Question

Problem 26.84 A resistor with 760 omega is connected to the plates of a charged capacitor with capacitance 4.76 muF . Just before the connection is made, the charge on the capacitor is 9.00 mC.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

here ,

R = 760 Ohm

C = 4.76 uF

Q = 9 mC

Now , energy stored in cacacitor

U0 = 0.5 Q^2/C

U0 = 0.5 * (0.009)^2/(4.76 *10^-6)

U0 = 8.5 J

the energy initially stored in the capacitor is 8.5 J

B)

as V = Q/C

V = 0.009/4.76 *10^-6

V =1890.8 V

Power = V^2/R

Power = 1890.8^2/(760)

Power = 4703.9 W

the power dissipated in resistor is 4703.9 W

C)

For half value of energy ,

V = V/sqrt(2)

V = 1890.8/sqrt(2)

V = 1336.8 V

Now ,

Power = V^2/R

Power = 1336.8^2/760

Power = 2351.4 W

the power dissipated is 2351.4 W