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Question
An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor. (π = 3.142)
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Solution
Data: L = 2H, i0 = 0.25A, f = 60 Hz, n = 3.142
ωL = 2πfL = 2(3.142)(60)(2) = 754.1 Ω
The effective potential difference across the inductor = `omega"Li"_"rms" = omega"L" "i"_0/sqrt2`
`= ((754.1)(0.25))/1.414 = 133.3 "V"`
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