Question

A resistor with 830? is connected to the plates of a charged capacitor with capacitance 4.60?F . Just before the connection is made, the charge on the capacitor is 7.00mC

Part A

What is the energy initially stored in the capacitor?

Part B

What is the electrical power dissipated in the resistor just after the connection is made?

Part C

What is the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in part A?

Answer 1

Capacitance, C = 4.60 x 10^-6 F

Resistance, R = 830 ohms

Charge, Qo = 7.0 x 10^-3 C

(a)

Initial energy stored, Uo = (Qo)^2 / 2 C

= ( 7.0 x 10^-3 )^2 / ( 2 * 4.60 x 10^-6)

= 5.326 J

(b)

Current through the resistor, Io = Qo / R C

= 7.0 x 10^-3 / ( 830 * 4.6 x 10^-6 )

= 1.83 A

Power dissipated, Po = Io^2 R

                                 = (1.83)^2 * 830

                                 = 2779.59 W

(c)

U = (1/2) Uo

( Q^2 / 2 C ) = (1/2) ( Qo^2 / 2 C)

Q^2 = Qo^2 / 2

Power, P = I^2 R

                = (Q / RC)^2 R

  

                = (1/2) (Qo / RC)^2 R

                = Po / 2

                = 2779.59 / 2

                = 1389.8 W