Question

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?

Short answer

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

Step-by-step solution

Voltage

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:

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

Equivalent Resistance

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

$R_{eq}=R_1+R_2$

total power dissipation

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.