Question

An electric bulb is designed to consume 55 W when operated at 110 volts. It is connected to a 220 V, 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the bulb gets correct voltage?

Answer 1

Power consumed by the electric bulb, P = 55 W
Voltage at which the bulb is operated, V= 110 V
Voltage of the line, V = 220 V
Frequency of the source, v = 50 Hz
P =$\frac{V^2}{R}$,
where R = resistance of electric bulb
$\therefore R=\frac{V^2}{P}$
$=\frac{110\times 110}{55}=220\mathrm{\Omega }$
If L is the inductance of the coil, then total reactance of the circuit (Z) is given by,
Z = $\sqrt{R^2+{\left(\omega L\right)}^2}$
=$\sqrt{{\left(220\right)}^2+{\left(100\mathrm{\pi }L\right)}^2}$ $\left(\because \omega =2\mathrm{\pi }f\right)$
Here, $\omega$ = angular frequency of the circuit
Now, current through the bulb, $I=\frac{V}{Z}$
$\therefore$ Voltage drop across the bulb, V = $\frac{V}{Z}\times R$
As per question,
$$\begin{gathered} 110=\frac{220\times 220}{\sqrt{{\left(220\right)}^2+{\left(100\mathrm{\pi }L\right)}^2}} \\ 110=\frac{{\left(220\right)}^2}{\sqrt{{\left(220\right)}^2+{\left(100\mathrm{\pi }L\right)}^2}} \\ \Rightarrow 220\times 2=\sqrt{{\left(220\right)}^2+{\left(100\mathrm{\pi }L\right)}^2} \\ \Rightarrow {\left(220\right)}^2+{\left(100\mathrm{\pi }L\right)}^2={\left(440\right)}^2 \\ \Rightarrow 48400+{10}^4{\mathrm{\pi }}^2{L}^2=193600 \\ \Rightarrow {10}^4{\mathrm{\pi }}^2{L}^2=193600-48400 \\ \Rightarrow L^2=\frac{145200}{{\mathrm{\pi }}^2\times {10}^4} \\ =1.4726 \\ \Rightarrow L=1.2135\cong 1.2\mathrm{H} \end{gathered}$$