JEE Main 2026 Apr 5 Shift 1, Physics Q19: LR, CR, LCR Circuits

JEE Main2026Apr 5, Shift 1Physics

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 $ω$ is

A
$\frac{6}{7}$
B
$\frac{3}{5}$
C
$\frac{7}{8}$
D
$\frac{3}{4}$

Solution

Let $X = 1/ (ω C)$ be the capacitive reactance at the original frequency. Then $I = \frac{V}{R ^{2} + X ^{2}}$ .

At $ω /4$ , the reactance becomes $4 X$ , and the current is $\frac{I}{3} = \frac{V}{R ^{2} + 16 X ^{2}}$ .

Dividing: $3 = \frac{R ^{2} + 16 X ^{2}}{R ^{2} + X ^{2}} \Rightarrow 9 (R^{2} + X^{2}) = R^{2} + 16 X^{2} \Rightarrow 8 R^{2} = 7 X^{2}$ .

Therefore $\frac{R}{X} = \frac{7}{8}$ .

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JEE Main 2026 Apr 5 Shift 1, Physics Q19: LR, CR, LCR Circuits

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