Q 12-07-035JEE MainJEE Main 2026 (5 Apr, Shift 1)Hard
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
Answer: (C) $\sqrt{\dfrac{7}{8}}$
Let $X$ be the capacitive reactance at $\omega$; at $\omega/4$ it becomes $4X$. The current falls to $1/3$, so the impedance triples:
$$R^2 + 16X^2 = 9(R^2 + X^2) \;\Rightarrow\; 7X^2 = 8R^2 \;\Rightarrow\; \frac{R}{X} = \sqrt{\frac{7}{8}}$$
Solution by Sreeraj P, M.Sc Physics