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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: - A body oscillates with amplitude of 10 cm in a horizontal platform.
- B is doped in Si or Ge, then we will get
- The current in a self-inductance L=40 mH is to be increased uniformly from 1A to 11 A
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Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body oscillates with amplitude of 10 cm in a horizontal platform.
- B is doped in Si or Ge, then we will get
- The current in a self-inductance L=40 mH is to be increased uniformly from 1A to 11 A
- Two coherent sources of intensity ratio α interfere. In interference pattern,
- A 2 kg mass starts from rest on an inclined smooth surface with inclination
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