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  4. An electric bulb of 100 W -300 V is connected with an AC supply of 500 V and (150/π) Hz. The required inductance to save the electric bulb is

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Q. An electric bulb of $100\, W -300 \, V$ is connected with an AC supply of $500 \, V$ and $\frac{150}{\pi} \, Hz$. The required inductance to save the electric bulb is

Alternating Current

Solution:

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$V^{2}+(300)^{2}=(500)^{2}$
$V^{2}=(400)^{2}$
$V=400$
and $I=\frac{100}{300}=\frac{1}{3} A$
$V=I X_{L}$
$400=\left(\frac{1}{3}\right) X_{L}$
$X_{L}=1200 \,\Omega$
$(2 \pi f) L=1200 \,\Omega$
So $L=4 \,H$