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:

Options

• 10 cos (500 t)

• −10 cos (500 t)

• 10 sin (500 t)

• −10 sin (500 t)

Answer 1

−10 cos (500 t)

Explanation:

In a pure inductive circuit current always lags behind the emf by `π/2`.

If v(t) = v₀ sin ωt then I = `"I"_0 sin(ω"t" - π/2)`

Now, given v(t) = 100 sin (500 t)

and I₀ = `"E"_0/(ω"L") = 100/(500 xx 0.02)` ...[∴ L = 0.02 H]

I₀ = `10 sin (500 "t" - π/2)` = −10 cos (500 t)