Chapter 28, Problem 16P

(a)

To determine

The potential difference across each resistor in terms of $\epsilon$.

Answer to Problem 16P

The potential difference across resistor $R_1$, $R_2$, $R_3$ and $R_4$ is $\frac{\epsilon }{3}$, $\frac{2\epsilon }{3}$, $\frac{4\epsilon }{9}$ and $\frac{2\epsilon }{3}$ respectively.

Explanation of Solution

Consider the circuit diagram as shown below.

Figure-(1)

Write the expression to calculate the resistance in loop 1.

    $R_{\text{s}}=R_2+R_3$

Here, $R_{\text{s}}$ is the equivalent resistance in series, $R_2$ and $R_3$ are the resistors.

Substitute $2R$ for $R_2$ and $4R$ for $R_3$ in the above equation to calculate $R_{\text{p}}$.

    $\begin{array}{c}{R}_{\text{s}}=2R+4R\\ =6R\end{array}$

Write the expression to calculate the equivalent resistance in loop 1.

    $\frac{1}{R_{\text{eq1}}}=\frac{1}{R_{\text{s}}}+\frac{1}{R_4}$

Here, $R_{\text{eq1}}$ is the equivalent resistance in loop 1, $R_{\text{s}}$ and $R_4$ are the resistors.

Substitute $\left(6R\right)$ for $R_{\text{p}}$ and $3R$ for $R_4$ in the above equation to calculate $R_{\text{eq1}}$.

    $\begin{array}{c}\frac{1}{R_{\text{eq1}}}=\frac{1}{6R}+\frac{1}{3R}\\ R_{\text{eq1}}=\frac{\left(6R\right)\left(3R\right)}{6R+3R}\\ =\frac{18R^2}{9R}\\ =2R\end{array}$

Write the expression to calculate the equivalent current of the circuit.

    $R_{\text{eq}}=R_{\text{eq1}}+R_1$

Here, $R_{\text{eq}}$ is the equivalent resistance in the circuit, $R_{\text{eq1}}$ is the equivalent resistance in loop 1 and $R_1$ is the resistor.

Substitute $2R$ for $R_{\text{eq1}}$ and $R$ for $R_1$ in the above equation to calculate $R_{\text{eq}}$.

    $\begin{array}{c}{R}_{\text{eq}}=2R+R\\ =3R\end{array}$

Write the expression to calculate the emf of the battery.

    $\begin{array}{l}\epsilon =I{R}_{\text{eq}}\\ I=\frac{\epsilon }{R_{\text{eq}}}\end{array}$

Here, $I$ is the current, $\epsilon$ is the emf and $R_{\text{eq}}$ is the equivalent resistance.

Substitute $3R$ for $R_{\text{eq}}$ in the above equation to calculate $I$.

    $I=\frac{\epsilon }{3R}$

Write the expression to calculate the potential drop across resistor $1$.

    $V_1=I{R}_1$

Here, $V_1$ is the potential drop across resistor $1$ and $I$ is the current.

Substitute $\left(\frac{\epsilon }{3R}\right)$ for $I$ in the above equation to calculate $V_1$.

    $\begin{array}{c}{V}_1=\left(\frac{\epsilon }{3R}\right)R\\ =\frac{\epsilon }{3}\end{array}$

Write the expression to calculate the potential drop across $R_4$.

    $V_4=\epsilon -V_1$

Here, $V_4$ is the potential drop across resistor $R_4$.

Substitute $\left(\frac{\epsilon }{3}\right)$ for $V_1$ in the above equation to calculate $V_4$.

    $\begin{array}{c}{V}_4=\epsilon -\frac{\epsilon }{3}\\ =\frac{2\epsilon }{3}\end{array}$

Since resistors $R_2$ and $R_3$ are connected in series the potential drop across resistor $R_4$ will be shared among resistors $R_2$ and $R_3$.

Write the expression to calculate the potential drop across resistor $R_2$.

    $V_2=\frac{1}{3}{V}_4$

Here, $V_2$ is the potential drops across resistor $R_2$.

Substitute $\left(\frac{2\epsilon }{3}\right)$ for $V_4$ in the above equation to calculate $V_2$.

    $\begin{array}{c}{V}_2=\frac{1}{3}\left(\frac{2\epsilon }{3}\right)\\ =\frac{2\epsilon }{9}\end{array}$

Write the expression to calculate the potential drop across resistor $R_3$.

    $V_3=\frac{2}{3}{V}_4$

Here, $V_3$ is the potential drop across resistor $R_3$.

Conclusion:

Substitute $\left(\frac{2\epsilon }{3}\right)$ in the above equation to calculate $V_3$.

    $\begin{array}{c}{V}_3=\frac{2}{3}\left(\frac{2\epsilon }{3}\right)\\ =\frac{4\epsilon }{9}\end{array}$

Therefore, the potential difference across resistor $R_1$, $R_2$, $R_3$ and $R_4$ is $\frac{\epsilon }{3}$, $\frac{2\epsilon }{3}$, $\frac{4\epsilon }{9}$ and $\frac{2\epsilon }{3}$ respectively.

(b)

To determine

The current in each resistor in terms of $I$.

Answer to Problem 16P

The current through resistor $R_1$ is $I$, $R_2$ is $\frac{I}{3}$, $R_3$ is $\frac{I}{3}$ and $R_4$ is $\frac{2I}{3}$.

Explanation of Solution

Write the expression to calculate the current through resistor $R_1$.

    $I_1=\frac{V_1}{R_1}$

Here, $I_1$ is the current through resistor $R_1$.

Substitute $\left(\frac{\epsilon }{3}\right)$ for $V_1$ and $3R$ for $R$ in the above equation to calculate $I_1$.

    $\begin{array}{c}{I}_1=\frac{\left(\frac{\epsilon }{3}\right)}{3R}\\ I_1=\frac{\epsilon }{3R}\\ I_1=I\end{array}$

Write the expression to calculate the current through resistor $R_2$.

    $I_2=\frac{V_2}{R_2}$

Here $I_2$ is the current through resistor $R_2$.

Substitute $\left(\frac{2\epsilon }{9}\right)$ for $V_2$ and $2R$ for $R_2$.

    $\begin{array}{c}{I}_2=\frac{\left(\frac{2\epsilon }{9}\right)}{2R}\\ =\frac{1}{3}\left(\frac{\epsilon }{3r}\right)\\ =\frac{I}{3}\end{array}$

Write the expression to calculate the current through resistor $R_3$.

    $I_3=\frac{V_3}{R_3}$

Here, $I_3$ is the current through resistor $I_3$.

Substitute $\left(\frac{4\epsilon }{9}\right)$ for $V_3$ and $4R$ for $R_3$.

    $\begin{array}{c}{I}_3=\frac{\left(\frac{4\epsilon }{9}\right)}{4R}\\ =\frac{1}{3}\left(\frac{\epsilon }{3r}\right)\\ =\frac{I}{3}\end{array}$

Write the expression to calculate the current through resistor $R_4$.

    $I_4=\frac{V_4}{R_4}$

Here, $I_4$ is the current through resistor $R_4$.

Conclusion:

Substitute $\frac{2\epsilon }{3}$ for $V_4$ and $3R$ for $R_4$ in the above equation to calculate $I_4$.

    $\begin{array}{c}{I}_4=\frac{\frac{2\epsilon }{3}}{3R}\\ =\frac{2}{3}\left(\frac{\epsilon }{3R}\right)\\ =\frac{2I}{3}\end{array}$

Therefore, the current through resistor $R_1$ is $I$, $R_2$ is $\frac{I}{3}$, $R_3$ is $\frac{I}{3}$ and $R_4$ is $\frac{2I}{3}$.

(c)

To determine

The current in each of the resistors if $R_3$ is increased.

Answer to Problem 16P

Current through resistors $R_1$, $R_2$ and $R_3$ decreases while current through resistor $R_4$ increases.

Explanation of Solution

If the resistance $R_3$ increases, the equivalent resistance of the total circuit increases. Due to this the current through resistor $R_1$ decreases. As the current decreases the potential drop also decreases. Resistors $R_2$, $R_3$ and $R_4$ are connected in parallel so potential drop increases across these resistors.

As a result the current through resistor $R_2$ and $R_3$ decreases when the current across the resistor $R_3$ is increased and current through resistor $R_4$ increases.

(d)

To determine

The new values of the current in each resistor in terms of $I$ if $R_3\to \infty$.

Answer to Problem 16P

The current through resistors $R_1$, $R_2$, $R_3$ and $R_4$ is $\frac{3I}{4}$, $0$, $0$ and $\frac{3I}{4}$ respectively.

Explanation of Solution

Write the expression to calculate the equivalent resistance.

    $R_{\text{eq}}=R_1+R_4$

Substitute $R$ for $R_1$ and $3R$ for $R_1$ in the above equation to calculate $R_{\text{eq}}$.

    $\begin{array}{c}{R}_{\text{eq}}=R+3R\\ =4R\end{array}$

Write the expression to calculate the new current.

    $I\text{'}=\frac{\epsilon }{R_{\text{eq}}}$

Here, $I\text{'}$ is the new current.

Substitute $4R$ for $R_{\text{eq}}$ and $3IR$ for $\epsilon$.

    $\begin{array}{c}I\text{'}=\frac{3IR}{4R}\\ =\frac{3I}{4}\end{array}$

Since, resistors $R_1$ and $R_2$ are connected in series, the current across the two resistors remains the same.

    $I_1^{\text{'}}=\frac{3I}{4}$

Write the expression to calculate the current across resistor $R_4$.

    $I_4^{\text{'}}=I_2^{\text{'}}$

Conclusion:

Substitute the value in the above expression to calculate $I_4^{\text{'}}$.

    $I_4^{\text{'}}=\frac{3I}{4}$

Therefore, the current through resistors $R_1$, $R_2$, $R_3$ and $R_4$ is $\frac{3I}{4}$, $0$, $0$ and $\frac{3I}{4}$ respectively.