Figure shows five $\mathbf{5}\mathbf{.00}\mathit{\Omega }$ resistors. Find the equivalent resistance between points

(a) F and H and

(b) F and G . (Hint: For each pair of points, imagine that a battery is connected across the pair.)

The resistance of each resistor is $R=5.00\text{Ω}$

Figure 27-34 is the circuit diagram of five resistors.

You can use the concept of the equivalent resistance of the series and the parallel circuits. Using those equations, you can find the equivalent resistance.

For series:

${\mathit{R}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathit{R}}_{\mathbf{1}}\mathbf{+}{\mathit{R}}_{\mathbf{2}}\mathbf{+}{\mathit{R}}_{\mathbf{3}}$

For parallel:

$\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{e}\mathbf{q}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{3}}}$

The equivalent resistance between points F and H:

Above the points F and H, there are two resistors which are in series with each other. So, their total resistance will be

$R+R=2R$

Similarly, below F and H, there is the same structure. So, the total resistance is

and these two resistances are parallel to the resistance between F and H , so you can write,

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{2R}+\frac{1}{2R}+\frac{1}{R}\\ =\frac{2}{2R}+\frac{1}{R}\\ =\frac{1}{R}+\frac{1}{R}\\ =\frac{2}{R}\end{array}$

$\begin{array}{c}{R}_{eq}=\frac{R}{2}\\ =\frac{5.00\Omega }{2}\\ =2.50\Omega \end{array}$

Hence, the required equivalent resistor is $2.50\text{Ω}$.

The equivalent resistance between F and G:

First, find the total resistance of the upper triangle where 2R is parallel to R ,

$\begin{array}{c}\frac{1}{R_T}=\frac{1}{2R}+\frac{1}{R}\\ =\frac{3R}{2R^2}\end{array}$

$\begin{array}{c}{R}_T=\frac{2R^2}{3R}\\ =\frac{2}{3}R\end{array}$

Now, this total resistance is in series with the resistance between H and G , so you get,

$\begin{array}{c}{R}_t=\frac{2}{3}R+R\\ =\frac{5R}{3}\end{array}$

As this total resistance is parallel to the resistance between F and G, you can write

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\frac{5}{3}R}+\frac{1}{R}\\ =\frac{3}{5R}+\frac{1}{R}\\ =\frac{(3R+5R)}{5R^2}\end{array}$

$\begin{array}{c}{R}_{eq}=\frac{5R^2}{3R+5R}\\ =\frac{5R^2}{8R}\\ =\frac{5}{8}R\end{array}$

Substitute known value in the above equation.

$\begin{array}{c}{R}_{eq}=\frac{5}{8}(5.00\Omega )\\ =\frac{25}{8}\Omega \\ =3.12\Omega \end{array}$

Hence, the required equivalent resistor is $3.12\text{Ω}$.