Question

A $12\Omega$ resistor and an inductor of $\frac{0.05}{\pi }\text{H}$ with negligible resistance are connected in series. Across this combination, an alternating voltage of $130\text{V},50\text{Hz}$ is connected. Calculate the rms value of alternating current in the circuit.

• A. $10\text{A}$
• B. $15\text{A}$
• C. $18\text{A}$
• D. $12\text{A}$

Answer 1

The correct option is A $10\text{A}$

The impedance of the circuit is given by,

$Z=\sqrt{R^2+{\omega }^2{L}^2}=\sqrt{R^2+(2\pi fL{)}^2}$

$\Rightarrow Z=\sqrt{{12}^2+{(2\times \pi \times 50\times \frac{0.05}{\pi })}^2}$

$\Rightarrow Z=13\Omega$

Now, rms value of alternating current in the circuit,

$i_{rms}=\frac{V_{rms}}{Z}=\frac{130}{13}=10\text{A}$