What is the equivalent resistance between points a and b in FIGURE EX28.27?

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

In case we have 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 equivalent resistance for the parallel connection is

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\hspace{0.17em}\frac{1}{R_2}+....+\hspace{0.17em}\frac{1}{R_N}$ (1)

There are three resistors that are connected in parallel, so let us find the equivalent resistance for this combination using equation (1) by

$$\begin{gathered} \frac{1}{R_{1,eq}}=\frac{1}{100\Omega }+\hspace{0.17em}\frac{1}{100\Omega }+\hspace{0.17em}\frac{1}{100\Omega } \\ \frac{1}{R_{1,eq}}={\left(\frac{1}{100\Omega }+\hspace{0.17em}\frac{1}{100\Omega }+\hspace{0.17em}\frac{1}{100\Omega }\right)}^{-1} \\ \frac{1}{R_{1,eq}}=\frac{100}{3}\Omega \end{gathered}$$

There are two resistors $100\Omega$and $100\Omega$are connected in parallel, so let us find the equivalent resistance $R_{2,eq}$for this combination using equation (1) by

$$\begin{gathered} \frac{1}{R_{2,eq}}=\frac{1}{100\Omega }+\hspace{0.17em}\frac{1}{100\Omega } \\ \frac{1}{R_{2,eq}}={\left(\frac{1}{100\Omega }+\hspace{0.17em}\frac{1}{100\Omega }\right)}^{-1} \\ \frac{1}{R_{2,eq}}=\frac{100}{2}\Omega \end{gathered}$$

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 in the form

$E_{eq}=R_1+\hspace{0.17em}{R}_2$ (2)

The two combinations $R_{2,eq}$and $R_{1,eq}$are connected in series with the resistance $100\Omega$. So, we can use equation (2) to get the equivalent resistance between points a and b by

role="math" localid="1648832883654" $$\begin{gathered} R_{ab}=R_{2,eq}+R_{1,eq}+100\Omega \\ =\frac{100}{3}\Omega +\hspace{0.17em}\frac{100}{2}\Omega +\hspace{0.17em}100\Omega \\ =183.3\Omega \end{gathered}$$