Chapter 18, Problem 88P

(a)

To determine

Draw the equivalent circuit with one resistor and one capacitor.

Answer to Problem 88P

The equivalent circuit has been drawn.

Explanation of Solution

The resistor $R_1$ and $R_2$ are 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 capacitance is

    $C\text{'}=C_1+C_2$        (II)

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

    $C_{eq}=\left[\frac{1}{C\text{'}}+\frac{1}{C_3}\right]$        (III)

Substitute the value of $C\text{'}$ in the above equation

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

Conclusion:

Substitute $25\Omega$ for $R_1$, $33\Omega$ for $R_2$ in (I)

    $\begin{array}{c}{R}_{eq}=25\Omega +33\Omega \\ =58\Omega \end{array}$

Substitute $\text{12μF}$ for $C_1$, $\text{23μF}$ for $C_2$ and $\text{46μF}$ for $C_2$ in(IV)

    $\begin{array}{c}{C}_{eq}={\left[\frac{1}{12\text{μF}+2\text{3μF}}+\frac{1}{46\text{μF}}\right]}^{-1}\\ =20\text{μF}\end{array}$

The equivalent circuit with the equivalent resistance and capacitance values is shown below.   

    fig

(b)

To determine

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

Answer to Problem 88P

The charge on the capacitor is $\underset{\_}{\text{4}{\text{.1×10}}^{\text{-5}}\text{C}}$ and current on the resistor is zero.

Explanation of Solution

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

Write the equation for charge on equivalent capacitor

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

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

Conclusion:

Substitute $\text{6V}$ for $\Delta V$, $20\times {10}^{-6}\text{F}$ for $C_{eq}$ in (V)

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

This charge is same across $C_3$. Then the voltage across $C_3$

    $\begin{array}{c}{V}_3=\frac{Q_3}{C_3}\\ =\frac{1.2\times {10}^{-4}\text{C}}{46\times {10}^{-6}\text{F}}\\ =2.61\text{V}\end{array}$

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

    $\text{6V-2}\text{.61V=3}\text{.39V}$

Then the charge on the capacitor $C_1$ is,

    $\begin{array}{c}{Q}_1=C_1{V}_1\\ =\left({\text{12×10}}^{\text{-6}}\text{F}\right)\left(3.39\text{V}\right)\\ =4.1\times {10}^{-5}\text{C}\end{array}$

The charge on the capacitor is $\underset{\_}{\text{4}{\text{.1×10}}^{\text{-5}}\text{C}}$ and current on the resistor is zero.

(c)

To determine

Time constant of the circuit.

.

Answer to Problem 88P

Time constant of the circuit is $\underset{\_}{\text{1}\text{.2}\text{ms}}$

Explanation of Solution

Write the equation for time constant

    $\tau =RC$              

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

Conclusion:    

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

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

Time constant of the circuit is $\underset{\_}{\text{1}\text{.2}\text{ms}}$