91Ó°ÊÓ

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

The two points and there are resistors connected at both ends of the points, we called that the resistors are connected in parallel. This is obvious when the resistors are aligned side by side. The potential difference across the resistors is the same in parallel connection.

The equation shows the equivalent resistance for the parallel connection in the form

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

The two resistors $60\mathrm{Î\copyright }$and $40\mathrm{Î\copyright }$and are connected in parallel, so find the equivalent resistance$R_{1,eq}$for this combination using equation (1) by

$$\begin{gathered} \frac{1}{R_{1,eq}}=\frac{1}{60\mathrm{Î\copyright }}+\frac{1}{40\mathrm{Î\copyright }} \\ {R}_{1,eq}={\left(\frac{1}{60\mathrm{Î\copyright }}+ \frac{1}{40\mathrm{Î\copyright }}\right)}^{-1} \\ {R}_{1,eq}=24\mathrm{Î\copyright } \end{gathered}$$

There are three resistors $60\mathrm{Î\copyright },60\mathrm{Î\copyright }and45\mathrm{Î\copyright }$ , and are connected in parallel, so let us find the equivalent resistance $R_{2,eq}$for this combination using equation (1) by

$$\begin{gathered} R_{2,eq}=\frac{1}{60\mathrm{Î\copyright }}+\frac{1}{40\mathrm{Î\copyright }}+\frac{1}{45\mathrm{Î\copyright }} \\ {R}_{2,eq}={\left(\frac{1}{60\mathrm{Î\copyright }}+\frac{1}{40\mathrm{Î\copyright }}+\frac{1}{45\mathrm{Î\copyright }}\right)}^{-1} \\ {R}_{2,eq}=18\mathrm{Î\copyright } \end{gathered}$$

In these combinations, we conclude that both combinations are in series, so we get the equivalent resistance in the left branch by

$$\begin{gathered} R_{left}=R_{2,eq}+R_{1,eq} \\ =18\mathrm{Î\copyright }+ 24\mathrm{Î\copyright } \\ =42\mathrm{Î\copyright } \end{gathered}$$

In the right branch the resistance is $R_{right}=21\mathrm{Î\copyright }$. The two resistors $R_{right}$and $R_{left}$are connected in parallel, so find the equivalent resistance $R_{ab}$for this combination using equation (1) by

$$\begin{gathered} \frac{1}{R_{ab}}=\frac{1}{R_{right}}+\frac{1}{R_{left}} \\ \frac{1}{R_{ab}}=\frac{1}{21\mathrm{Î\copyright }}+\frac{1}{42\mathrm{Î\copyright }} \\ {R}_{ab}={\left(\frac{1}{21\mathrm{Î\copyright }}+\frac{1}{42\mathrm{Î\copyright }}\right)}^{-1} \\ {R}_{ab}=14\mathrm{Î\copyright } \end{gathered}$$