A 0.75μF capacitor is charged to 70 V. It is then connected in series with a 55Ω resistor and a 140 Ω resistor and allowed to discharge completely. Part A How much energy is dissipated by the 55Ω resistor?

Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.

Energy dissipates in 55Ω resistor is given by V²/R

Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.

So for 55ohms, using voltage divider rule

V=R1 / (R1+R2) * Vt

V=55 / (55+140) * 70

V=19.74Volts is across the 55ohms resistor.

Then, energy loss will be

E=V²/R

E=19.74²/55

E=7.09J

7.09J of heat is dissipated by the 55ohms resistor