### Maximizing the power of load in an electric circuit

Let the EMF, internal resistance, load and the current of a circuit be E, r, R and I respectively.

E = I(R + r)

I = E/(R + r)

The power of the load, the external resistor in this case, is as follows:

P = VI

P = E^{2}R/(R + r)^{2}

Let's differentiate P with respect to R

dP/dR = E^{2/} / (R + r)^{2} - 2E^{2}R / (R + r)^{3}

When P is maximum,

dP/dR = 0

E^{2/} / (R + r)^{2} - 2E^{2}R / (R + r)^{3} = 0

R + r = 2R

R = r

As you can see, the power is maximum, when the external resistance is equal to the internal resistance of the battery or cell.

E.g.

**R = r**

E = 12V R = 6 r = 6

Power = (12/12)^{2} X 6 = 6W

**R > r**

E = 12 R = 12 r = 6

Power = (12/18)^{2} X 12 = 5.3W

**R < r**

E = 12 R = 4 r = 6

Power = (12/10)^{2} X 4 = 5.7 W

You can practise the above with the following interactive simulation:

For a comprehensive tutorial on electricity, for GCSE, IGCSE, IB, A Level, GCE OL and GCE AL, please use this link.

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