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In an ac circuit an alternating voltage e = 200 √2 sin 100 t volts is connected to a capacitor of capacity 1 µF. The r.m.s. value of the current in the circuit is
Options
(a) 10 mA
(b) 100 mA
(c) 200 mA
(d) 20 mA
Correct Answer:
20 mA
Explanation:
Vᵣₘₛ = 200 √2 / √2 = 200 V
Iᵣₘₛ = Vᵣₘₛ / Xc = 200/1 / 100 x 10⁻⁶
= 2 x 10⁻² = 20 mA
Related Questions: - If alternating source of primary coil on transformer is replaced by a Laclanche cell
- The power factor of an AC circuit having resistance R and inductance L connected
- The charge flowing through a resistance R varies with time t as Q=at-bt²,
- A capacitor of 10 µF charged upto 250 volt is connected in parallel with another
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Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If alternating source of primary coil on transformer is replaced by a Laclanche cell
- The power factor of an AC circuit having resistance R and inductance L connected
- The charge flowing through a resistance R varies with time t as Q=at-bt²,
- A capacitor of 10 µF charged upto 250 volt is connected in parallel with another
- The radius of germanium nuclide is measured to be twice the radius of ₄⁹Be.
Topics: Alternating Current (96)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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10^-6 is wrong it will be 10*10^-6=10^-5 so correct answer 200mA
V(rms) = 200 √2 / √2 = 200 V
I(rms) = V(rms) / Xc = (200/1) / (100 x 10⁻⁶)
= 2 x 10⁻² = 20 mA