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

⇦

⇨

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. The damping force on an oscillator is directly proportional to the velocity.
2. When the speed of electron beam used in Young’s double slit experiment is increased,
3. If the band gap between valence band and conduction band in a material is 5.0 eV,
4. What is the nature of Gaussian surface involved in Gauss’s law of electrostatics?
5. What determines the nature of the path followed by the particle?

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