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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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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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