Chapter 28, Problem 43P

(a)

To determine

The potential difference across the capacitor.

Answer to Problem 43P

The potential difference across the capacitor is $6.00\text{V}$.

Explanation of Solution

Let the resistance across $1.00\Omega$ be $R_1$, across $4.00\Omega$ be $R_2$, across $8.00\Omega$ be $R_3$ and across $2.00\Omega$ be $R_4$.

When the circuit is connected with a voltage source, the capacitor starts to charge. So, it acts as an open circuit as shown in figure below.

Figure (1)

Write the expression for the equivalent resistance when the resistors are connected in series.

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

Here, the equivalent resistance is $R_{\text{eq}}$ and the resistors connected in series are $R_1$ and $R_2$.

Write the expression to current through the series connection..

    $I_1=\frac{V}{R_{\text{eq}}}$                                                                                                                    (II)

Here, $I_1$ is the current flowing through the resistors in series and $V$ is the potential difference of the battery.

Write the expression to determine the potential difference $V_1$ in the circuit shown in figure (1)

    $V_1=V-\left(I_1{R}_1\right)$                                                                                                        (III)

Write the expression to determine the potential difference $V_2$ in the circuit shown in figure (1)

 using the voltage division rule.

    $V_2=\left(\frac{R_4}{R_3+R_4}\right)V$                                                                                                    (IV)

Write the expression to determine the potential difference across the capacitor.

    $V_{\text{C}}=V_1-V_2$                                                                                                              (V)

Here, $V_{\text{C}}$ is the expression to determine the potential difference across the capacitor.

Conclusion:

Substitute $1.00\Omega$ for $R_1$ and $4.00\Omega$ for $R_2$ in equation (I) to solve for $R_{\text{eq}}$.

    $\begin{array}{c}{R}_{\text{eq}}=1.00\Omega +4.00\Omega \\ =5.00\Omega \end{array}$

Substitute $5.00\Omega$ for $R_{\text{eq}}$ and $10.0\text{V}$ for $V$ in equation (II) to solve for $I_1$.

    $\begin{array}{c}{I}_1=\frac{10.0\text{V}}{5.00\Omega }\\ =2.00\text{A}\end{array}$

Substitute $1.00\Omega$ for $R_1$, $10\text{V}$ for $V$ and $2.00\text{A}$ for $I_1$ in equation (III) to solve for $V_1$.

    $\begin{array}{c}{V}_1=10\text{V}-\left(\left(2.00\text{A}\right)1.00\Omega \right)\\ =8\text{V}\end{array}$

Substitute $2.00\Omega$ for $R_4$, $10\text{V}$ for $V$ and $8.00\Omega$ for $R_3$ in equation (IV) to solve for $V_2$.

    $\begin{array}{c}{V}_2=\left(\frac{2.00\Omega }{8.00\Omega +2.00\Omega }\right)\left(10\text{V}\right)\\ =2\text{V}\end{array}$

Substitute $2\text{V}$ for $V_2$ and $8\text{V}$ for $V_1$ in equation (V) to solve for $V_{\text{C}}$.

    $\begin{array}{c}{V}_{\text{C}}=8\text{V}-2\text{V}\\ =6.\text{00}\text{V}\end{array}$

Therefore, the potential difference across the capacitor is $6.00\text{V}$.

(b)

To determine

The time interval at which the capacitor discharges to one tenth of its initial voltage.

Answer to Problem 43P

The time interval at which the capacitor discharges to one tenth of its initial voltage is $8.29\mu \text{s}$.

Explanation of Solution

When the battery is disconnected, the capacitor starts discharging acting as a voltage source.

The circuit diagram is shown below after the battery is disconnected.

Figure (2)

The resistors $R_1$ and $R_3$ are in series.

Write the expression for the equivalent resistance when the resistors are connected in series.

    $R_{\text{eq}}{}^{\prime }=R_1+R_3$                                                                                                       (VI)

Here, the equivalent resistance is $R_{\text{eq}}{}^{\prime }$.

Also, the resistors $R_2$ and $R_4$ are in series.

Write the expression for the equivalent resistance when the resistors are connected in series.

    $R_{\text{eq}}{}^{\prime \text{}\prime }=R_2+R_4$                                                                                                    (VII)

Here, the equivalent resistance is $R_{\text{eq}}{}^{\prime \text{}\prime }$.

Now the resistors $R_{\text{eq}}{}^{\prime }$ and $R_{\text{eq}}{}^{\prime \text{}\prime }$ are in parallel.

Write the expression for the equivalent resistance when the resistors are connected in parallel.

    $R_{\text{P}}=R_{\text{eq}}{}^{\prime }+R_{\text{eq}}{}^{\prime \text{}\prime }$                                                                                                 (VIII)

Here, the equivalent resistance for the parallel connection is $R_{\text{P}}$.

Write the expression to calculate the time interval at which the capacitor discharges to one tenth of its initial voltage.

    $V\left(t\right)=V_0{e}^{-\left(\frac{t}{R_{\text{P}}C}\right)}$                                                                                                   (IX)

Here, $V\left(t\right)$ is voltage across the capacitor after disconnection, $V_0$ is the initial voltage and $t$ is the time interval.

Substitute $\frac{V_0}{10}$ for $V\left(t\right)$ in equation (IX).

    $\begin{array}{c}\frac{V_0}{10}=V_0{e}^{-\left(\frac{t}{R_{\text{P}}C}\right)}\\ 1=10e^{-\left(\frac{t}{R_{\text{P}}C}\right)}\\ \ln 1=\ln 10+\left(-\left(\frac{t}{R_{\text{P}}C}\right)\right)\\ \left(\frac{t}{R_{\text{P}}C}\right)=2.302\end{array}$                                                                                (X)

Conclusion:

Substitute $1.00\Omega$ for $R_1$ and $8.00\Omega$ for $R_3$ in equation (VI) to solve for $R_{\text{eq}}{}^{\prime }$.

    $\begin{array}{c}{R}_{\text{eq}}{}^{\prime }=1.00\Omega +8.00\Omega \\ =9.00\Omega \end{array}$

Substitute $2.00\Omega$ for $R_1$ and $4.00\Omega$ for $R_3$ in equation (VII) to solve for $R_{\text{eq}}{}^{\prime \text{}\prime }$.

    $\begin{array}{c}{R}_{\text{eq}}{}^{\prime \text{}\prime }=2.00\Omega +4.00\Omega \\ =6.00\Omega \end{array}$

Substitute $9.00\Omega$ for $R_{\text{eq}}{}^{\prime }$ and $6.00\Omega$ for $R_{\text{eq}}{}^{\prime \text{}\prime }$ in equation (VIII) to solve for $R_{\text{P}}$.

    $\begin{array}{c}\frac{1}{R_{\text{P}}}=\frac{1}{9.00\Omega }+\frac{1}{6.00\Omega }\\ R_{\text{P}}=\frac{\left(9.00\Omega \right)\left(6.00\Omega \right)}{9.00\Omega +6.00\Omega }\\ =3.60\Omega \end{array}$

Substitute $3.60\Omega$ for $R_{\text{P}}$, and $1.00\mu \text{F}$ for $C$ in equation (X) to solve for $t$.

    $\begin{array}{c}\left(\frac{t}{\left(3.60\Omega \right)\left(1.00\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)}\right)=2.302\\ t=8.29\times {10}^{-6}\text{s}\left(\frac{{10}^6\mu \text{s}}{1\text{s}}\right)\\ t=8.29\mu \text{s}\end{array}$

Therefore, the time interval at which the capacitor discharge to one tenth of its initial voltage is $8.29\mu \text{s}$.