Chapter 17, Problem 71E

(a)

To determine

The current in the EMF in the circuit shown in Figure 17.31.

Answer to Problem 71E

The current in the EMF in the circuit shown in Figure 17.31 is $\underset{\_}{\epsilon /r}$.

Explanation of Solution

The resistor network is shown in Figure 17.31.

The resistances are labelled as $R_1$, $R_2$, $R_3$, $R_4$, $R_5$, $R_6$, and $R_7$ as shown in Figure 1.

The resistors $R_1$, $R_2$, and $R_3$ are in parallel, so that their total resistance can be found out as,

  $\frac{1}{R_{1,2,3}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$        (I)

The resistor $R_4$ and the combination $R_{1,2,3}$ are in series. Hence their total resistance can be found out as,

  $R_{1,2,3,4}=R_{1,2,3}+R_4$        (II)

The resistors $R_5$, $R_6$, are in parallel. So their net resistance is given by,

  $R_{5,6}=\frac{R_5{R}_6}{R_5+R_6}$        (III)

The resistor $R_7$ and $R_{5,6}$ are in series. Hence their total resistance is,

  $R_{5,6,7}=R_{5,6}+R_7$        (IV)

The combinations $R_{1,2,3,4}$ and $R_{4,5,7}$ are in parallel. Thus, the equivalent resistance of the network can be found out as,

  $R_{\text{equivalent}}=\frac{R_{1,2,3,4}{R}_{5,6,7}}{R_{1,2,3,4}+R_{5,6,7}}$        (V)

Write the expression for the current in the EMF.

  $I=\frac{\epsilon }{R_{\text{equivalent}}}$        (VI)

Here, $I$ is the current, and $\epsilon$ is the EMF.

Conclusion:

Substitute $2r$ for $R_1$, $2r$ for $R_2$, and $r$ for $R_3$ in equation (I) and solve for $R_{1,2,3}$.

  $\begin{array}{c}\frac{1}{R_{1,2,3}}=\frac{1}{2r}+\frac{1}{2r}+\frac{1}{r}\\ \frac{1}{R_{1,2,3}}=\frac{4}{2r}\\ R_{1,2,3}=\frac{r}{2}\end{array}$

Substitute $\frac{r}{2}$ for $R_{1,2,3}$, and $\frac{3}{2}r$ for $R_4$ in equation (II) to find $R_{1,2,3,4}$.

  $\begin{array}{c}{R}_{1,2,3,4}=\frac{r}{2}+\frac{3}{2}r\\ =2r\end{array}$

Substitute $r$ for $R_5$ and $R_6$ in equation (III) to find $R_{5,6}$.

  $\begin{array}{c}{R}_{5,6}=\frac{\left(r\right)\left(r\right)}{r+r}\\ =\frac{r}{2}\end{array}$

Substitute $\frac{r}{2}$ for $R_{5,6}$, and $\frac{3}{2}r$ for $R_7$ in equation (IV) to find $R_{5,6,7}$.

  $\begin{array}{c}{R}_{5,6,7}=\frac{r}{2}+\frac{3}{2}r\\ =2r\end{array}$

Substitute $2r$ for $R_{1,2,3,4}$, and $2r$ for $R_{5,6,7}$ in equation (V) to find $R_{\text{equivalent}}$.

  $\begin{array}{c}{R}_{\text{equivalent}}=\frac{\left(2r\right)\left(2r\right)}{2r+2r}\\ =r\end{array}$

Substitute $r$ for $R$ in equation (VI) to find $I$.

  $I=\frac{\epsilon }{r}$

Therefore, the current in the EMF in the circuit shown in Figure 17.31 is $\underset{\_}{\epsilon /r}$.

(b)

To determine

The current in the lower $\frac{3}{2}r$ resistor.

Answer to Problem 71E

The current in the lower $\frac{3}{2}r$ resistor is $\underset{\_}{\epsilon /2r}$.

Explanation of Solution

The lower $\frac{3}{2}r$ resistor is represented by $R_7$ in the Figure 1.

Since the lower and upper branches of the network are connected in parallel, the potential difference across both the branches are same and equal to $\epsilon$. Since $R_{5,6}$ and $R_7$ are in series, the current through both will be same. It can be computed as,

  $I_7=\frac{\epsilon }{R_{5,6,7}}$        (VII)

Here, $I_7$ is the current through $R_7$.

Conclusion:

Substitute $2r$ for $R_{5,6,7}$ in equation (VII) to find $I_7$.

  $I_7=\frac{\epsilon }{2r}$

Therefore, the current in the lower $\frac{3}{2}r$ resistor is $\underset{\_}{\epsilon /2r}$.