Question

A coil of resistance $300\Omega$ and inductance $1.0H$ is connected across an alternating voltage of peak value $300V$ of frequency $\frac{300}{2\pi }Hz.$ Calculate the peak potential difference across inductor.

• A. $150\sqrt{2}V$
• B. $150V$
• C. $100\sqrt{2}V$
• D. $100V$

Answer 1

The correct option is A $150\sqrt{2}V$

The given $LR$ is represented in the below diagram.


The individual phasor diagrams for the resistor and inductor can be drawn as


$V_{0R}=i_{0}R$ and $V_{0L}=i_0{X}_L$

Combining the two phasor diagrams, we get


$V_0=\sqrt{V_{0R}^2+V_{0L}^2}$
$V_0=i_0\sqrt{R^2+X_L^2}=i_{0}Z$

where $Z$ (impedance of circuit) $=\sqrt{R^2+X_L^2}$
$=\sqrt{R^2+{\omega }^2{L}^2}$
$=\sqrt{R^2+(2\pi f{)}^2{L}^2}$
$=\sqrt{{300}^2+{(2\pi \times \left(\frac{300}{2\pi }\right))}^2\times 1^2}$
$=\sqrt{{300}^2+{300}^2}=300\sqrt{2}\Omega$

Peak current $i_0=\frac{V_0}{Z}=\frac{300}{300\sqrt{2}}$
$=\frac{1}{\sqrt{2}}A$

Therefore, peak voltage across inductor $=X_L\times i_0$
$=\frac{1}{\sqrt{2}}\times 300=\frac{300}{\sqrt{2}}V$
$=150\sqrt{2}V$