Question

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.)

Short answer

1. The equivalent resistor is $R_{eq}=2.50\text{Ω}$.
2. The equivalent resistor is $R_{eq}=3.12\text{Ω}$.

Step-by-step solution

Given data:

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

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

Understanding the concept:

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:

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

(a) Calculate the equivalent resistance between points F  and H:

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 $2.50\text{Ω}$.

(b) Calculate the equivalent resistance between points   F and  G:

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 $3.12\text{Ω}$.