Chapter 23, Problem 73A

To determine

The equivalent resistance of the circuit.

Answer to Problem 73A

The equivalent resistance of the circuit is $45\Omega$.

Explanation of Solution

Given:

A circuit consisting of two parallel branch of resistors, connected in series with another resistor across a battery. The branch currents are marked as $I_1$ , $I_2$ and $I_3$ and all the resistance values are 30-Ω.

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.

  

Figure 1

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

  $\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{30}+\frac{1}{30}\\ R_{eq}=15\text{-}\Omega \end{array}$

Now, this equivalent resistance is in series with another 30-Ω resistance in (2),the net equivalent resistance of the circuit can be calculated as,

  $\begin{array}{c}{R}_{feq}=30+R_{eq}\\ =30+15\\ =45\Omega \end{array}$

Conclusion:

Therefore, the equivalent resistance of the circuit is $45\Omega$ .