⇦ | ![]() | ⇨ |
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:
- The resistance of a bulb filament is 100 Ω at a temperature of 100⁰C.
- The moment of inertia of a uniform circular disc of radius R and mass M about an axis
- A boat is sent across a river with a velocity of 8 km/h.If the resultant velocity
- A polarised light of intensity I₀ is passed through another polariser whose pass axis
- The velocity of a particle performing simple harmonic motion, when it passes
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