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Question:

An alternative voltage given by V = 140(sin314t) is connected across a pure resistor of 50Ω. Find (i) The frequency of the source. (ii) The rms current through the resistor.

Solution:

(i) If V = 140sin314t
Comparing it with V = V₀sinωt
ω = 314
or 2πf = 314
or f = 314/2π = 314/(2 × 3.14) = 50Hz
(ii) I₀ = V₀/R = 140/50 = 2.8A
therefore Irms = I₀/√2 = 2.8/√2 = 1.98A