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. If 75% of the radioactive reaction is completed in 2 hrs, what would be
2. Dependence of intensity of gravitational field (E) of earth with distance(r)
3. A potentiometer wire of length 100cm has a resistance of 10Ω. It is connected in series
4. A current of 2A flows through a 2Ω resistor when connected across a battery
5. A drum of radius R and mass M, rolls down without slipping along an inclined plane

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