JEE Main 2026 · 5 Apr · Shift 1 · Physics Q44

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PhysicsSingle correct+4 / -1Question 44 of 75

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

\sqrt{\dfrac67}

\sqrt{\dfrac35}

\sqrt{\dfrac78}

\sqrt{\dfrac34}

Answer and worked solution

Official answer: C

Impedances. At ω: $Z_1^2=R^2+X^2$ . At ω/4 the capacitive reactance becomes 4X: $Z_2^2=R^2+16X^2$ .

Current falls to I/3, so $Z_2=3Z_1$ :
$R^2+16X^2=9R^2+9X^2\ \Rightarrow\ 7X^2=8R^2.$

Ratio: $\dfrac RX=\sqrt{\dfrac78}$ .

(C)

Topics:Reactance and impedance

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JEE Main 2026 · 5 Apr · Shift 1, Physics question 44