An a.c. source of angular frequency ω is connected across a resistor R and a capacitor C in series. The current is observed as I . Now the frequency of the source is changed to ω/4 , (keeping the voltage unchanged) the

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Quantrex Academy · Accepted Answer

Correct answer: C. (7)/(8)

The current in the RC circuit at frequency

\omega

is given by

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

where

X_C = \dfrac{1}{\omega C}

is the capacitive reactance.

When the frequency is changed to

\omega/4

, the new reactance becomes

X_C' = \dfrac{1}{(\omega/4)C} = 4X_C

.

The new current is given as

I/3

, so

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

Substituting the expression for

I

, we get

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

Squaring both sides, we obtain

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

9R^2 + 9X_C^2 = R^2 + 16X_C^2

8R^2 = 7X_C^2

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

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

Answer:

\sqrt{\dfrac{7}{8}}

An a.c. source of angular frequency ω is connected across a...

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