Question

Question: A resistor with resistance is connected to a battery that has emf and internal resistance . For what two values of will the power dissipated in the resistor be ?

Short answer

Answer

The two values of are$0.2\Omega$ and$0.8\Omega$ .

Step-by-step solution

Define the ohm’s law, resistance (R) and power (P).

Consider the formula for the Ohm’s law.

V = I R

Here, I is current in ampere (A), is resistance in ohms $\left(\Omega \right)$and V is the potential difference volt (V) .

Consider the power is the product of potential difference and the current and is given as follows:

$P=I^{2}R$

Consider the current if the emf is , the internal resistance is and the load resistance is Ris as follow:

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

Determine the two resistances.

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

$$\begin{gathered} \epsilon =12.0\text{V} \\ r=40\Omega \\ P=80.0\text{W} \end{gathered}$$

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 and in the equation.

role="math" localid="1655957149434" $$\begin{gathered} R^2+2R\left(0.4\right)-\frac{{\left(12\right)}^{2}R}{80}+{\left(0.4\right)}^2=0 \\ {R}^2-R+0.16=0 \\ {R}^2-0.2R-0.8R+0.16=0 \\ \left(R-0.2\right)\left(R-0.8\right)=0 \end{gathered}$$

Solve further as,

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

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