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 charge stored by the capacitor.

Initially, the initial charge stored by the capacitor by the capacitor is provided and from the property of the capacitor, initially capacitor behave like short circuit $t = {0^ + }$ and at the steady state the charge stored by the capacitor is equal to the source voltage if this is connected in parallel with the circuit. If it is not connected with parallel with the circuit. Finally, the voltage across the capacitor is calculated by using final charge stored by the capacitor.

Fundamentals

The expression of the time constant in the series RC circuit is expresses as follows,

$\tau = RC$

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 expresses 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 expresses as follows,

$\tau = RC$

Substitute $1.5{\rm{ k}}\Omega$for$R$and $20{\rm{ }}\mu {\rm{F}}$for$C$in the above expression of the time constant of the circuit.

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

Calculate the time to take reduced the charge across the capacitor. The expression of the final charge stored by the capacitor is expresses as follows,

$Q = {Q_0}{e^{ - t/\tau }}$

Rearrange the above expression in terms of time$t$.

$\frac{Q}{{{Q_0}}} = {e^{ - t/\tau }}$

Take natural log both side.

$t = - \tau \ln \left( {\frac{Q}{{{Q_0}}}} \right)$

Substitute $25{\rm{ }}\mu {\rm{C}}$for${Q_0}$and $10{\rm{ }}\mu {\rm{C}}$for$Q$and $0.03{\rm{ s}}$for$\tau$in the above expression of the time$t$.

$\begin{array}{c}\\t = - \left( {0.03{\rm{ s}}} \right)\ln \left( {\frac{{{\rm{10 }}\mu {\rm{C}}}}{{25{\rm{ }}\mu {\rm{C}}}}} \right)\\\\ = 0.027{\rm{ s}}\\\end{array}$

Ans:

The time taken to reduce the charge of the capacitor to $10{\rm{ }}\mu {\rm{C}}$is equal to$0.027{\rm{ s}}$.