Question

A 20 μF capacitor initially charged to 25 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the capacitor's charge to 10 μC?

 

Answer 1

Concept and reason

The concept required to solve this problem is the charge stored by the capacitor. Initially, the initial charge stored by the capacitor by the capacitor is provided. From the capacitor's property, initially, the capacitor behaves like a short circuit \(t=0^{+}\). At the steady-state, the capacitor's charge is equal to the source voltage if this is connected in parallel with the circuit. Suppose it is not connected in parallel with the circuit. Finally, the voltage across the capacitor is calculated by using the final charge stored by the capacitor.

Fundamentals

The expression of the time constant in the series \(\mathrm{RC}\) circuit is expressed as follows, \(\tau=R C\)

Here, \(R\) is the equivalent resistance, \(C\) is the equivalent capacitance, and \(\tau\) is the time constant. The expression of the final charge stored by the capacitor is expressed as follows, \(Q=Q_{0} e^{-t / \tau}\)

Here, \(Q\) is the final charge stored by the capacitor, and \(Q_{0}\) is the initial charge stored by the capacitor.

 

Calculate the time constant of the circuit.

The expression of the time constant of the circuit is expressed as follows, \(\tau=R C\)

Substitute \(1.5 \mathrm{k} \Omega\) for \(R\) and \(20 \mu \mathrm{Ffor} C\) in the above expression of the time constant of the circuit.

$$ \begin{array}{c} \tau=(1.5 \mathrm{k} \Omega)(20 \mu \mathrm{F}) \\ =(1.5 \mathrm{k} \Omega)\left(\frac{1 \times 10^{3} \Omega}{1 \mathrm{k} \Omega}\right)(20 \mu \mathrm{F})\left(\frac{1 \times 10^{-6} \mathrm{~F}}{1 \mu \mathrm{F}}\right) \\ =0.03 \mathrm{sec} \end{array} $$

Form the expression of the time constant, the time constant of the \(R C\) circuit is the product of the resistance and the capacitance of the circuit.

 

Calculate the time to take reduced the charge across the capacitor. The expression of the final charge stored by the capacitor is expressed as follows, \(Q=Q_{0} e^{-t / \tau}\)

Rearrange the above expression in terms of timet. \(\frac{Q}{Q_{0}}=e^{-t / \tau}\)

Take natural log both side. \(t=-\tau \ln \left(\frac{Q}{Q_{0}}\right)\)

Substitute \(25 \mu\) Cfor \(Q_{0}\) and \(10 \mu\) Cfor \(Q\) and 0.03 sfor \(\tau\) in the above expression of the timet.

\(t=-(0.03 \mathrm{~s}) \ln \left(\frac{10 \mu \mathrm{C}}{25 \mu \mathrm{C}}\right)\)

\(=0.027 \mathrm{~s}\)

The time taken to reduce the charge of the capacitor to \(10 \mu \mathrm{Cis}\) equal to \(0.027 \mathrm{~s}\).

The charge stored by the capacitor depends on the initial value of the capacitor and the time constant of the circuit. To calculate the final charge stored by the capacitor, the capacitor uses the final expression of the capacitor's charge.

The time taken to reduce the charge of the capacitor to \(10 \mu\) Cis equal to 0.027 s.