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

Question

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 (A) 2 H (B) (1/2) H (C) 4 H (D) (1/4) H

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0b3d29a0479bd7b01c7c2942dced0077-.png" /> V2+(300)2=(500)2 V2=(400)2 V=400 and I=(100/300)=(1/3) A V=I XL 400=((1/3)) XL XL=1200 Ω (2 π f) L=1200 Ω So L=4 H

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

$2 H$

$\frac{1}{2} H$

$4 H$

$\frac{1}{4} H$

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

Q. An electric bulb of 100W−300V is connected with an AC supply of 500V and π150​Hz. The required inductance to save the electric bulb is

2H

21​H

4H

41​H