Chapter 17, Problem 70E

To determine

The equivalent resistance of the network shown in Figure 17.30.

Answer to Problem 70E

The equivalent resistance of the network shown in Figure 17.30 is $\underset{\_}{4r}$.

Explanation of Solution

The resistor network is shown in Figure 17.30.

The resistances are labelled as $R_1$, $R_2$, $R_3$, $R_4$, $R_5$, $R_6$, $R_7$, $R_8$, and $R_9$ as shown in Figure 1.

The resistors $R_7$, $R_8$, and $R_9$ are in series, so that their total resistance is given by,

  $R_{7,8,9}=R_7+R_8+R_9$        (I)

The resistor $R_6$ and the combination $R_{7,8,9}$ are in parallel with total resistance,

  $R_{6,7,8,9}=\frac{R_6{R}_{7,8,9}}{R_6+R_{7,8,9}}$        (II)

The resistors $R_4$, $R_5$, and $R_{6,7,8,9}$ are in series. So their net resistance is given by,

  $R_{4,5,6,7,8,9}=R_4+R_5+R_{6,7,8,9}$        (III)

The resistor $R_3$ and $R_{4,5,6,7,8,9}$ are in parallel, with total resistance,

  $R_{3,4,5,6,7,8,9}=\frac{R_3{R}_{4,5,6,7,8,9}}{R_3+R_{4,5,6,7,8,9}}$        (IV)

The resistors $R_1$, $R_2$, and $R_{3,4,5,6,7,8,9}$ are in series. Thus, the equivalent resistance of the network is given by,

  $R_{\text{equivalent}}=R_1+R_2+R_{3,4,5,6,7,8,9}$        (V)

Conclusion:

Substitute $2r$ for $R_7$, $R_8$, and $R_9$ in equation (I) to find $R_{7,8,9}$.

  $\begin{array}{c}{R}_{7,8,9}=2r+2r+2r\\ =6r\end{array}$

Substitute $3r$ for $R_6$, and $6_r$ for $R_{7,8,9}$ in equation (II) to find $R_{6,7,8,9}$.

  $\begin{array}{c}{R}_{6,7,8,9}=\frac{\left(3r\right)\left(6r\right)}{3r+6r}\\ =2r\end{array}$

Substitute $2r$ for $R_4$, $R_5$, and $R_{6,7,8,9}$ in equation (III) to find $R_{4,5,6,7,8,9}$.

  $\begin{array}{c}{R}_{4,5,6,7,8,9}=2r+2r+2r\\ =6r\end{array}$

Substitute $3r$ for $R_3$,and $6r$ for $R_{4,5,6,7,8,9}$ in equation (IV) to find $R_{3,4,5,6,7,8,9}$.

  $\begin{array}{c}{R}_{3,4,5,6,7,8,9}=\frac{\left(3r\right)\left(6r\right)}{3r+6r}\\ =2r\end{array}$

Substitute $r$ for $R_1$ and $R_2$, and $2r$ for $R_{3,4,5,6,7,8,9}$ in equation (V) to find $R_{\text{equivalent}}$.

  $\begin{array}{c}{R}_{\text{equivalent}}=r+r+2r\\ =4r\end{array}$

Therefore, the equivalent resistance of the network shown in Figure 17.30 is $\underset{\_}{4r}$.