Question

A 0.50uF capacitor is charged to 100V . It is then connected in series with a 35? resistor and a 150? resistor and allowed to discharge completely.

How much energy is dissipated by the 35? resistor?

Answer 1

V1 be the voltage across 35ohm and V2 be the voltage across 150 ohm resistor.

So 100V is connected in series with 35 ohm and 150 ohm.

voltage across 35 ohm is

V1=(R1/R1+R2)V

V1=(35/35+150)*100

V1=(35/185)*100

V1=18.91V

P=V^2/R

P=(18.91)^2/35

P=357.75/35

P=10.22

Answer 2

100 v and discharges compltle ly

so v = 100v

r = 150+35 = 185 ohms

I = 100/185 = 0.5405A

P = i^2 r = 10.226W

Answer 3

energy stored in the capacitor=(1/2)*C*V^2=0.5*0.5*10^-6*100^2=0.0025J

this much energy will be dissipated across the resistors in propertion to their value

so energy dissipated across 35 ohm resistor=(35/(35+150))*0.0025J=0.000473J