Question

A resistor with 870.0 ω is connected to the plates of a charged capacitor with capacitance 4.80 μf. just before the connection is made, the charge on the capacitor is 9.50 mc. a) What is the energy initially stored in the capacitor?

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

a) The initial energy stored in the capacitor is 2.58 * 10^-4 J. b) The electrical power dissipated in the resistor just after the connection is made is 1.03 * 10^-5 W. c) The electrical power dissipated in the resistor when the energy stored in the capacitor is halved is 2.57 * 10^-6 W.

Step-by-step explanation:

a) The initial energy stored in the capacitor can be calculated using the formula:

E = (1/2) * C * V^2, where E is the energy, C is the capacitance, and V is the voltage. Plugging in the values, we have: E = (1/2) * (4.80 * 10^-6 F) * (9.50 * 10^-3 C)^2 = 2.58 * 10^-4 J.

b) The electrical power dissipated in the resistor just after the connection is made can be calculated using the formula:

P = (V^2) / R, where P is the power, V is the voltage, and R is the resistance. Plugging in the values, we have: P = (9.50 * 10^-3 C)^2 / 870.0 Ω = 1.03 * 10^-5 W.

c) 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) can be calculated using the formula:

P = (V^2) / R, where P is the power, V is the voltage, and R is the resistance. Since the energy stored in the capacitor has decreased to half, the voltage across the capacitor will also decrease to half. Plugging in the values, we have: P = ((9.50 * 10^-3 C) / 2)^2 / 870.0 Ω = 2.57 * 10^-6 W.