Question

What is the equivalent resistance between points a and b in FIGURE EX28.25?

Short answer

The equivalent resistance between points a and b is$14\Omega$.

Step-by-step solution

Given Information

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

Simplify

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

$$\begin{gathered} \frac{1}{R_{1,eq}}=\frac{1}{60\Omega }+\frac{1}{40\Omega } \\ {R}_{1,eq}={\left(\frac{1}{60\Omega }+\hspace{0.17em}\frac{1}{40\Omega }\right)}^{-1} \\ {R}_{1,eq}=24\Omega \end{gathered}$$

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

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

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

$$\begin{gathered} R_{left}=R_{2,eq}+R_{1,eq} \\ =18\Omega +\hspace{0.17em}24\Omega \\ =42\Omega \end{gathered}$$

Calculation

In the right branch the resistance is $R_{right}=21\Omega$. 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

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