An alternating voltage given as, V=100√2 sin100t V is applied to a capacitor

An alternating voltage given as, V=100√2 sin100t V is applied to a capacitor of 1 μF. The current reading of the ammeter will be equal to …….. mA

Options

(a) 20
(b) 10
(c) 40
(d) 80

Correct Answer:

10

Explanation:

Given: V = 100√2 sin 100t, ω = 100 and C = 1μF

The peak voltage , V₀ = 100√2

Therefore, rms voltage, V(rms0 = V₀ / √2 = 100√2 / √2 = 100 V

Current reading of ammeter, I = V(rms) / Xc

I = V(rms) / [1/ωC] = 100 V / [1/(100 × 1 × 10⁻⁶]

= 100 × 100 × 10⁻⁶ = 10⁻² A

= 10⁻² × (10 / 10) = 10 × 10⁻³ A = 10 mA

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