Chapter 19, Problem 22P

To determine

The maximum voltage applied across the whole network.

Answer to Problem 22P

The maximum voltage applied to network is $67.08\text{V}$ .

Explanation of Solution

Given info:

The resistance of the first resistor in parallel is, $R_1=2.8\text{kΩ}=2.8\text{kΩ}\times \frac{1000\Omega }{1\text{kΩ}}=2800\Omega$ .

The resistance of the second resistor in parallel is, $R_2=2.1\text{kΩ}=2.1\text{kΩ}\times \frac{1000\Omega }{1\text{kΩ}}=2100\Omega$ .

The resistance of the resistor in series is, $R_3=1.8\text{kΩ}=1.8\text{kΩ}\times \frac{1000\Omega }{1\text{kΩ}}=1800\Omega$ .

The power of each resistor is, $P_{1,2,3}=\frac{1}{2}\text{W}$ .

The number of resistors in network is, $n=3$ .

Formula Used:

The expression to calculate the equivalent resistance of the network is,

  $R=\frac{R_1{R}_2}{R_1+R_2}+R_3$

The expression to calculate the maximum voltage applied to network is,

  $V=\sqrt{n\cdot P\cdot R}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}R=\frac{\left(2800\Omega \right)\left(2100\Omega \right)}{2800\Omega +2100\Omega }+1800\Omega \\ =3000\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}V=\sqrt{\left(3\right)\left(\frac{1}{2}\text{W}\right)\left(3000\Omega \right)}\\ =67.08\text{V}\end{array}$

Conclusion:

Therefore, the maximum voltage applied to network is $67.08\text{V}$ .