A resistor with 810 is connected to the plates of a charged capacitor with capac

Question

A resistor with 810 is connected to the plates of a charged capacitor with capacitance 4.32 F . Just before the connection is made, the charge on the capacitor is 7.90 mC 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?

Explanation / Answer

C = 4.32 x 10^-6 F

R = 810ohms

Qo = 7.9 x 10^-3 C

PART A

Uo = (Qo)^2 / 2 C

= ( 7.9*10^-3 )^2 / ( 2*4.32*10^-6) = 7.223 J

PART B

Io = Qo / R C = 7.9*10^-3 / ( 810*4.32*10^-6 ) = 2.2576 A

Po = Io^2 R = (2.2576)^2 * 810 = 4128.37 W

PART C

U = (1/2) Uo

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

Q^2 = Qo^2 / 2

P = I^2 R = (Q / RC)^2 R

                = (1/2) (Qo / RC)^2 R

                = Po / 2

                = 4128.37 / 2

                = 2064.18 W

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.