⇦ | ![]() | ⇨ |
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:
- An object placed at 20 cm in front of a concave mirror produces three times
- A light emitting diode(LED) has a voltage drop of 2 volt across it
- Identify the wrong statement with reference to solar cell
- The angle between the vectors (i+j) and (j+k) is
- A wheel of mass 10 kg has a moment of inertia of 10 kgm² about its own axis
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
Leave a Reply