Alternating Current

JEE Main level, single correct, medium

54 of 190

Question 54 of 190

An a.c. source of angular frequency ω\omega is connected across a resistor RR and a capacitor CC in series. The current is observed as II. Now the frequency of the source is changed to ω/4\omega/4, (keeping the voltage unchanged) the current is found to be I/3I/3. The ratio of resistance to reactance at frequency ω\omega is

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Answer: (C) 78\sqrt{\dfrac{7}{8}}

The current in the RC circuit at frequency ω\omega is given by

I=VR2+XC2I = \dfrac{V}{\sqrt{R^2 + X_C^2}}

where XC=1ωCX_C = \dfrac{1}{\omega C} is the capacitive reactance.

When the frequency is changed to ω/4\omega/4, the new reactance becomes XC′=1(ω/4)C=4XCX_C' = \dfrac{1}{(\omega/4)C} = 4X_C.

The new current is given as I/3I/3, so

I3=VR2+(4XC)2\dfrac{I}{3} = \dfrac{V}{\sqrt{R^2 + (4X_C)^2}}

Substituting the expression for II, we get

V3R2+XC2=VR2+16XC2\dfrac{V}{3\sqrt{R^2 + X_C^2}} = \dfrac{V}{\sqrt{R^2 + 16X_C^2}}

Squaring both sides, we obtain

19(R2+XC2)=1R2+16XC2\dfrac{1}{9(R^2 + X_C^2)} = \dfrac{1}{R^2 + 16X_C^2}

9R2+9XC2=R2+16XC29R^2 + 9X_C^2 = R^2 + 16X_C^2

8R2=7XC28R^2 = 7X_C^2

R2XC2=78\dfrac{R^2}{X_C^2} = \dfrac{7}{8}

RXC=78\dfrac{R}{X_C} = \sqrt{\dfrac{7}{8}}

Answer: 78\sqrt{\dfrac{7}{8}}

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