Chapter 19, Problem 20P

To determine

The net resistance of the circuit.

Answer to Problem 20P

The net resistance of the circuit is $\text{4}\text{.6}\text{kΩ}$ .

Explanation of Solution

Given info:

The value of resistance of each resistor is,

  $R_{1,2,3,4,5,6}=2.8\text{kΩ}=2.8\text{kΩ}\times \frac{1000\Omega }{1\text{kΩ}}=2800\Omega$

Formula Used:

The expression to calculate the net resistance of the circuit is,

  $R_e=\frac{1}{\frac{1}{\frac{1}{\frac{1}{R_1+R_2}+\frac{1}{R_3}}+\frac{1}{R_4}}+\frac{1}{R_5}}+R_6$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_e=\frac{1}{\frac{1}{\frac{1}{\frac{1}{2.8\text{kΩ}+2.8\text{kΩ}}+\frac{1}{2.8\text{kΩ}}}+\frac{1}{2.8\text{kΩ}}}+\frac{1}{2.8\text{kΩ}}}+2.8\text{kΩ}\\ =\text{4}\text{.6}\text{kΩ}\end{array}$

Conclusion:

Therefore, the net resistance of the circuit is $\text{4}\text{.6}\text{kΩ}$ .