An AC source of angular frequency ω is connected across a resistor R and a capacitor C in series. The current is I. The source frequency is then reduced to ω /4 while the rms voltage is kept unchanged, and the current becomes I /3. If $X_{c}$ is the capacitive reactance at the original frequency ω, what is R / $X_{c}$ ?
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Question

An AC source of angular frequency ω is connected across a resistor R and a capacitor C in series. The current is I. The source frequency is then reduced to ω /4 while the rms voltage is kept unchanged, and the current becomes I /3. If $X_{c}$ is the capacitive reactance at the original frequency ω, what is R / $X_{c}$ ?

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G1 Physics · Accepted Answer

At the original frequency, I = V ÷ √[R²+X(C)²]. When the angular frequency becomes ω ÷ 4, capacitive reactance becomes four times larger: X'(C) = 4X(C). The new current is I ÷ 3 = V ÷ √[R²+16X(C)²]. Therefore, √[R²+16X(C)²] = 3√[R²+X(C)²]. Squaring: R²+16X(C)² = 9R²+9X(C)². Hence, 7X(C)² = 8R². Therefore, R ÷ X(C) = √(7 ÷ 8). Correct option: C.