Q124 / 180 · AIIMS 2018 (26 May) · Physics
An electric bulb has a rated power of $50 \mathrm{~W}$ at $100 \mathrm{~V}$. If it is used on an AC source of $200 \mathrm{~V}$, $50 \mathrm{~Hz}$, a choke has to be used in series with it. This choke should have an inductance of
- A$1 \mathrm{mH}$
- B$0.1 \mathrm{mH}$
- C$0.1 \mathrm{H}$
- Correct answerD$1.1 \mathrm{H}$
Solution
Here, $P=50 \mathrm{~W}, V=100 \mathrm{~V}$
$I=\frac{P}{V}=\frac{50}{100}=0.5 \mathrm{~A}, R=\frac{V}{I}=\frac{100}{0.5}=200 \Omega$
Let $L$ be the inductance of the choke coil
$$
\begin{aligned}
& \therefore \quad I_v=\frac{E_v}{Z} \text { or } Z=\frac{E_v}{I_v}=\frac{200}{0.5}=400 \Omega \\
& =100 \sqrt{12} \Omega \\
&
\end{aligned}
$$
Now, $X_L=\sqrt{Z^2-R^2}=\sqrt{400^2-200^2}$
$=100 \sqrt{12} \Omega$
or, $\omega L=100 \times 2 \sqrt{3} \Omega$
$\begin{aligned} \Rightarrow L=\frac{200 \sqrt{3}}{\omega}=\frac{200 \sqrt{3}}{2 \pi v}=\frac{200 \sqrt{3}}{100 \pi} & =\frac{2 \times 1.732}{3.14} \\ & =1.1 \mathrm{H}\end{aligned}$
Hence, the correct answer is option: 'D'.