An electric bulb of 100 ,W -300 ,V is connected with an AC supply of 500 ,V and frac{150}{π} ,Hz .

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7.Alternating Current

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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 ..........$H$

A

$2$

B

$\frac{1}{2}$

C

$4$

D

$\frac{1}{4}$

Solution

(c)

$V^2+(300)^2=(500)^2$

$V^2=(400)^2$

$V=400$

and $I=\frac{100}{300}=\frac{1}{3} \,A$

$V=X_L$

$400=\left(\frac{1}{3}\right) X_L$

$X_L=1200 \,\Omega$

$(2 \pi f) L=1200 \,\Omega$

So $L=4 \,H$

Standard 12

Physics

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