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Alternating Current MCQs for NEET — Physics Questions with Answers

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In an LR-circuit, the inductive reactance is equal to the resistance R of the circuit. An e.m.f. $E = E_{0} \cos (ω t)$ applied to the circuit. The power consumed in the circuit is:

$\frac{E_{0}^{2}}{R}$

$\frac{E_{0}^{2}}{2 R}$

$\frac{E_{0}^{2}}{4 R}$

$\frac{E_{0}^{2}}{8 R}$

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Explanation

$P = E_{r m s} i_{r m s} \cos ϕ = \frac{E_{0}}{\sqrt{2}} \times \frac{i_{0}}{\sqrt{2}} \times \frac{R}{Z}$

⇒ $\frac{E_{0}}{\sqrt{2}} \times \frac{E_{0}}{Z \sqrt{2}} \times \frac{R}{Z} \Rightarrow P = \frac{E_{0}^{2} R}{2 Z^{2}}$

Given $X_{L} = R$ so, $Z = \sqrt{2} R \Rightarrow P = \frac{E_{0}^{2}}{4 R}$

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One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has a self-inductance of value: (f = 50 Hz)

0.052 H

2.42 H

16.2 mH

1.62 mH

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An ac source of angular frequency ω is fed across a resistor r and a capacitor C in series. The current registered is I. If now the frequency of the source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency ω.

$\sqrt{\frac{3}{5}}$

$\sqrt{\frac{2}{5}}$

$\sqrt{\frac{1}{5}}$

$\sqrt{\frac{4}{5}}$

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Explanation

At angular frequency ω, the current in RC circuit is given by

$i_{r m s} = \frac{V_{r m s}}{\sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}}$ ......(i)

Also $\frac{i_{r m s}}{2} = \frac{V_{r m s}}{\sqrt{R^{2} + {(\frac{1}{\frac{ω}{3} C})}^{2}}} = \frac{V_{r m s}}{\sqrt{R^{2} + \frac{9}{ω^{2} C^{2}}}}$ ......(ii)

From equation (i) and (ii) we get

$3 R^{2} = \frac{5}{ω^{2} C^{2}} \Rightarrow \frac{\frac{1}{ω C}}{R} = \sqrt{\frac{3}{5}}$

⇒ $\frac{X_{C}}{R} = \sqrt{\frac{3}{5}}$

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

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

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

$\sqrt{5} X_{C}, \tan^{- 1} (2)$

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

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Explanation

$X_{L} = R, X_{C} = R / 2$

$∴ \tan ϕ = \frac{X_{L} - X_{C}}{R} = \frac{R - \frac{R}{2}}{R} = \frac{1}{2}$

$\Rightarrow ϕ = \tan^{- 1} (1 / 2)$

Also $Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{R^{2} + \frac{R^{2}}{4}} = \frac{\sqrt{5}}{2} R$

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A filament bulb (500 W,100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is

230 $Ω$

46 $Ω$

26 $Ω$

13 $Ω$

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Explanation

then $P_{r a t e d} = \frac{V_{r a t e d}^{2}}{R}$

$∴$ Current in the bulb, $i = \frac{P}{V}$

$i = \frac{500}{100} = 5 A$

$∴$ Resistance of bulb, $R_{b} = \frac{100 \times 100}{500} = 20 Ω$

$∵$ Resistance R is connected in series.

$∴$ Current $i = \frac{E}{R_{n e t}} = \frac{230}{R + R_{0}}$

$\Rightarrow R + 20 = \frac{230}{5} = 46 ∴ R = 26 Ω$

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Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication ?

$R = 20 Ω, L = 1.5 H, C = 35 μ F$

$R = 25 Ω, L = 2.5 H, C = 45 μ F$

$R = 15 Ω, L = 3.5 H, C = 30 μ F$

$R = 25 Ω, L = 1.5 H, C = 45 μ F$

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Explanation

Now, quality factor is given by Q= $\frac{1}{R} \sqrt{\frac{L}{C}}$

From the given options highest value of Q is associated with $R = 15 Ω$ , L=3.5H and C=30 $μ F$

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The potential differences across the resistance, capacitance and inductance are 80V, 40Vand 100V respectively in an L-C-R circuit. The power factor of this circuit is

0.4

0.5

0.8

1.0

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Explanation

$= \cos ϕ = \frac{R}{Z} = \frac{I R}{I Z} = \frac{80}{\sqrt{{(X_{L} - X_{C})}^{2} + R^{2}}}$

= $\frac{80 I}{\sqrt{{(l X_{L} - l X_{C})}^{2} + {(l R)}^{2}}} = \frac{80}{\sqrt{{(100 - 40)}^{2} + {(80)}^{2}}} = \frac{80}{\sqrt{{(60)}^{2} + {(80)}^{2}}} = \frac{80}{100} = 0.8$

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A 100 $Ω$ resistance and a capacitor of 100 $Ω$ reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is

2.2 A

11A

4.4A

11 $\sqrt{2}$ A

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Explanation

(a) The impedance of the R-C circuit, Z= $\sqrt{R^{2} + X_{c}^{2}}$

where, R= 100 $Ω$ and $X_{C}$ =100 $Ω$

$\Rightarrow Z = \sqrt{{(100)}^{2} + {(100)}^{2}} = 100 \sqrt{2} Ω$

The peak value of the current,

$l_{m a x} = \frac{V_{m a x}}{Z} = \frac{220 \sqrt{2}}{100 \sqrt{2}} = 2.2 A$

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An inductor 20 mH, a capacitor 50μF, and a resistor 40Ω are connected in series across a source of emf V=10sin340t. The power loss in the AC circuit is:

0.67 W

0.78W

0.89 W

0.46 W

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Explanation

(d)
$G i v e n : L = 20 m H, C = 50 μ F a n d V = 10 \sin 340 t P o w e r l o s s i n A C c i r c u i t, P_{a v} = {I_{v}}^{2} R = {(\frac{E_{v}}{Z})}^{2} R P_{a v} = {(\frac{\frac{10}{\sqrt{2}}}{\sqrt{{(40)}^{2} + {(340 \times 20 \times 10^{- 3} - \frac{1}{340 \times 50 \times 10^{- 6}})}^{2}}})}^{2} \times 40 P_{a v} = 0.46 W$

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A coil of self-inductance L is connected in series with a bulb B and an AC source. The brightness of the bulb decreases when

frequency of the AC source is decreased

the number of turns in the coil is reduced

a capacitance of reactance X_c=X_L is included in the same circuit

an iron rod is inserted in the coil

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Explanation

(d) As Z=√R²+X²L=√R²+(2πfvL)²

As I=V/Z, P=I²R

i.e., V↑,L↑=>Z↑,I↓ and P↓

$A s w e k n o w; Z = \sqrt{R^{2} + {X_{L}}^{2}} = = \sqrt{R^{2} + {(2 πfVL)}^{2}} A s V a n d L i n c r e a s e s, Z a l s o i n c r e a s e s . \Rightarrow I = \frac{V}{Z} d e c r e a s e s a n d P = I^{2} R a l s o d e c r e a s e s .$

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