Chapter 35, Problem 52A

Consider the combination series and parallel circuit shown here.

a. Identify the parallel part of the circuit. What is the equivalent resistance of this part? In other words, what single resistance could replace this part of the circuit and not change the total current from the battery?

b. What is the equivalent resistance of all the resistors? In other words, what single resistance could replace the whole circuit without changing the current produced by the battery?

To determine

The parallel part of the circuit, equivalent resistance of parallel part.

Answer to Problem 52A

Both $10\Omega$ resistance in the circuit forms the parallel part of the circuit and the equivalent resistance of the part is $5\Omega$ .

Explanation of Solution

Given info:

First resistance in series in the circuit, $R_1=20\Omega$

Second resistance in series in the circuit, $R_1=20\Omega$

Resistance of the both parallel resistor, $R_3=R_4=10\Omega$

Both resistance of $10\Omega$ in the circuit are in the parallel connection. Both resistances of $10\Omega$ forms the parallel part of the circuit.

Formula Used:

Write the expression to calculate the equivalent resistance of the parallel part of the circuit.

  $R_p=\frac{R_3\cdot R_4}{R_3+R_4}$

Calculation:

Substitute given values in the above expression.

  $\begin{array}{c}{R}_p=\frac{\left(10\Omega \right)\left(10\Omega \right)}{10\Omega +10\Omega }\\ =5\Omega \end{array}$

Conclusion:

Both $10\Omega$ resistance in the circuit forms the parallel part of the circuit and the equivalent resistance of the part is $5\Omega$ .

To determine

The equivalent resistance of the entire resistor in the circuit.

Answer to Problem 52A

The equivalent resistance of all the resistor in the circuit is $40\Omega$ .

Explanation of Solution

Given info:

First resistance in series in the circuit, $R_1=20\Omega$

Second resistance in series in the circuit, $R_2=15\Omega$

Equivalent resistance of the parallel part, $R_p=5\Omega$

Formula Used:

Write the expression to calculate the equivalent resistance of all the resistor in circuit.

  $R=R_1+R_p+R_2$

Calculation:

Substitute given values in the above expression.

  $\begin{array}{c}R=20\Omega +5\Omega +15\Omega \\ =40\Omega \end{array}$

Conclusion:

The equivalent resistance of all the resistor in the circuit is $40\Omega$ .