The alternating current in a circuit is given by I=50 sin314t. The peak value

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The alternating current in a circuit is given by I=50 sin314t. The peak value and frequency of the current are

Options

(a) I₀=25 A and f=100 Hz
(b) I₀=50 A and f=50 Hz
(c) I₀=50 A and f=100 Hz
(d) I₀=25 A and f=50 Hz

Correct Answer:

I₀=50 A and f=50 Hz

Explanation:

From standard equation, we have I = I₀ sin ωt —–(i)

Given, I = 50 sin 31 4t —-(ii)

Comparing equation (i) and (ii), we get I₀ = 50 A, ω = 2πf = 314

⇒ f = 314 / (2 × 3.14) = 50 Hz

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Topics: Alternating Current (96)
Subject: Physics (2479)

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