Question

A coil of negligible resistance is connected in series with a 90 $\Omega$ resistor across a 120 V, 60 Hz line. An ac voltmeter reads 90 V across the resistance, then the inductance of the coil is approximately

• A. 0.2 H
• B. 0.3 H
• C. 0.4 H
• D. 0.7 H

Answer 1

The correct option is A 0.2 H

Using Ohm's law

$V=IR$

where V is potential difference across resistance and I is current in it.

As both inductor and resistor are connected in series the current will be same in both the inductor and resistor, which is given as

$I=\frac{120}{\sqrt{R^2+L^2{\omega }^2}}$

From given data $V=90V$ and $R=90\Omega$

$90=\frac{R}{\sqrt{R^2+L^2{\omega }^2}}120$

$\frac{1}{120}=\frac{1}{\sqrt{{90}^2+L^2{\left(2\pi 60\right)}^2}}$

$(120{)}^2=(90{)}^2+{120}^2{\pi }^2{L}^2\Rightarrow L^2=\frac{{120}^2-{90}^2}{{120}^2{\pi }^2}$

$L=\sqrt{\frac{30\times 210}{{120}^2{\pi }^2}}$

$L=0.21H$