Chapter 18, Problem 94P

(a)

To determine

Draw the equivalent circuit with one resistor and one capacitor.

Answer to Problem 94P

The equivalent circuit with one resistor and one capacitor is shown below.

Explanation of Solution

The resistor $R_1$ and $R_2$ in series then the equivalent resistance is,

$R_{eq}=R_1+R_2$ (I)

The capacitors $C_1$ and $C_2$ are in parallel then the equivalent capacitor is,

$C\text{'}=C_1+C_2$

This capacitor is in series with $C_3$. Then the final equivalent capacitance is,

$C_{eq}={\left[\frac{1}{C\text{'}}+\frac{1}{C_3}\right]}^{-1}$

Substitute the value of $C\text{'}$ in the above equation to get the final equivalent capacitance.

$C_{eq}={\left[\frac{1}{C_1+C_2}+\frac{1}{C_3}\right]}^{-1}$ (II)

Conclusion:

Substitute $25\Omega$ for $R_1$ and $\text{33 }\Omega$ for $R_1$ in equation I.

$R_{eq}=25\Omega +\text{33 }\Omega =58\Omega$

Substitute $12\text{ μF}$ for $C_1$ , $23\text{ μF}$ for $C_2$ and $\text{46 μF}$ for $C_3$ in equation II.

$\begin{array}{c}{C}_{eq}={\left[\frac{1}{12\text{ μF}+23\text{ μF}}+\frac{1}{\text{46 μF}}\right]}^{-1}\\ =20\text{ μF}\end{array}$

The equivalent resistance and capacitance values are used to draw the equivalent circuit and shown below.

(b)

To determine

Find the charge on the capacitor $C_1$ and current on the resistor $R_1$.

Answer to Problem 94P

The charge on the capacitor $C_1$ is $\text{4}\text{.1}\times {\text{10}}^{-5}\text{ C}$ and current on the resistor $R_1$ is $0$. 

Explanation of Solution

The current in the resistor with fully charged capacitor is zero. Since there is no flow of charge from the battery to the capacitor.

Write the equation for charge on equivalent capacitor,

$Q=C_{eq}\left(\Delta V\right)$ (III)

Here, $Q$ is the charge, $C_{eq}$ equivalent capacitance and $\Delta V$ is the potential difference.

Conclusion:

Substitute $\text{6}\text{.0 V}$ for $\Delta V$ and $\text{20}\text{.0}\times {\text{10}}^{-6}\text{ F}$ for $C_{eq}$ in equation III.

$\begin{array}{c}Q=\left(\text{20}\text{.0}\times {\text{10}}^{-6}\text{ F}\right)\left(\text{6}\text{.0 V}\right)\\ =1.2\times {10}^{-4}\text{ C}\end{array}$

This charge is same on the charge of capacitor $C_3$ . Then find the voltage drop across $C_3$.

$V_3=\frac{Q_3}{C_3}$

Substitute $1.2\times {10}^{-4}\text{ C}$ for $Q_3$ and $\text{46}\text{.0}\times {\text{10}}^{-6}\text{ F}$ for $C_3$ in the above equation.

$\begin{array}{c}{V}_3=\frac{1.2\times {10}^{-4}\text{ C}}{\text{46}\text{.0}\times {\text{10}}^{-6}\text{ F}}\\ =2.61\text{V}\end{array}$

Then the voltage drop across the capacitors $C_1$ and $C_2$ is,

$6.0\text{ V}-2.61\text{ V}=3.39\text{V}$

Then the charge on the capacitor $C_1$ is,

$Q_1=C_1{V}_1$

Substitute $3.39\text{V}$ for $V_1$ and  $\text{12}\times {\text{10}}^{-6}\text{ F}$ for $C_1$ in the above equation.

$\begin{array}{c}{Q}_1=\left(\text{12}\times {\text{10}}^{-6}\text{ F}\right)\left(3.39\text{V}\right)\\ =4.1\times {10}^{-5}\text{ C}\end{array}$

Therefore, charge on the capacitor $C_1$ is $\text{4}\text{.1}\times {\text{10}}^{-5}\text{ C}$ and current on the resistor $R_1$ is $0$.

(c)

To determine

Find the time constant of the circuit.

Answer to Problem 94P

The time constant of the circuit is $\text{1}\text{.2 ms}$.

Explanation of Solution

Write the equation for time constant,

$\tau =RC$ (IV)

Here, $\tau$ is the time constant, $R$ is the resistance and $C$ is the current.

Conclusion:

Substitute $\text{58 }\Omega$ for $R$  and $\text{20}\text{.0}\times {\text{10}}^{-6}\text{ F}$ for $C$ in equation IV.

$\begin{array}{c}\tau =\left(\text{58 }\Omega \right)\left(\text{20}\text{.0}\times {\text{10}}^{-6}\text{ F}\right)\\ =1.2\text{ ms}\end{array}$

Therefore, the time constant of the circuit is $\text{1}\text{.2 ms}$.

(d)

To determine

Find the time at which the voltage across the three capacitors $50\%$ of its final value.

Answer to Problem 94P

The time at which the voltage across the three capacitors $50\%$ of its final value is $\text{8}\text{.4}\times {\text{10}}^{-4}\text{ s}$.

Explanation of Solution

Write the equation for voltage across the capacitor.

$V=\epsilon \left(1-e^{-t/\left(RC\right)}\right)$ (V)

Here $V$ is the voltage across the capacitor, $\epsilon$ is the emf or potential difference, $t$ is the time, $R$ is the resistance and $C$ is the capacitance.

The voltage across the capacitor is equal to $50\%$ of the emf.

Then, the voltage is $V=0.5\epsilon$ , substitute this in equation V.

$\begin{array}{c}0.5\epsilon =\epsilon \left(1-e^{-t/\left(RC\right)}\right)\\ 0.5=1-e^{-t/\left(RC\right)}\end{array}$

Rewrite the above equation to get time $t$ .

$\begin{array}{c}{e}^{-t/\left(RC\right)}=0.50\\ -\frac{t}{RC}=\ln 0.50\end{array}$

$t=-RC\ln 0.50$ (VI)

Conclusion:

Substitute $\text{58 }\Omega$ for $R$  and $\text{20}\text{.0}\times {\text{10}}^{-6}\text{ F}$ for $C$ in equation VI.

$\begin{array}{c}t=\left(\text{58 }\Omega \right)\left(\text{20}\text{.0}\times {\text{10}}^{-6}\text{ F}\right)\ln 0.50\\ =8.0\times {10}^{-4}\text{ s}\end{array}$

Therefore, the time at which the voltage across the three capacitors $50\%$ of its final value is $\text{8}\text{.4}\times {\text{10}}^{-4}\text{ s}$.