Question

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

Answer 1

Dear Student,

$$\begin{gathered} {\mathrm{R}}_1=25\mathrm{ohm} \\ \mathrm{Power}\mathrm{dissipates},{\mathrm{P}}_1=36\mathrm{W} \\ {\mathrm{R}}_2=15\mathrm{ohm} \\ \mathrm{We}\mathrm{know}\mathrm{that}, \\ \mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}} \\ {\mathrm{V}}^2=\mathrm{PR} \\ \mathrm{V}=\sqrt{\mathrm{PR}} \\ \mathrm{V}=\sqrt{{\mathrm{P}}_1{\mathrm{R}}_1} \\ \mathrm{V}=\sqrt{36\times 25} \\ \mathrm{V}=\sqrt{900}=30\mathrm{V} \\ \mathrm{Now},{\mathrm{R}}_2\mathrm{is}\mathrm{connected}\mathrm{in}\mathrm{series}\mathrm{with}{\mathrm{R}}_1.\mathrm{So}, \\ \mathrm{R}={\mathrm{R}}_1+{\mathrm{R}}_2 \\ \mathrm{R}=25+15=40\mathrm{ohm} \\ \mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}} \\ \mathrm{P}=\frac{{\left(30\right)}^2}{40} \\ \mathrm{P}=\frac{900}{40}=22.5\mathrm{W} \end{gathered}$$