An alternating voltage given as, V=100√2 sin100t V is applied to a capacitor

An Alternating Voltage Given As V1002 Sin100t V Is Applied Physics Question

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

Related Questions:

  1. Two identical long conducting wires AOB and COD are placed at right angle
  2. What is the difference between soft and hard X-ray?
  3. A thin and circular disc of mass M and radius R is rotating in a horizontal plane
  4. A body of mass m is thrown upwards at an angle θ with the horizontal with velocity
  5. Four blocks of same mass connected by cords are pulled by a force F

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.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*