Question

The switch in the figure has been closed for a very long time.

Part A What is the charge on the capacitor?

Part B The switch is opened at . t= 0 s At what time has the charge on the capacitor decreased to 29% of its initial value?

Answer 1

Concepts and reason

The concepts used to solve this problem are combination of series resistance, Ohm’s law and expression of charge on the capacitor, and the charge on the capacitor after time $t$ .

First calculate the equivalent resistance and after that calculate the current in the circuit, next calculate the potential difference, finally calculate the charge across the capacitor, in part (B) calculate the time taken to charge on the capacitor decrease to $29\%$ of its initial value.

Fundamentals

The equivalent resistance of series resistance is as follows.

${R_{{\rm{eq}}}} = {R_1} + {R_2}$

Here ${R_1}$ and ${R_2}$ are two resistances connected in series.

From Ohm’s law,

$V = IR$

Here, $I$ is current, $R$ is resistance, and $V$ is potential difference.

The charge across the capacitor is,

$Q = CV$

Here $Q$ is charge, $C$ is capacitance and $V$ is potential difference.

The charge on the capacitor after time $t$ is as follows,

$Q = {Q_{{\rm{max}}}}{e^{\frac{{ - t}}{{RC}}}}$

Here, ${Q_{{\rm{max}}}}$ is maximum charge, $R$ is resistance, $t$ is time taken by capacitor, $C$ is capacitance.

As the switch closed for long time the capacitor behaves like open circuit, so there is no current flow from the capacitor.

The below image shows the condition of circuit at $t = \infty$ .

Calculate the equivalent resistance by using series resistance formula.

The equivalent resistance of series resistance is,

${R_{{\rm{eq}}}} = {R_1} + {R_2}$

Here ${R_1}$ and ${R_2}$ are two resistances connected in series.

Substitute $60{\rm{ }}\Omega$ for ${R_1}$ and $40{\rm{ }}\Omega$ for ${R_2}$ .

$\begin{array}{c}\\{R_{{\rm{eq}}}} = 60{\rm{ }}\Omega + 40{\rm{ }}\Omega \\\\ = 100{\rm{ }}\Omega \\\end{array}$

Now calculate the current by using Ohm’s law.

From Ohm’s law,

$V = IR$

Here, $I$ is current, $R$ is resistance, and $V$ is potential difference.

Rearrange the equation $V = IR$ for current $I$ .

$I = \frac{V}{R}$

Substitute $100.0{\rm{ V}}$ for $V$ , and $100{\rm{ }}\Omega$ for ${R_{{\rm{eq}}}}$ .

$\begin{array}{c}\\I = \frac{{100.0{\rm{ V}}}}{{100{\rm{ }}\Omega }}\\\\ = 1.0{\rm{ A}}\\\end{array}$

Now calculate the voltage across the $40{\rm{ }}\Omega$ resistor.

$\begin{array}{c}\\V' = I\left( {40{\rm{ }}\Omega } \right)\\\\ = \left( {1.0{\rm{ A}}} \right)\left( {40{\rm{ }}\Omega } \right)\\\\ = 40{\rm{ V}}\\\end{array}$

Now the maximum on the capacitor is,

The charge across the capacitor is,

${Q_{{\rm{max}}}} = CV$

Here ${Q_{{\rm{max}}}}$ is maximum charge, $C$ is capacitance and $V$ is potential difference.

Substitute $2.0{\rm{ \mu F}}$ for $C$ and $40{\rm{ V}}$ for $V'$ .

$\begin{array}{c}\\{Q_{{\rm{max}}}} = \left( {2.0{\rm{ \mu F}}} \right)\left( {40{\rm{ V}}} \right)\\\\ = 80.0{\rm{ \mu C}}\\\end{array}$

The below image shows the condition of circuit at $t = 0$ .

Calculate the equivalent resistance $t = 0$ when switch open.

The equivalent resistance of series resistance is,

${R_{{\rm{eq}}}} = {R_1} + {R_2}$

Here ${R_1}$ and ${R_2}$ are two resistances connected in series.

Substitute $10{\rm{ }}\Omega$ for ${R_1}$ and $40{\rm{ }}\Omega$ for ${R_2}$ .

$\begin{array}{c}\\{R_{{\rm{eq}}}} = 10{\rm{ }}\Omega + 40{\rm{ }}\Omega \\\\ = 50{\rm{ }}\Omega \\\end{array}$

Now calculate the charge on capacitor after time $t$ when the charge on the capacitor decreased to $29{\rm{\% }}$ of the initial value.

The charge on the capacitor after time $t$ is,

$Q = {Q_{{\rm{max}}}}{e^{\frac{{ - t}}{{RC}}}}$

Here, ${Q_{{\rm{max}}}}$ is maximum charge, $R$ is resistance, $t$ is time taken by capacitor, $C$ is capacitance.

Rearrange the equation $Q = {Q_{{\rm{max}}}}{e^{\frac{{ - t}}{{RC}}}}$ for $t$ .

$t = - \left( {RC} \right)\ln \left( {\frac{Q}{{{Q_{{\rm{max}}}}}}} \right)$

Substitute $50{\rm{ }}\Omega$ for $R$ , $2.0{\rm{ \mu F}}$ for $C$ , $0.29{Q_{{\rm{max}}}}$ for $Q$ .

$\begin{array}{c}\\t = - \left( {\left( {50{\rm{ }}\Omega } \right)\left( {2.0{\rm{ \mu F}}} \right)} \right)\ln \left( {\frac{{0.29{Q_{{\rm{max}}}}}}{{{Q_{{\rm{max}}}}}}} \right)\\\\ = - \left( {\left( {50{\rm{ }}\Omega } \right)\left( {2.0{\rm{ \mu F}}} \right)\left( {\frac{{{{10}^{ - 6}}{\rm{ F}}}}{{1{\rm{ F}}}}} \right)} \right)\ln \left( {0.29} \right)\\\\ = - \left( {\left( {50{\rm{ }}\Omega } \right)\left( {2.0 \times {{10}^{ - 6}}{\rm{ F}}} \right)} \right)\left( { - 1.23} \right)\\\\ = 1.24 \times {10^{ - 4}}{\rm{ s}}\\\end{array}$

Convert second in to millisecond.

$\begin{array}{c}\\t = \left( {1.24 \times {{10}^{ - 4}}{\rm{ s}}} \right)\left( {\frac{{{{10}^3}{\rm{ ms}}}}{{1{\rm{ s}}}}} \right)\\\\ = 0.124{\rm{ ms}}\\\end{array}$

Ans: Part A

Charge on the capacitor is $80.0{\rm{ \mu C}}$ .

Part B

The required time is $0.124{\rm{ ms}}$ .