Question

A 12.0-μF capacitor is charged to a potential of 50.0V and then discharged through a 175-Ω resistor How long does it take the capacitor to lose (a) half of its charge and (b) half of its stored energy? 

Answer 1

Concepts and reason

The main concepts used to solve the problem are relation between charge, capacitance, and voltage, decaying charge expression, and decaying energy expression.

Initially, use the relation between charge, capacitance, and voltage to calculate the charge. Then, use the condition given to calculate the final charge on the capacitor. Later, use the decay charge equation to calculate the time.

Finally, use the decaying energy equation to calculate the time taken by capacitor to lose half of its stored energy.

Fundamentals

The relation between charge, capacitor, and voltage is,

$Q=CV $

Here, Q is the charge, C is the capacitance, and V is the voltage.

The charge stored in the capacitor is expressed as follows:

Here, is the initial charge, is the final charge, t is the time, R is the resistance, and C is the capacitance.

The energy stored in the capacitor is expressed as follows:

Here, is the initial energy, is the final energy, t is the time, R is the resistance, and C is the capacitance.

(a)

Calculate the time does it take the capacitor to lose half of its charge.

The relation between charge, capacitor, and voltage is,

$Q=CV $

Here, Q is the charge, C is the capacitance, and V is the voltage.

Substitute for C and for V in expression $Q=CV $.

The final charge on the capacitor is half of the initial charge.

Here, Q is the final charge and is the initial charge.

Substitute for in expression .

The charge stored in the capacitor is expressed as follows:

Here, is the initial charge, is the final charge, t is the time, R is the resistance, and C is the capacitance.

Rearrange the equation and solve for time .

Take natural log both sides of equation .

Substitute for in expression .

Substitute for and for in expression .

Or

(b)

Calculate the time taken by the capacitor to lose half of its stored energy.

The energy stored in the capacitor is expressed as follows:

Here, is the initial energy, is the final energy, t is the time, R is the resistance, and C is the capacitance.

The final energy on the capacitor is half of the initial energy.

Here, U is the final charge and is the initial charge.

Substitute for U in expression .

Take natural log both sides in the equation .

Rewrite the equation for time t.

Substitute for and for in equation .

Or

Ans: Part a

The time taken by the capacitor to lose half of its charge is .

Part b

The time taken by the capacitor to lose half of its energy is .