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Alternating Current MCQs for NEET — Physics Questions with Answers

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In an ac circuit, a resistance of R ohm is connected in series with an inductance L. If phase angle between voltage and current be 45°, the value of inductive reactance will be

$\frac{R}{4}$

$\frac{R}{2}$

Cannot be found with the given data

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Explanation

$\tan ϕ = \frac{X_{L}}{R}$

$\Rightarrow \tan 45^{o} = \frac{X_{L}}{R} = 1 \Rightarrow X_{L} = R$

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The phase difference between the current and voltage of LCR circuit in series combination at resonance is

π/2

–π

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Explanation

At resonance LCR series circuit behaves as pure resistive circuit. For resistive circuit $ϕ = 0^{o}$

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In a series resonant circuit, the ac voltage across resistance R, inductance L and capacitance C are 5 V, 10 V and 10 V respectively. The ac voltage applied to the circuit will be

20 V

10 V

5 V

25 V

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Explanation

$V = \sqrt{V_{R}^{2} + {(V_{L} - V_{C})}^{2}} = \sqrt{{(5)}^{2} + {(10 - 10)}^{2}} = 5 V o l t$

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In a series LCR circuit, resistance R = 10Ω and the impedance Z = 20Ω. The phase difference between the current and the voltage is

30°

45°

60°

90°

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Explanation

$\cos ϕ = \frac{R}{Z} = \frac{10}{20} = \frac{1}{2} \Rightarrow ϕ = 60^{o}$

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In an ac circuit the reactance of a coil is $\sqrt{3}$ times its resistance, the phase difference between the voltage across the coil to the current through the coil will be

π/3

π/2

π/4

π/6

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Explanation

$\tan ϕ = \frac{X_{L}}{R} = \frac{\sqrt{3} R}{R} = \sqrt{3}$ ⇒ $ϕ = 60^{o} = π / 3$

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The capacity of a pure capacitor is 1 farad. In dc circuits, its effective resistance will be

Zero

Infinite

1 ohm

1/2 ohm

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Explanation

$X_{C} = \frac{1}{2 π ν C} = \frac{1}{0} = \infty$

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The power factor of an ac circuit having resistance (R) and inductance (L) connected in series and an angular velocity ω is

$R / ω L$

$R / {(R^{2} + ω^{2} L^{2})}^{1 / 2}$

$ω L / R$

$R / {(R^{2} - ω^{2} L^{2})}^{1 / 2}$

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Explanation

$\cos ϕ = \frac{R}{Z} = \frac{R}{{(R^{2} + ω L^{2})}^{1 / 2}}$

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An inductor of inductance L and resistor of resistance R are joined in series and connected by a source of frequency ω. The power dissipated in the circuit is :

$\frac{(R^{2} + ω^{2} L^{2})}{V}$

$\frac{V^{2} R}{(R^{2} + ω^{2} L^{2})}$

$\frac{V}{(R^{2} + ω^{2} L^{2})}$

$\frac{\sqrt{R^{2} + ω^{2} L^{2}}}{V^{2}}$

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Explanation

$P = V i \cos ϕ = V (\frac{V}{Z}) (\frac{R}{Z}) = \frac{V^{2} R}{Z^{2}}$

$= \frac{V^{2} R}{(R^{2} + ω^{2} L^{2})}$

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In a LCR circuit the pd between the terminals of the inductance is 60 V, between the terminals of the capacitor is 30V and that between the terminals of resistance is 40V. the supply voltage will be equal to ……

50 V

70 V

130 V

10 V

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Explanation

$V = \sqrt{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}$

$= \sqrt{{(40)}^{2} + {(60 - 30)}^{2}} = 50 V$

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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

$\frac{1}{2 π f (2 π f L + R)}$

$\frac{1}{π f (2 π f L + R)}$

$\frac{1}{2 π f (2 π f L - R)}$

$\frac{1}{π f (2 π f L - R)}$

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Explanation

$\tan ϕ = \frac{X_{C} - X_{L}}{R}$

⇒ $\tan 45^{o} = \frac{\frac{1}{2 π f C} - 2 π f L}{R}$

$\Rightarrow C = \frac{1}{2 π f (2 π f L + R)}$

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