Chapter 24, Problem 107A

To determine

The equivalent resistance of the given network.

Answer to Problem 107A

The equivalent resistance of the network is $\text{128}\text{.33 }\Omega$.

Explanation of Solution

Given:

A network consisting of three parallel branches of resistors, connected in series with a set of two resistors in series. All the resistance values are $55\Omega$ .

Formula used:

The equivalent resistance of a parallel branch of two resistances is given by,

  $\frac{1}{R_{eq}}=\frac{1}{R_A}+\frac{1}{R_B}$ …… (1)

Now, the equivalent for any two resistances in series, the equivalent resistance is given by,

  $R_{eq}=R_A+R_B$ …… (2)

Calculation:

Consider the circuit shown in Figure 1.

  

Here, the equivalent resistance of the parallel branches can be calculated using (1) as,

  $\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{55}+\frac{1}{55}+\frac{1}{55}\\ R_{eqp}=18.33\Omega \end{array}$

The equivalent of series resistances is calculated using (2) as,

  $\begin{array}{c}{R}_{eqs}=55+55\\ =110\Omega \end{array}$

Now, this equivalent resistance of the parallel network is in series with equivalent of series resistance, the net equivalent resistance of the network can be calculated as,

  $\begin{array}{c}{R}_{eq}=R_{eqs}+R_{eqp}\\ =110+18.33\\ =128.33\text{}\Omega \end{array}$

Conclusion:

Therefore, equivalent resistance of the network is $128.33\text{}\Omega$ .