Chapter 22.1, Problem 22SSC

To determine

Change in power.

Answer to Problem 22SSC

  $\Delta P=4\text{W}$ increase.

Explanation of Solution

Given:

Source voltage, $V=12\text{V}$

Resistance of circuit, $R_1=12\Omega$

Decreased resistance of circuit, $R_2=9\Omega$

Formula Used:

Power dissipation in resistance is given by,

  $P=\frac{V^2}{R}$

Where, $P$ is power dissipated, $V$ is voltage across resistance and $R$ is resistance.

Calculation:

For resistance $R_1$ , the power dissipated is

  $\begin{array}{c}{P}_1=\frac{V^2}{R_1}\\ =\frac{{12}^2}{12}\\ =12\text{W}\end{array}$

For resistance $R_2$ , the power dissipated is

  $\begin{array}{c}{P}_2=\frac{V^2}{R_2}\\ =\frac{{12}^2}{9}\\ =16\text{W}\end{array}$

So, change in power is

  $\begin{array}{c}\Delta P=P_2-P_1\\ =16-12\\ =4\text{W}\end{array}$

Conclusion:

Therefore, the change in power is $\Delta P=4\text{W}$ that is power is increased.