A resistor with R₁ = 25.0 Ω is connected to a battery that has negligible internal resistance and electrical energy is dissipated by R₁ at a rate of 36.0 W. If a second resistor with R₂ = 15.0 Ω is connected in series with R₁, what is the total rate at which electrical energy is dissipated by the two resistors?

Given Data :

• R₁ = 25.0 Ω
• R₂ = 15.0 Ω
• P₁= 36.0 W

To find the dissipated power by the two resistors, the battery has zero internal resistance, this means that the voltage drop by R; is the same voltage of the battery. So, use R; to get the voltage of the battery before connecting R where the dissipated power P is related to the voltage by the equation as:

$$\begin{gathered} P_1=\frac{V^2}{R_1} \\ V=\sqrt{P_1{R}_1} \end{gathered}$$

Calculate V as:

$$\begin{gathered} \mathrm{V}=\sqrt{{\mathrm{P}}_1{\mathrm{R}}_1} \\ =\sqrt{\left(36\mathrm{W}\right)\left(25\mathrm{\Omega }\right)} \\ =30\mathrm{V} \end{gathered}$$

After connecting R; the voltage drop in the battery is due to the equivalent resistance of R₁ and R₂ which are in series and its combination is given by

$R_{eq}=R_1+R_2$

The total dissipated power by the two resistors will be:

$$\begin{gathered} P_1=\frac{V^2}{R_{eq}} \\ =\frac{V^2}{R_1+R_2} \\ =\frac{{\left(30V\right)}^2}{25\Omega +15\Omega } \\ =22.5W \end{gathered}$$

Therefore, the total rate at which electrical energy is dissipated by the two resistors is 22.5 W.