Question

An electric bulb of $100\,{\text{W}} - {\text{300 V}}$ is connected with an AC supply of ${\text{500 V}}$ and $\dfrac{{150}}{\pi }{\text{ Hz}}$. The required inductance to save the electric bulb is:
A. ${\text{2}}\,{\text{H}}$
B. $\dfrac{1}{2}\,{\text{H}}$
C. $4\,{\text{H}}$
D. $\dfrac{1}{4}\,{\text{H}}$

Answer 1

Use the formula for the potential difference across inductor in terms of inductive reactance. Use the formula for inductive reactance in terms of angular frequency and inductance. Also use the formula for the power and angular frequency. Determine the excess potential difference that will be across the inductor and do the needful calculations.

Formula used:
The potential difference is given by
${V_L} = I{X_L}$ …… (1)
Here, ${V_L}$ is the potential difference across the inductor, $I$ is the current and ${X_L}$ is the inductive reactance.
The inductive reactance ${X_L}$ is given by
${X_L} = \omega L$ …… (2)
Here, $\omega$ is the angular frequency and $L$ is the inductance.
The power $P$ is given by
$P = IV$ …… (3)
Here, $I$ is the current and $V$ is the potential difference.
The angular frequency $\omega$ is given by
$\omega = 2\pi f$
Here, $f$ is the linear frequency.

Complete step by step answer:
The operating voltage and power of the electric bulb are ${\text{300 V}}$ and $100\,{\text{W}}$ respectively.
$P = 100\,{\text{W}}$
The AC supply has the potential difference of ${\text{500 V}}$ and the linear frequency $\dfrac{{150}}{\pi }{\text{ Hz}}$.
$f = \dfrac{{150}}{\pi }{\text{ Hz}}$
The excess voltage ${V_L}$ applied to the bulb will appear across the inductor to save the bulb.
We can determine this excess voltage ${V_L}$ by taking the difference between the potential difference of AC supply and the bulb.
${V_L} = {\text{500 V}} - 3{\text{00 V}} = {\text{200 V}}$
We can determine the potential difference across the inductor using equation (1).
Substitute $\dfrac{P}{V}$ for $I$ and $\omega L$ for ${X_L}$ in equation (1).
${V_L} = \dfrac{P}{V}\omega L$
Here, $V$ is the potential difference across the AC supply.
Substitute $2\pi f$ for $\omega$ in the above equation.
${V_L} = \dfrac{P}{V}\left( {2\pi f} \right)L$
$\Rightarrow {V_L} = \dfrac{{2\pi fPL}}{V}$
Rearrange the above equation for the inductance $L$.
$\Rightarrow L = \dfrac{{{V_L}V}}{{2\pi fP}}$
Substitute ${\text{200}}\,{\text{V}}$ for ${V_L}$, ${\text{500}}\,{\text{V}}$ for $V$, $\dfrac{{150}}{\pi }{\text{ Hz}}$ for $f$ and $100\,{\text{W}}$ for $P$ in the above equation.
$\Rightarrow L = \dfrac{{\left( {{\text{200}}\,{\text{V}}} \right)\left( {{\text{500 V}}} \right)}}{{2\pi \left( {\dfrac{{150}}{\pi }{\text{ Hz}}} \right)\left( {100\,{\text{W}}} \right)}}$
$\Rightarrow L = \dfrac{{\left( {{\text{200}}\,{\text{V}}} \right)\left( {{\text{500 V}}} \right)}}{{2\left( {150{\text{ Hz}}} \right)\left( {100\,{\text{W}}} \right)}}$
$\Rightarrow L = 3.33\,{\text{H}}$
$\Rightarrow L \approx 4\,{\text{H}}$

Therefore, the inductance of the inductor should be $4\,{\text{H}}$ to save the electric bulb.

So, the correct answer is “Option C”.

Note:
The students may assume that the potential difference across the inductor is the same as the potential difference of the AC supply or the electric bulb. The electric bulb needs less potential difference than the potential difference of the AC supply. Hence, the potential difference is equal to the difference between potential difference of AC supply and bulb drops across the inductor.