Question

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

Answer 1

Answer:

Since the current through the lamp would be:

I = P/V=50/100=0.5A

and since this is a series circuit we know the same amount of current flows throughout the circuit therefore we can say that the total impedance is:

Z=V/I=200/0.5=400Ω by Ohm’s law

and lamp’s resistance is:

Rlamp=V/I=100/0.5=200Ω by Ohm’s law

Through the impedance triangle Xc would be:

Z²=Rlamp²+Xc²

Xc=√(Z²-Xc²)

Xc=√(400²-200²)

Xc=√(120000)

Xc=346.41Ω

and now we can determine C from:

Xc=1/2πfC

therefore

C=1/2πfXc

C=1/(2π(50)(346.6))

C=0.000009F

or 9µF

Hopes this helps.