An alternating voltage E = 100√(2) sin (50t) is connected to a 2 μ F capacitor through an a.c. ammeter. The ammeter reading will be (a) 10 mA (b) 5 mA (c) 20 mA (d) 30 mA

Answer: (a). Ammeter reads rms current: Irms = ErmsXC = Erms ω C. Erms = (100√(2))/(√(2)) = 100 V, ω = 50 rad/s. Irms = 100 × 50 × 2 × 10⁻⁶ = 10⁻² A = 10 mA So the answer is (a) 10 mA.

An alternating voltage E = 100√(2) sin (50t) is connected to a 2 μ F… | MHT CET 2025 (19 Apr, Shift 1)

Q22MHT CET 2025 · 19 APR · SHIFT 1Easy

Pure R, L and C Circuits

An alternating voltage $E = 100\sqrt{2} \sin(50t)$ is connected to a $2\ \mu F$ capacitor through an a.c. ammeter. The ammeter reading will be

Show answer and solution

Answer: (a)

Ammeter reads rms current: $I_{rms} = \frac{E_{rms}}{X_C} = E_{rms}\,\omega C$ .
$E_{rms} = \frac{100\sqrt{2}}{\sqrt{2}} = 100$ V, $\omega = 50$ rad/s.
$I_{rms} = 100 \times 50 \times 2 \times 10^{-6} = 10^{-2}$ A $= 10$ mA
So the answer is (a) 10 mA.

Report a mistake

Pure R, L and C Circuits question from MHT CET 2025 (19 Apr, Shift 1)

More questions on Pure R, L and C Circuits