Q193 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Physics
An alternating voltage given as, $E = 100\sqrt{2}\sin(100t)$ volt is applied to a capacitor of $1\ \mu\text{F}$. The current reading of the ammeter will be equal to
- A$20\ \text{mA}$
- Correct answerB$10\ \text{mA}$
- C$40\ \text{mA}$
- D$80\ \text{mA}$
Solution
Given alternating voltage $E = 100\sqrt{2}\sin(100t)$ V.
Peak voltage $E_{0} = 100\sqrt{2}$ V.
RMS voltage $E_{\text{rms}} = \dfrac{E_{0}}{\sqrt{2}} = 100$ V.
Angular frequency $\omega = 100$ rad/s.
Capacitance $C = 1\ \mu\text{F} = 10^{-6}$ F.
Capacitive reactance $X_{C} = \dfrac{1}{\omega C} = \dfrac{1}{100 \times 10^{-6}} = 10^{4}\ \Omega$.
The ammeter reads the RMS value of the current.
$I_{\text{rms}} = \dfrac{E_{\text{rms}}}{X_{C}} = \dfrac{100}{10^{4}} = 10^{-2}$ A $= 10$ mA.
Answer: $10\ \text{mA}$
Peak voltage $E_{0} = 100\sqrt{2}$ V.
RMS voltage $E_{\text{rms}} = \dfrac{E_{0}}{\sqrt{2}} = 100$ V.
Angular frequency $\omega = 100$ rad/s.
Capacitance $C = 1\ \mu\text{F} = 10^{-6}$ F.
Capacitive reactance $X_{C} = \dfrac{1}{\omega C} = \dfrac{1}{100 \times 10^{-6}} = 10^{4}\ \Omega$.
The ammeter reads the RMS value of the current.
$I_{\text{rms}} = \dfrac{E_{\text{rms}}}{X_{C}} = \dfrac{100}{10^{4}} = 10^{-2}$ A $= 10$ mA.
Answer: $10\ \text{mA}$
Hence, the correct answer is option: 'B'.