Chapter 25, Problem 99A

To determine

To calculate: Total resistance.

Answer to Problem 99A

The total resistance of the three resistors connected in parallel and then series which is connected to two resistors in parallel is $71\Omega$ .

Explanation of Solution

Given:

From the given figure,

Resistance of three resistors in parallel $\left(R_3\right)$$=$$85\Omega$

Resistance of two resistors in parallel $\left(R_2\right)$$=$$85\Omega$

Formula Used:

Calculation:

Total resistance of the resistors in parallel:

  $\frac{1}{R_{3inparallel}}=\frac{1}{85\Omega }+\frac{1}{85\Omega }+\frac{1}{85\Omega }$

  $R_{3inparallel}=28.3\Omega$

Now calculating the resistance of the two resistors in parallel:

  $\begin{array}{l}\frac{1}{R_{2inparallel}}=\frac{1}{85\Omega }+\frac{1}{85\Omega }\\ R_{2inparallel}=42.5\Omega \end{array}$

To calculate the total resistance add $R_{2inparallel}$ and $R_{3inparallel}$

  $R=R_{3inparallel}+R_{2inparallel}$

  $=28.3\Omega +42.5\Omega$

  $R$$=71\Omega$

Conclusion:

Hence, the total resistance of the three resistors connected in parallel and then series which is connected to two resistors in parallel is $71\Omega$ .