Chapter 28, Problem 70AP

(a)

To determine

The equilibrium charge on the capacitor as a function of $R$ for the given circuit.

Answer to Problem 70AP

The equilibrium charge on the capacitor as a function of $R$ for the given circuit is $\left(\left(14.4\times {10}^{-6}\text{F}\right)-\left(\frac{\left(\text{15}\times {\text{10}}^{-6}R\right)\text{F}}{2.00\Omega +R}\right)\right)$.

Explanation of Solution

Let the resistance across $3.00\Omega$ be $R_1$, across $80.0\Omega$ be $R_2$ and across $2.00\Omega$ be $R_3$.

The resistors $R_1$ and $R_2$ are in series connection.

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 across $R_2$.

    $V_1=\left(I_1{R}_2\right)$                                                                                                              (III)

Here, $V_1$ is the potential difference across resistor $R_2$.

The resistors $R_3$ and $R$ are in series connection.

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

    $R_{\text{eq}}{}^{\prime }=R_3+R$                                                                                                           (IV)

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

Write the expression to current through the series connection.

    $I_2=\frac{V}{R_{\text{eq}}{}^{\prime }}$                                                                                                                   (V)

Here, $I_2$ is the current flowing through the resistors in series.

Write the expression to determine the potential difference across $R$.

    $V_2=\left(I_{2}R\right)$                                                                                                               (VI)

Here, $V_2$ is the potential difference across resistor $R$.

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

    $V_{\text{C}}=\left|V_1-V_2\right|$                                                                                                          (VII)

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

Write the expression to calculate the amount of charge stored in the capacitor.

    $Q=V_{\text{C}}C$                                                                                                               (VIII)

Here, $Q$ is the amount of charge stored in the capacitor and $C$ is capacitance of the capacitor.

Conclusion:

Substitute $3.00\Omega$ for $R_1$ and $80.0\Omega$ for $R_2$ in equation (I) to solve for $R_{\text{eq}}$.

    $\begin{array}{c}{R}_{\text{eq}}=3.00\Omega +80.0\Omega \\ =83.0\Omega \end{array}$

Substitute $83.0\Omega$ for $R_{\text{eq}}$ and $5.00\text{V}$ for $V$ in equation (II) to solve for $I_1$.

    $\begin{array}{c}{I}_1=\frac{5.0\text{V}}{83.0\Omega }\\ =0.06\text{A}\end{array}$

Substitute $80.0\Omega$ for $R_2$, and $0.06\text{A}$ for $I_1$ in equation (III) to solve for $V_1$

    $\begin{array}{c}{V}_1=\left(\left(0.06\text{A}\right)80.0\Omega \right)\\ =4.80\text{V}\end{array}$

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

    $R_{\text{eq}}{}^{\prime }=2.00\Omega +R$

Substitute $2.00\Omega +R$ for $R_{\text{eq}}{}^{\prime }$ and $5.00\text{V}$ for $V$ in equation (V) to solve for $I_2$.

    $I_2=\frac{5.00\text{V}}{2.00\Omega +R}$

Substitute $\frac{5.00\text{V}}{2.00\Omega +R}$ for $I_2$ in equation (VI) to solve for $V_2$.

    $V_2=\left(\frac{5.00\text{V}}{2.00\Omega +R}\left(R\right)\right)$

Substitute $\left(\frac{5.00\text{V}}{2.00\Omega +R}\left(R\right)\right)$ for $V_2$ and $4.80\text{V}$ for $V_1$ in equation (VII) to solve for $V_{\text{C}}$.

    $V_{\text{C}}=\left|4.80\text{V}-\left(\frac{5.00\text{V}}{2.00\Omega +R}\left(R\right)\right)\right|$

Substitute $\left|4.80\text{V}-\left(\frac{5.00\text{V}}{2.00\Omega +R}\left(R\right)\right)\right|$ for $V_{\text{C}}$ and $3.00\mu \text{F}$ for $C$ in equation (VIII) to solve for $Q$.

    $\begin{array}{c}Q=\left|4.80\text{V}-\left(\frac{5.00\text{V}}{2.00\Omega +R}\left(R\right)\right)\right|\left(3.00\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)\\ =\left(\left(14.4\times {10}^{-6}\text{F}\right)-\left(\frac{\left(\text{15}\times {\text{10}}^{-6}R\right)\text{F}}{2.00\Omega +R}\right)\right)\end{array}$

Therefore, the equilibrium charge on the capacitor as a function of $R$ for the given circuit is $\left(\left(14.4\times {10}^{-6}\text{F}\right)-\left(\frac{\left(\text{15}\times {\text{10}}^{-6}R\right)\text{F}}{2.00\Omega +R}\right)\right)$.

(b)

To determine

The charge when $R=10\Omega$.

Answer to Problem 70AP

The charge when $R=10\Omega$ is $1.9\mu \text{C}$.

Explanation of Solution

Write the expression for the equilibrium charge on the capacitor as a function of $R$ for the given circuit.

    $Q=\left(\left(14.4\times {10}^{-6}\text{F}\right)-\left(\frac{\left(\text{15}\times {\text{10}}^{-6}R\right)\text{F}}{2.00\Omega +R}\right)\right)$                                                                (IX)

Conclusion:

Substitute $10.0\Omega$ for $R$ in equation (IX) to solve for $Q$.

    $\begin{array}{c}Q=\left(\left(14.4\times {10}^{-6}\text{F}\right)-\left(\frac{\left(\left(\text{15}\times {\text{10}}^{-6}\text{F}\right)\left(10.0\Omega \right)\right)}{2.00\Omega +10.0\Omega }\right)\right)\\ =1.9\times {10}^{-6}\text{C}\times \frac{{10}^6\mu \text{C}}{1\text{C}}\\ =1.9\mu \text{C}\end{array}$

Therefore, the charge when $R=10\Omega$ is $1.9\mu \text{C}$.

(c)

To determine

Whether the charge on the capacitor can be zero and the value of $R$ when the charge is $0$.

Answer to Problem 70AP

Yes, the charge on the capacitor can be zero when the value of $R$ is $48\Omega$.

Explanation of Solution

Write the expression for the equilibrium charge on the capacitor as a function of $R$ for the given circuit.

    $Q=\left(\left(14.4\times {10}^{-6}\text{F}\right)-\left(\frac{\left(\text{15}\times {\text{10}}^{-6}R\right)\text{F}}{2.00\Omega +R}\right)\right)$                                                                 (X)

Conclusion:

Yes, the charge on the capacitor can be zero.

Substitute $0$ for $Q$ in equation (X) to solve for $R$.

    $\begin{array}{c}0=\left(\left(14.4\times {10}^{-6}\text{F}\right)-\left(\frac{\left(\text{15}\times {\text{10}}^{-6}R\right)\text{F}}{2.00\Omega +R}\right)\right)\\ \left(\frac{\left(\text{15}\times {\text{10}}^{-6}R\right)\text{F}}{2.00\Omega +R}\right)=\left(14.4\times {10}^{-6}\text{F}\right)\\ \left(\text{15}\times {\text{10}}^{-6}R\right)\text{F}=\left(2.00\Omega +R\right)\left(14.4\times {10}^{-6}\text{F}\right)\\ \left(15-14.4\right)\times {10}^{-6}R=28.8\times {10}^{-6}\end{array}$

Solve further.

    $\begin{array}{c}0.6R=28.8\\ R=48\Omega \end{array}$

Therefore, yes, the charge on the capacitor can be zero when the value of $R$ is $48\Omega$.

(d)

To determine

The maximum possible value of the magnitude of charge and the value of $R$ at which this is achieved.

Answer to Problem 70AP

The maximum possible value of the magnitude of charge is $14.4\mu \text{C}$ and the value of $R$ at which this is achieved is $0.00\Omega$.

Explanation of Solution

Write the expression for the equilibrium charge on the capacitor as a function of $R$ for the given circuit.

    $Q=\left(\left(14.4\times {10}^{-6}\text{F}\right)-\left(\frac{\left(\text{15}\times {\text{10}}^{-6}R\right)\text{F}}{2.00\Omega +R}\right)\right)$                                                                 (XI)

It is clear from equation (XI) that the maximum charge can be achieved when the term containing $R$ is reduced to zero.

This can be achieved by substituting zero for $R$.

Conclusion:

Substitute $0$ for $R$ in equation (XI) to solve for $Q_{\max }$.

    $\begin{array}{c}{Q}_{\max }=\left(\left(14.4\times {10}^{-6}\text{F}\right)-\left(\frac{\left(\text{15}\times {\text{10}}^{-6}\left(0\right)\right)\text{F}}{2.00\Omega +\left(0\right)}\right)\right)\\ =\left(14.4\times {10}^{-6}\text{C}\times \frac{{10}^6\mu \text{C}}{1\text{C}}\right)\\ =\left(14.4\mu \text{C}\right)\end{array}$

Therefore, The maximum possible value of magnitude of charge is $14.4\mu \text{C}$ and the value of $R$ at which this is achieved is $0.00\Omega$.

(e)

To determine

Whether it is experimentally meaningful to take $R=\infty$ and if yes, then the magnitude of charge if $R=\infty$.

Answer to Problem 70AP

It is experimentally not meaningful to take $R=\infty$.

Explanation of Solution

Write the expression for the potential difference across $R$.

    $V_2=\left(\frac{5.00\text{V}}{2.00\Omega +R}\left(R\right)\right)$                                                                                           (XII)

Conclusion:

Substitute $\infty$ for $R$ in equation (XII) to solve for $V_2$.

    $\begin{array}{c}{V}_2=\left(\frac{5.00\text{V}}{2.00\Omega +\infty }\left(\infty \right)\right)\\ V_2=\infty \end{array}$

Thus, this infinite value of voltage across the resistor $R$ shall lead to an infinite charge stored in the capacitor which is not possible.

Therefore, it is experimentally not meaningful to take $R=\infty$.