An alternating voltage V = 200√(2) sin (100t) volt is connected to a 1 μ F capacitor through an a.c. ammeter. The reading of the ammeter will be (a) 10 mA (b) 20 mA (c) 40 mA (d) 80 mA

Answer: (b). XC = (1)/(ω C) = 1100 × 10⁻⁶ = 10⁴ Ω Vrms = (200√(2))/(√(2)) = 200 V Irms = (200)/(10⁴) = 20 mA So the answer is (b) 20 mA.

An alternating voltage V = 200√(2) sin (100t) volt is connected to a… | MHT CET 2026 (13 May, Shift 2)

Q10MHT CET 2026 · 13 MAY · SHIFT 2Easy

Pure R, L and C Circuits

An alternating voltage $V = 200\sqrt{2} \sin(100t)$ volt is connected to a $1\ \mu$ F capacitor through an a.c. ammeter. The reading of the ammeter will be

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Answer: (b)

$X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-6}} = 10^4\ \Omega$
$V_{rms} = \frac{200\sqrt{2}}{\sqrt{2}} = 200$ V
$I_{rms} = \frac{200}{10^4} = 20$ mA
So the answer is (b) 20 mA.

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Pure R, L and C Circuits question from MHT CET 2026 (13 May, Shift 2)

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