Alternating Current

Practice Questions:

Question 31: moderate

In an LCR series circuit the potential differences across the resistance, capacitance and inductance are \(80\text{ V}\), \(40\text{ V}\) and \(100\text{ V}\) respectively. The power factor of this circuit is:

1. \(0.8\)

2. \(1.0\)

3. \(0.4\)

4. \(0.5\)

View Answer

Total voltage in series LCR is \(V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{80^2 + (100 - 40)^2} = 100\text{ V}\). Power factor is \(\cos\phi = \frac{V_R}{V} = \frac{80}{100} = 0.8\).

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Question 32: easy

The peak value of the alternating current given by \(I = 4 \sin\omega t + 4 \sin(\omega t + 2\pi/3)\) is:

1. \(4\sqrt{2}\)

2. \(2\sqrt{2}\)

3. \(8\)

4. \(4\)

View Answer

The two currents can be vectorially added with phase difference \(\phi = 2\pi/3 = 120^\circ\). Resultant amplitude is \(I_0 = \sqrt{4^2 + 4^2 + 2 \times 4 \times 4 \cos 120^\circ} = 4\).

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Question 33: easy

In LCR series circuit the values of \(L\), \(C\), \(R\) and \(E_0\) are \(0.01\text{ H}\), \(10^{-5}\text{ F}\), \(25\ \Omega\) and \(220\text{ V}\) respectively. The value of current flowing in the circuit at \(f = 0\) and \(f = \infty\) will respectively be:

1. \(8\text{ A}\) and \(0\text{ A}\)

2. \(0\text{ A}\) and \(0\text{ A}\)

3. \(8\text{ A}\) and \(8\text{ A}\)

4. \(0\text{ A}\) and \(8\text{ A}\)

View Answer

At \(f = 0\), the capacitor acts as an open circuit (\(X_C = \infty\)). At \(f = \infty\), the inductor acts as an open circuit (\(X_L = \infty\)). Therefore, current is zero in both cases.

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Question 34: easy

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:

1. Bulb will give more intense light

2. Bulb will give less intense light

3. Bulb will give light of same intensity as before

4. Bulb will stop radiating light

View Answer

As frequency increases, capacitive reactance \(X_C = \frac{1}{2\pi f C}\) decreases. This decreases the total impedance \(Z = \sqrt{R^2 + X_C^2}\), thereby increasing the current, causing the bulb to glow more brightly.

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Question 35: easy

Assertion (A): Average power consumed in an AC circuit is equal to average power consumed by resistors in the circuit.

Reason (R): Average power consumed by capacitor and inductor is zero.

1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

2. Both Assertion and Reason are true but Reason is not correct explanation of Assertion.

3. Assertion is true but Reason is false.

4. Assertion and Reason are false.

View Answer

Average power dissipated in an AC circuit is given by \(P_{avg} = V_{rms} I_{rms} \cos \phi = I_{rms}^2 R\). Perfect inductor and capacitor have phase angle \(90^\circ\), resulting in zero power consumption.

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Question 36: easy

The phase difference between voltage and current in series LCR circuit at half power frequency is:

1. \(0\)

2. \(\pi/4\)

3. \(\pi/2\)

4. \(\pi\)

View Answer

At half-power frequencies, the power factor is \(\cos\phi = \frac{1}{\sqrt{2}}\), which gives \(\phi = \pi/4\).

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Question 37: easy

For a series RLC circuit \(R = X_L = 2X_C\). The impedance of the circuit and phase difference between \(V\) and \(i\) will be:

1. \(\frac{\sqrt{5}R}{2}\text{, } \tan^{-1}(2)\)

2. \(\frac{\sqrt{5}R}{2}\text{, } \tan^{-1}(1/2)\)

3. \(\sqrt{5}X_C\text{, } \tan^{-1}(2)\)

4. \(\sqrt{5}R\text{, } \tan^{-1}(1/2)\)

View Answer

Impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (R - R/2)^2} = \frac{\sqrt{5}R}{2}\). Phase difference \(\tan\phi = \frac{X_L - X_C}{R} = 1/2 \Rightarrow \phi = \tan^{-1}(1/2)\).

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Question 38: easy

An L-C-R series circuit with \(100\ \Omega\) resistance is connected to an ac source of \(100\text{ V}\) and angular frequency \(300\text{ rad/s}\). When the capacitance is removed, the current lags behind the voltage by \(45^\circ\). When the inductance is removed, the current leads the voltage by \(45^\circ\). The current flowing in the original circuit will be:

1. \(1\text{ A}\)

2. \(1.5\text{ A}\)

3. \(2\text{ A}\)

4. \(3\text{ A}\)

View Answer

Removing C gives \(X_L = R\) and removing L gives \(X_C = R\). Hence, \(X_L = X_C\), indicating resonance. Impedance at resonance is \(Z = R = 100\ \Omega\). Current \(I = V/Z = 100/100 = 1\text{ A}\).

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Question 39: easy

In a circuit \(L\), \(C\) and \(R\) are connected in series with an alternating voltage source of frequency \(f\). If the current lags the voltage by \(45^\circ\), then the value of \(L\) will be

1. \(\frac{2\pi f C R + 1}{4\pi^2 f^2 C}\)

2. \(\frac{2\pi f C R - 1}{C}\)

3. \(\frac{2\pi f C R + 1}{2\pi f C}\)

4. \(\frac{2\pi f C R + 1}{C}\)

View Answer

Since current lags voltage, \(tan(45^\circ) = \frac{X_L - X_C}{R} = 1 ⇒ X_L - X_C = R\). Substituting \(X_L = 2\pi f L\) and \(X_C = \frac{1}{2\pi f C}\), we get \(2\pi f L = R + \frac{1}{2\pi f C} = \frac{2\pi f C R + 1}{2\pi f C}\). Thus, \(L = \frac{2\pi f C R + 1}{4\pi^2 f^2 C}\).

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Question 40: easy

Statement A: In a pure capacitive ac circuit, the phase difference between current and voltage is \(\pi/2\).

Statement B: In a pure inductive ac circuit, the phase difference between current and voltage is \(\pi\).

1. Both statement (A) and statement (B) are correct

2. Both statement (A) and statement (B) are incorrect

3. Statement (A) is correct and statement (B) is incorrect

4. Statement (A) is incorrect and statement (B) is correct

View Answer

In a purely capacitive circuit, current leads the voltage by \(\pi/2\). In a purely inductive circuit, current lags the voltage by \(\pi/2\). Hence, Statement A is correct while Statement B is incorrect.

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