Q105 / 150 · MHT CET 2024 (03 May Shift 1) · Physics
An alternating voltage $v=200 \sqrt{2} \sin (100 t)$ is connected to a $1 \mu \mathrm{~F}$ capacitor through an a. c. ammeter. The reading of the ammeter shall be
- A10 mA
- Correct answerB20 mA
- C40 mA
- D80 mA
Solution
Alternating voltage: $\mathrm{e}=200 \sqrt{2} \sin (100 \mathrm{t})$ volt Comparing with $\mathrm{e}=\mathrm{e}_0 \sin \omega t$
$\omega=100 \mathrm{rad} / \mathrm{s}, \mathrm{e}_0=200 \sqrt{2}$
Capacitive reactance, $\begin{aligned} & \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 10^{-6}} \Omega=10^4 \Omega \\ & \mathrm{I}_0=\frac{\mathrm{e}_0}{\mathrm{X}_{\mathrm{C}}} \\ & \mathrm{I}_0=\frac{200 \sqrt{2}}{10^4} \\ & \mathrm{I}_0=2 \sqrt{2} \times 10^{-2} \mathrm{~A} \\ & \mathrm{I}_{\mathrm{rmi}}=\frac{\mathrm{I}_0}{\sqrt{2}}=\frac{2 \sqrt{2} \times 10^{-2}}{\sqrt{2}}=2 \times 10^{-2} \cdot \mathrm{~A}=20 \mathrm{~mA} \end{aligned}$
Capacitive reactance, $\begin{aligned} & \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 10^{-6}} \Omega=10^4 \Omega \\ & \mathrm{I}_0=\frac{\mathrm{e}_0}{\mathrm{X}_{\mathrm{C}}} \\ & \mathrm{I}_0=\frac{200 \sqrt{2}}{10^4} \\ & \mathrm{I}_0=2 \sqrt{2} \times 10^{-2} \mathrm{~A} \\ & \mathrm{I}_{\mathrm{rmi}}=\frac{\mathrm{I}_0}{\sqrt{2}}=\frac{2 \sqrt{2} \times 10^{-2}}{\sqrt{2}}=2 \times 10^{-2} \cdot \mathrm{~A}=20 \mathrm{~mA} \end{aligned}$
Hence, the correct answer is option: 'B'.