Question

A $12\Omega$ resistance and an inductance of $\frac{0.05}{\pi }\text{H}$ with negligible resistance are connected in series. Across the ends of this circuit a $130\text{V}$ alternating voltage of frequency $50\text{Hz}$ is connected. Calculate the alternating current in the circuit and the potential difference across the resistance.

• A. $10\text{A},50\text{V}$
• B. $10\text{A},120\text{V}$
• C. $5\text{A},60\text{V}$
• D. $5\text{A},120\text{V}$

Answer 1

The correct option is B $10\text{A},120\text{V}$

The impedance of the circuit is given by

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

$=\sqrt{[R^2+(2\pi fL{)}^2]}=\sqrt{(12{)}^2+{\{2\times 3.14\times 50\times \frac{0.05}{3.14}\}}^2}$

$=\sqrt{(144+25)}=13\Omega$

Current in the circuit

$i=\frac{V}{Z}=\frac{130}{13}=10\text{A}$

Potential difference across resistance

$V_R=iR=10\times 12=120\text{V}$

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