Question

A bulb of rated values $60\text{V},10\text{W}$ is connected in series with a source of 100 V and $50\text{Hz}$. The coefficient of self induction of a coil to be connected in series for its operation will be -

• A. 1.53 H
• B. 2.15 H
• C. 3.27 H
• D. 3.89 H

Answer 1

The correct option is A 1.53 H

Current through the inductance (L) = current through the bulb
$I_{bulb}=\frac{P}{V}=\frac{10}{60}=\frac{1}{6}A=I_L$
Voltage across the bulb $V_{bulb}=V_R=60V$
By using $V=\sqrt{V_R^2+V_L^2}$
$V_L=\sqrt{{100}^2-{60}^2}=80\text{volt}$
$\Rightarrow I{X}_L=80\Rightarrow I\left(2\pi fL\right)=80$
$\Rightarrow \frac{1}{6}(2\pi \times 50\times L)=80\Rightarrow L=1.53\text{H}$