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) $\sqrt{\dfrac{7}{8}}$
Concept. $I=\dfrac{V}{\sqrt{R^2+X_C^2}}$ with $X_C\propto\dfrac1\omega$: at $\omega/4$, $X_C$ becomes $4X$.
Step 1. Equation. Current $\dfrac13$ means impedance $\times3$: $R^2+16X^2=9(R^2+X^2)$.
Step 2. Solve. $7X^2=8R^2\Rightarrow\dfrac RX=\sqrt{\dfrac78}$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
