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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 particle moving in a straight line with uniform acceleration is observed
- Which of the following, is a vector quantity?
- A step-down transformer has 50 turns on secondary and 1000 turns on primary winding.
- The velocity of radio-waves is 3×10⁵ km/s. The frequency corresponding to wavelength
- Surface tension of a solution is 30 x 10⁻² N/m. The radius of the soap bubble is 5 cm
Topics: Alternating Current (96)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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