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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
- An electric dipole placed in a non-uniform electric field experiences
- A particle moves along the x-axis from x=0 to x=5 m under the influence of a force
- Two capacitors having capacitances C₁ and C₂ are charged with 120V and 200V batteries
- The displacement of a particle varies according to the relation x=4 (cos π t + sin π t)
- If potential(in volts) in a region is expressed as V(x,y,z) = 6xy – y + 2yz,
Topics: Alternating Current (96)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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