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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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Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A motor-cyclist drives a motor cycle in a vertical circle. His minimum velocity
- Three capacitors 3μF,6μF and 6μF are connected in series to a source of 120V.
- A particle is released from rest from a tower of height h, i.e,t₁ : t₂ : t₃ is
- A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 secs
- A beam of light of wavelength 600 nm from a distant source falls
Topics: Alternating Current (96)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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