Figure 27-79 shows three$\mathbf{20}\mathbf{.0}\mathbf{}\mathbf{\text{\hspace{0.17em}Ω}}$resistors. Find the equivalent resistance between points (a), (b), and (c). (Hint: Imagine that a battery is connected between a given pair of points.)

The resistance value of the three resistors,$R=20.0\Omega$

Using the concept of equivalent resistance for the series and parallel combination of resistors in the circuit, the resistance between points A and B can be calculated.

Again, the current through a conductor is high, and its resistance can be neglected. Thus, the resistance value of a conducting wire is zero.

Formulae:

The equivalent resistance for a series combination,

$R_{eq}=\sum _1^n{R}_i$ (i)

The equivalent resistance for a parallel combination,

$R_{eq}=\sum _1^n\frac{1}{R_i}$ (ii)

Between points A and B, as the three resistors are in parallel connection, the equivalent resistance can be given using equation (ii) as follows:

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}\\ =\frac{3}{R}\\ R_{eq}=\frac{R}{3}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{R}_{eq}=\frac{20.0\Omega }{3}\\ =6.67\Omega \end{array}$

Hence, the value of equivalence resistance is.$6.67\Omega$

Between points A and C, as the three resistors are in parallel connection, the equivalent resistance can be given using equation (ii) as follows:

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}\\ =\frac{3}{R}\\ R_{eq}=\frac{R}{3}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{R}_{eq}=\frac{20.0\Omega }{3}\\ =6.67\Omega \end{array}$

Hence, the value of equivalence resistance is.$6.67\Omega$

Since points B and C are connected using a conducting wire, the equivalent resistance between these points is zero.

Hence, the value of equivalent resistance is.$0\Omega$