The sinusoidal voltage applied to the circuit below is V S (t) =
| The sinusoidal voltage applied to the circuit below is VS(t) = 10 cos(2t) V. The voltage VL(t) across the inductor is
A. –10 sin 2t V
B. 10 cos 2t V
C. 10 sin (2t – 45°) V
D. 10cos (2t – 45°) V
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
VS = 10∠0°, ω = 2 rad/s
Impedance, jωL = j (2) (1) = j2 Ω
VL(t) = 10cos (2t + 90°)
VL(t) = – 10 sin 2t V cos (θ + 90°) = – sin θ
