Question

When a dc voltage of 200V is applied to a coil of self-inductance ($2\sqrt{3}/\pi$) H, a current of 1 A flows through it. But by replacing dc source with ac source of 200 V, the current in the coil is reduced to 0.5 A. Then the frequency of ac supply is:

• A. $100Hz$
• B. $75Hz$
• C. $60Hz$
• D. $50Hz$

Answer 1

The correct option is D $50Hz$

Let, $R=$ resistance of coil

When DC connected,

$R=\frac{V}{I}=\frac{200}{1}=200$ $\Omega$

When AC source $200$ $V$ is applied, $I=\frac{200}{\sqrt{(200{)}^2+{\omega }^2{L}^2}}$..................(1)

Given, $I=0.5A$, $L=2\sqrt{3}/\pi$ and we know $\omega =2\pi f$.

Putting these values in (1) and solving for $f$.

We get, $f=50Hz$