Q5P - Introduction to Electrodynamics

Question

A battery of emf $\epsilon$and internal resistance r is hooked up to a variable "load" resistance,R . If you want to deliver the maximum possible power to the load, what resistance R should you choose? (You can't change e and R , of course.)

Step-by-Step Solution

Answer

The maximum power is delivered to the load when the internal resistance of the battery is equal to the load resistance.

Step 1: Write the given data from the question.

The emf of the battery is $\epsilon$ .

Internal resistance of the battery is $r$ .

Load resistance is R .

Step 2: Determine the value of the resistance to deliver the maximum power to load.

Consider the circuit diagram shown below.

 

 

The current in the above circuit is given by,

$\mathit{i}\mathbf{=}\frac{\mathbf{E}}{\mathbf{R}\mathbf{+}\mathbf{r}}$

 

The power deliver to the load is given by,

$\mathit{P}\mathbf{=}{\mathit{i}}^{\mathbf{2}}\mathit{R}$

 

Substitute $\frac{E}{R+r}$ for into above equation.

         $p={\left(\frac{E}{R+r}\right)}^{2}R$                                                     …… (1)

The maximum power can be delivered to the load under the condition $\frac{dP}{dR}=0$ .

Now differentiate the equation (1) with respect to R,

$$\begin{gathered} \frac{dP}{dR}=\frac{{\left(R+r\right)}^2{E}^2+E^{2}R\times 2\left(R\times r\right)}{{\left(R\times r\right)}^4} \\ \frac{dP}{dR}=\frac{{\left(R+r\right)}^2{E}^2+2E^2\left(R\times r\right)}{{\left(R\times r\right)}^4} \end{gathered}$$

Substitute the above equation equal to zero.

$$\begin{gathered} \frac{dP}{dR}=0 \\ \frac{{\left(R+r\right)}^2{E}^2-2E^2{R}^4\left(R+r\right)}{\left(R+r\right)}=0 \\ \left(R+r\right)E^2\left[R+r-2R\right]=0 \\ \left(R+r\right)E^2\left[r-R\right]=0 \end{gathered}$$

Solve further as,

$$\begin{gathered} r=R=0 \\ R=r \end{gathered}$$

 

Hence the maximum power is delivered to the load when the internal resistance of the battery is equal to the load resistance.