Question

The maximum power drawn out of the cell is given by (where r is internal resistance) :-

Answer 1

If r is the internal resistance, then from Ohm’s Law we know that,

E = I × (r+R)

So,

I = E/(r+R)

Now, we also know that P = (I^2) × R

Therefore, P = E^R/ (r + R) ^ 2

In case of maximum power, r = R,

Then, P (max) = E^2/ 4r.

Hence maximum power drawn out of the cell is given by E^2/ 4r.