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An electric bulb has a rated power of $50 \mathrm{~W}$ at $100 \mathrm{~V}$. If it is used on an AC source $200 \mathrm{~V}, 50$ $\mathrm{Hz}$, a choke has to be used in series with it. This choke should have an inductance of
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Verified Answer
The correct answer is:
$1.1 \mathrm{H}$
Resistance of bulb
$$
R=\frac{V^{2}}{P}=\frac{(100)^{2}}{50}=200 \Omega
$$
Current through bulb $(I)=\frac{V}{R}$
$$
=\frac{100}{200}=0.5 \mathrm{~A}
$$
In a circuit containing inductive reactance $\left(X_{L}\right)$ and resistance $(R)$, impedance $(Z)$ of the circuit is
$$
Z=\sqrt{R^{2}+\omega^{2} L^{2}}
$$
Here, $Z=\frac{200}{0.5}=400 \Omega$
Now, $X_{L}^{2}=Z^{2}-R^{2}=(400)^{2}-(200)^{2}$
$$
\begin{aligned}
(2 \pi f L)^{2} &=12 \times 10^{4} \\
L &=\frac{2 \sqrt{3} \times 100}{2 \pi \times 50}=\frac{2 \sqrt{3}}{\pi}=1.1 \mathrm{H}
\end{aligned}
$$
$$
R=\frac{V^{2}}{P}=\frac{(100)^{2}}{50}=200 \Omega
$$
Current through bulb $(I)=\frac{V}{R}$
$$
=\frac{100}{200}=0.5 \mathrm{~A}
$$
In a circuit containing inductive reactance $\left(X_{L}\right)$ and resistance $(R)$, impedance $(Z)$ of the circuit is
$$
Z=\sqrt{R^{2}+\omega^{2} L^{2}}
$$
Here, $Z=\frac{200}{0.5}=400 \Omega$
Now, $X_{L}^{2}=Z^{2}-R^{2}=(400)^{2}-(200)^{2}$
$$
\begin{aligned}
(2 \pi f L)^{2} &=12 \times 10^{4} \\
L &=\frac{2 \sqrt{3} \times 100}{2 \pi \times 50}=\frac{2 \sqrt{3}}{\pi}=1.1 \mathrm{H}
\end{aligned}
$$
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