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 = E²R/(R + r)²
Let's differentiate P with respect to R
dP/dR = E^2/ / (R + r)² - 2E²R / (R + r)³
When P is maximum, 

dP/dR = 0
E^2/ / (R + r)² - 2E²R / (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:

 

 

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