Question

A coil has negligible resistance and an inductive reactance of $20\Omega$ at $50\text{Hz}$. If an AC source of $200\text{V}$ and $100\text{Hz}$ frequency is connected across the coil, the rms current in the coil will be

• A. $2.0\text{A}$
• B. $5.0\text{A}$
• C. $7.0\text{A}$
• D. $10.0\text{A}$

Answer 1

The correct option is B $5.0\text{A}$

We know that,

$X_L=\omega L=2\pi fL$

$\Rightarrow 20=2\pi \left(50\right)L$

$\Rightarrow L=\frac{1}{5\pi }\text{H}$

When the coil is connected across the AC source,

$\therefore X_L^{\prime }=2\pi \left(100\right)\times \frac{1}{5\pi }=40\Omega$

Now,

$i_{rms}=\frac{V_{rms}}{X_L^{\prime }}=\frac{200}{40}=5.0\text{A}$

Hence, option $\left(B\right)$ is the correct answer.