⇦ | ![]() | ⇨ |
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 bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg.
- At place on earth, the vertical component of earth’s magnetic field is √3 times
- On the horizontal surface of a truck a block of mass 1 kg is placed (µ=0.6) and truck
- The current gain of a transistor in common base configuration is 0.96.
- An electric dipole of moment p is lying along a uniform electric field E. The work
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
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