question 45: easy

A series LCR circuit contains a capacitor of capacitance \(10^{-6}\text{ F}\) and an inductor of inductance \(10^{-4}\text{ H}\). The resonant frequency of the circuit will be

1. \[10^5\text{ Hz}\]

2. \[10\text{ Hz}\]

3. \[\frac{10^5}{2\pi}\text{ Hz}\]

4. \[\frac{10}{2\pi}\text{ Hz}\]

View Answer

The resonant frequency is \[f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10^{-4} \times 10^{-6}}} = \frac{10^5}{2\pi}\text{ Hz}\].

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Question 46: easy

Power consumed in A.C circuit is zero then the ac source could be connected to

1. Resistor only

2. Inductor only

3. Capacitor only

4. Both (2) and (3)

View Answer

The average power consumed in an AC circuit is given by \( P_{avg} = V_{rms} I_{rms} \cos \phi \). For a purely inductive or purely capacitive circuit, the phase difference \( \phi = 90^\circ \), which makes the power factor \( \cos \phi = 0 \), resulting in zero power consumption.

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Question 47: easy

Consider an a.c circuit having resistor of resistance \( R \) and an inductor having reactance \( X_L \). If voltage leads the current by an angle \( 60^\circ \), then

1. \( X_L = R \)

2. \( X_L = 2R \)

3. \( X_L = \sqrt{3}R \)

4. \( X_L = \sqrt{2}R \)

View Answer

In an inductive-resistive (RL) series circuit, the phase angle \( \phi \) is given by \( \tan \phi = \frac{X_L}{R} \). Since \( \phi = 60^\circ \), \( \tan 60^\circ = \frac{X_L}{R} \implies X_L = \sqrt{3}R \).

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Question 48: easy

A series LCR circuit contains a capacitor of capacitance \(10^{-6}\text{ F}\) and an inductor of inductance \(10^{-4}\text{ H}\). The resonant frequency of the circuit will be

1. \(10^5\text{ Hz}\)

2. \(10\text{ Hz}\)

3. \(\frac{10^5}{2\pi}\text{ Hz}\)

4. \(\frac{10}{2\pi}\text{ Hz}\)

View Answer

The resonant frequency of a series LCR circuit is given by \(f_0 = \frac{1}{2\pi\sqrt{LC}}\). Substituting values: \(f_0 = \frac{1}{2\pi\sqrt{10^{-4} \times 10^{-6}}} = \frac{10^5}{2\pi}\text{ Hz}\).

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Question 49: easy

Assertion (A): If an iron rod is inserted into a steady current carrying solenoid, the current in solenoid decreases.

Reason (R): Magnetic flux linked with solenoid increases.

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (3) (A) is true but (R) is false

4. (4) Both (A) and (R) are false

View Answer

Inserting an iron rod increases magnetic flux \( \Phi \). By Lenz's law, this induces an opposing \( \text{EMF} \), causing current \( \text{I} \) to decrease during insertion.

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Question 50: easy

Assertion (A): ac current flows through a bulb and a solenoid connected in series. If a soft iron core is inserted in the solenoid, the bulb glows much brighter.

Reason (R): The inductance of solenoid decreases on inserting soft iron core in it.

1. Both A & R are true and the (R) is the correct explanation of the (A)

2. Both A & R are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

Inserting a soft iron core increases solenoid inductance \(L\), thus increasing inductive reactance \(X_L = omega L\). This increases circuit impedance \(Z\), reducing current \(I = V/Z\) and making the bulb dimmer. Both Assertion (A) and Reason (R) are false.

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