Question

A resistor with 860 \Omega is connected to the plates of a charged capacitor with capacitance 5.02 \mu F. Just before the connection is made, the charge on the capacitor is 9.10 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 0.409 joules.

b) The electrical power dissipated in the resistor just after the connection is made is 0.478 watts.

c) The electrical power dissipated in the resistor when the energy stored in the capacitor has decreased to half the value calculated in part (a) is 0.119 watts.

Step-by-step explanation:

a) The initial energy stored in the capacitor
`(\(E_(initial)\))` can be calculated using the formula
`\(E_(initial) = (1)/(2) C V^2\),` where C is the capacitance and V is the voltage across the capacitor. Given the capacitance
`\(C = 5.02 \mu F\)` and the charge
`\(Q = 9.10 mC\)`, the voltage V across the capacitor is calculated as
`\(V = (Q)/(C)\).` Then, the energy stored in the capacitor is
`\(E_(initial) = (1)/(2) * 5.02 * 10^(-6) * (9.10 * 10^(-3))^2\), resulting in \(E_(initial) = 0.409 J\).\\`

b) The electrical power dissipated in the resistor P just after the connection is made can be found using Ohm's law,
`\(P = (V^2)/(R)\)`, where V is the voltage across the resistor and R is the resistance. The initial voltage across the capacitor is equal to the initial voltage across the resistor. So,
`\(P = (V^2)/(R) = ((9.10 * 10^(-3))^2)/(860)\), giving \(P = 0.478 W\).`

c) When the energy stored in the capacitor decreases to half the initial value, the power dissipated in the resistor can be found similarly by using the new energy value in the power formula. With
`\(E_(half) = (E_(initial))/(2)\), substitute \(E_(half)\)` into the energy formula to find the corresponding voltage across the capacitor. Then, calculate the power using the resistor's resistance, obtaining
`\(P = 0.119 W\).`

Answer 2

The initial energy stored in the capacitor is 2.071 × 10⁻⁷ J. The electrical power dissipated in the resistor just after the connection is made is 9.63 × 10⁻⁸ W. The electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the initially calculated value is 4.65 × 10⁻⁶ W.

Step-by-step explanation:

To find the initial energy stored in the capacitor, we can use the formula:

Energy = (1/2) × C × V²

where C is the capacitance and V is the initial voltage across the capacitor plates. Plugging in the given values, we get:

Energy = (1/2) × (5.02 × 10⁻⁶) × (9.10 × 10⁻³)² = 2.071 × 10⁻⁷ J

To find the electrical power dissipated in the resistor just after the connection is made, we can use the formula:

Power = (V²) / R

where V is the initial voltage across the capacitor plates and R is the resistance. Plugging in the given values, we get:

Power = (9.10 × 10⁻³)² / (860) = 9.63 × 10⁻⁸ W

To find the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the initially calculated value, we first need to find the new energy stored in the capacitor. Since the energy is halved, the new energy is:

New energy = (1/2) × (2.071 × 10⁻⁷) = 1.036 × 10⁻⁷ J

To find the new voltage across the capacitor plates, we can rearrange the formula for energy to solve for V:

V = √(2 × Energy / C)

Plugging in the new values, we get:

V = √(2 × (1.036 × 10⁻⁷) / (5.02 × 10⁻⁶)) = 0.020 V

Finally, we can use the new voltage and the resistance to calculate the power:

Power = (0.020²) / (860) = 4.65 × 10⁻⁶ W