A sinusoidal voltage V ( t ) = 100 sin ( 500 t ) is applied across a pure inductance of L = 0.02 H . The current through the coil is:

Question

TestHub · Accepted Answer

In a pure inductive circuit current always lags behind the emf by $\frac{π}{2}$ .
If $v (t) = v_{0} sin ω t$ then $I = I_{0} sin (ω t - \frac{π}{2})$ Now, given $v (t) = 100 sin (500 t)$ and $I_{0} = \frac{E _{0}}{ω L} = \frac{100}{500 \times 0.02} [∵ L = 0.02 H]$ $I_{0} = 10 sin (500 t - \frac{π}{2})$ $I_{0} = - 10 cos (500 t)$

Physics - EMI/AC Question with Solution | TestHub

Options:

Doubts & Discussion