Q137 / 200 · NEET 2024 · Physics

A $10 \mu \mathrm{F}$ capacitor is connected to a $210 \mathrm{~V}, 50 \mathrm{~Hz}$ source as shown in figure. The peak current in the circuit is nearly $(\pi=3.14)$ : Figure

  • Correct answer
    A
    0.93 A
  • B
    1.20 A
  • C
    $\quad 0.35 \mathrm{~A}$
  • D
    0.58 A

Solution

Capacitive Reactance $\begin{aligned} X_C & =\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \\ & =\frac{1000}{3.14} \end{aligned}$ $\begin{aligned} & V_{\mathrm{rms}}=210 \mathrm{~V} \\ & i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}=\frac{210}{X_C} \end{aligned}$ $\begin{aligned} \text { Peak current } & =\sqrt{2} i_{\mathrm{ms}}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932 \\ & \simeq 0.93 \mathrm{~A} \end{aligned}$

Hence, the correct answer is option: 'A'.

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