A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when:

Given:

• A cell with emf E and internal resistance r
• External resistance R

We need to find when power delivered to R is maximum.

Step 1: Current in the circuit

Total resistance = R+r

So, current:

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

Step 2: Power delivered to external resistance

Power in R is:

$$p=\ \ \ i^2R$$

Substitute i:

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

Step 3: Condition for maximum power

To find maximum power, we differentiate P with respect to R and set it equal to zero:

$$\ \frac{\ dP}{dR}=0$$

After differentiating and simplifying (this step involves standard calculus), we get:

R=r