Question

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

(1) 2H
(2)1/2H
(3) 4H
(4)1/4H​

Answer 1

Answer:

I kept a pic...ur option is (C)

Explanation:

Hope it definitely helps u...

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Answer 2

Answer:

3

Explanation:

The voltage at which the bulb operates is 300v And the aac supply has voltage 500v

Therefore, the difference of the voltage Vl=500−300=200v

Will appear across the inductance of the spred by the coil of the bulb.

Therefore Vl=ωL×I⇒ p/v = 100/500

p/v = 1/5 A

Thus, Vl =2πfL×I⇒200=2π× 150/π 1 1/5

⇒l= 200/60 =3.33≈4H