A resistor with resistance $\mathit{R}$is connected to a battery that has emf 12.0 V and internal resistance $\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{\Omega }$. For what two values of $\mathit{R}$ will the power dissipated in the resistor be 80.0 W ?

Consider the formula for the Ohm’s law.

$V=IR$

Here, $I$is current in ampere $\left(A\right)$,$R$ is resistance in ohms role="math" localid="1655724796140" $\left(\mathrm{\Omega }\right)$and $V$is the potential difference volt $\left(V\right)$.

Consider the power role="math" localid="1655724850722" $\mathit{\left(}\mathit{P}\mathit{\right)}$is the product of potential difference$\mathbf{\left(}\mathit{V}\mathbf{\right)}$ and the current$\mathbf{\left(}\mathit{I}\mathbf{\right)}$ and is given as follows:

$P=I^{2}R$

Consider the current if the emf is $\epsilon$, the internal resistance is $r$ and the load resistance is Ris as follow:

$\mathbf{I}\mathbf{=}\frac{\mathbf{\epsilon }}{\mathbf{R}\mathbf{+}\mathbf{r}}$

Consider the given, emf, internal resistance and the power consumed.

$\begin{array}{l}\epsilon =12.0\mathrm{V}\\ r=40\mathrm{\Omega }\\ P=80.0\mathrm{W}\end{array}$

Derive the equation to determine the resistance.

$$\begin{gathered} P=I^{2}R \\ P={\left[\frac{\epsilon }{R+r}\right]}^{2}R \\ P=\frac{{\epsilon }^{2}R}{R^2+2Rr+r^2} \\ {R}^2=2Rr-\frac{{\epsilon }^{2}R}{P}+r^2=0 \end{gathered}$$

Substitute the values of $P,r\mathrm{and}\epsilon$in the equation.

$$\begin{gathered} R^2+2R\left(0.4\right)-\frac{{\left(12\right)}^{2}R}{80}+{\left(0.4\right)}^2=0 \\ {R}^2-R+0.16=0 \\ {R}^2-0.2R-0.8R+0.16=0 \\ \left(R-0.2\right)\left(R-0.8\right)=0 \end{gathered}$$

Solve further as,

$$\begin{gathered} R=0.2\mathrm{\Omega } \\ R=0.8\mathrm{\Omega } \end{gathered}$$

Hence, the two values of $R$are $0.2\mathrm{\Omega }$and $0.8\mathrm{\Omega }$will dissipated the power of 80.0 W .