Question

PLZZ ..... SOLVE IT GUYZZ............


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

Answer 1

hope this it will help u ....

Answer 2

Answer:

2H.

Explanation:

From the question we get that the power ratings of the lamp is 100W and the voltage is 300V. So, it will draw a current of 100W/300V = 1/3A since, i=P/V while supplied with a steady voltage of 300 V.

The inductor will suffer a potential drop of 500-300=200V.

Hence, from the above we will get the reactance of the inductor which will be  [R=V/I] 200/(1/3) which on solving we will get 600 ohms.

So, to find the inductive reactance we use the formulae of XL which is ωL or 2πfL, hence the value of L = XL/(2πf) . Since the value of reactance XL is 600.

So, 600/(2π150/π) which will be 600/(300*PI/PI) or 2H.