Maximizing the power of load in an electric circuit

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Maximizing the power of load in a circuit: R = r


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

Maximising the power of a load - external resistance = internal resistance


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 = E2/ / (R + r)2 - 2E2R / (R + r)3
When P is maximum, 

dP/dR = 0
E2/ / (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:


 

 

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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