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

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 : (A) 10 cos (500t) (B) —10 cos (500t) (C) 10 sin (500t) (D) -10 sin (500t)

Tardigrade Admin · 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 s in (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)

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

10 cos (500t)

—10 cos (500t)

10 sin (500t)

-10 sin (500t)

In a pure inductive circuit current always lags behind the emf by 2π​. If v(t)=v0​sinωt then I=I0​sin(ωt−2π​) Now, given v(t)=100sin(500t) and I0​=ωLE0​​=500×0.02100​[∵L=0.02H] I0​=10sin(500t−2π​) I0​=−10cos(500t)

Q. 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 :