An a.c. source of angular frequency $\omega$ 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 $\omega/4$, (keeping the voltage unchanged) the current is found to be $I/3$. The ratio of resistance to reactance at frequency $\omega$ is

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Alternating Current

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 current is found to be

I /3

. The ratio of resistance to reactance at frequency

ω

Answer: (C)

In a series

R C

circuit the current amplitude is

I = \frac{V _{0}}{R ^{2} + X _{C}^{2}}

with

X_{C} = \frac{1}{ω C}

. Reducing the frequency to

ω /4

raises the reactance to

4 X_{C}

, and the current falls to

I /3

\frac{I}{3} = \frac{V _{0}}{R ^{2} + 16 X _{C}^{2}}

. Dividing the two expressions,

3 = \frac{R ^{2} + 16 X _{C}^{2}}{R ^{2} + X _{C}^{2}}

. Squaring,

9 R^{2} + 9 X_{C}^{2} = R^{2} + 16 X_{C}^{2}

, so

8 R^{2} = 7 X_{C}^{2}

, giving

\frac{R}{X _{C}} = \frac{7}{8}

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option