What is the equivalent resistance between points a and b in FIGURE EX28.26?

We have to find the equivalent resistance between points a and b.

The resistors are connected one by one and with no junctions between them. they are called to be connected in series. For a series resistors, the equivalent resistance for their combination is given by equation

$R_{eq}=R_1+R_2...\left(1\right)$

As shown in the given figure, the two resistors $10\Omega and40\Omega$in the left branch are connected in series. So, use equation (1) to get the equivalent resistance in the left branch by

$R_{left}=R_1+R_2=10\Omega +40\Omega =50\Omega$

In right branch, the two resistors $55\Omega and20\Omega$are connected in series. So, use equation (1) to get the equivalent resistance in the right branch

$R_{right}=R_3+R_4=55\Omega +20\Omega =75\Omega$

The two resistors $R_{right}and{R}_{left}$are connected in parallel, so let us find the equivalent resistance $R_{eq}$for this combination using equation $\left(1\right)$by

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_{right}}+\hspace{0.17em}\frac{1}{R_{left}} \\ \frac{1}{R_{eq}}=\frac{1}{75\Omega }+\hspace{0.17em}\frac{1}{50\Omega } \\ {R}_{eq}={\left(\frac{1}{75\Omega }+\hspace{0.17em}\frac{1}{50\Omega }\right)}^{-1} \\ {R}_{eq}=30\Omega \end{gathered}$$

The resistance $R_5=10\Omega$at point (b) is in series with the combination $R_{eq}$, so we get equivalent resistance $R_{ab}$by

$R_{ab}=R_{eq}+R_5=30\Omega +10\Omega =40\Omega$