Question

an electric bulb rated 100W , 100 V has to be operated across 141.4 V , 50Hz A.C. supply . The capacitance of the capacitor which has to be connected in series with bulb do that the bulb will glow with full intensity ​

Answer 1

Answer:

given =

Power (p) = 100W

volt (v) =  100V

Frequency (f)= 50Hz

v2 = 141.4

to find = capicitance

solution,

Z² = R²+X²

the value of inductance is not given in quesstion,but wecan calculate the inductance by using

X= 2πfL

P = E²/R  = 484

I= Eamp =/R = 100/484 = 0.207

Z= E/I 220/0.207=1065

X= $\sqrt{Z^{2}-R^{2} }$  X= $\sqrt{1065^{2} -0.207^{2} }$

X=  948 Ω

X = 2πfL

948= 2×3.14×50×L

L= 3.02 at 50HZ.

the capistance of capicitor is 3.02 which is connected in series.