An a.c. voltage source V = V₀ sin ω t is connected across resistance ' R ' and capacitance ' C ' in series. It is given that R = (1)/(ω C) and the peak current is I₀. If the angular frequency of the voltage source is changed to ((ω)/(√(3))), then the new peak current in the circuit is (a) I₀/2 (b) (I₀)/(√(2)) (c) √(2)I₀ (d) √(3)I₀

Answer: (b). Initially XC = R, so Z = √(2)R and I₀ = (V₀)/(√(2)R). At (ω)/(√(3)), XC' = √(3)R, so Z' = √(R² + 3R²) = 2R. I₀' = V₀/2R = √(2)/2 I₀ = (I₀)/(√(2)) So the answer is (b) (I₀)/(√(2)).

An a.c. voltage source V = V₀ sin ω t is connected across resistance… | MHT CET 2024 (10 May, Shift 2)

Q26MHT CET 2024 · 10 MAY · SHIFT 2Medium

RL and RC Series Circuits

An a.c. voltage source $V = V_0 \sin \omega t$ is connected across resistance ' $R$ ' and capacitance ' $C$ ' in series. It is given that $R = \frac{1}{\omega C}$ and the peak current is $I_0$ . If the angular frequency of the voltage source is changed to $\left(\frac{\omega}{\sqrt{3}}\right)$ , then the new peak current in the circuit is

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Answer: (b)

Initially $X_C = R$ , so $Z = \sqrt{2}R$ and $I_0 = \frac{V_0}{\sqrt{2}R}$ .
At $\frac{\omega}{\sqrt{3}}$ , $X_C' = \sqrt{3}R$ , so $Z' = \sqrt{R^2 + 3R^2} = 2R$ .
$I_0' = \frac{V_0}{2R} = \frac{\sqrt{2}}{2} I_0 = \frac{I_0}{\sqrt{2}}$
So the answer is (b) $\frac{I_0}{\sqrt{2}}$ .

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RL and RC Series Circuits question from MHT CET 2024 (10 May, Shift 2)

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