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An electric bulb has a rated power of 50 W at 100 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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The correct answer is:
1.1 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
$Z=\sqrt{R^2+\omega^2 L^2}$ ...(i)
Here, $\quad Z=\frac{200}{0.5}=400 \Omega$
Now, $\quad X_L^2=Z^2-R^2$
$=(400)^2-(200)^2$
$(2 \pi f L)^2=12 \times 10^4$
$L=\frac{2 \sqrt{3} \times 100}{2 \pi \times 50}$
$=\frac{2 \sqrt{3}}{\pi}=1.1 \mathrm{H}$
$R=\frac{V^2}{P}=\frac{(100)^2}{50}$
$=200 \Omega$
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
$Z=\sqrt{R^2+\omega^2 L^2}$ ...(i)
Here, $\quad Z=\frac{200}{0.5}=400 \Omega$
Now, $\quad X_L^2=Z^2-R^2$
$=(400)^2-(200)^2$
$(2 \pi f L)^2=12 \times 10^4$
$L=\frac{2 \sqrt{3} \times 100}{2 \pi \times 50}$
$=\frac{2 \sqrt{3}}{\pi}=1.1 \mathrm{H}$
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