A)
B)
C)
D)
Correct Answer: D
Solution :
[d]: Current in LR circuit is \[I=\frac{{{V}_{0}}}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}\sin \left( \omega t-\frac{\pi }{2} \right),\] it is sinusoidal in nature.You need to login to perform this action.
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