# current electricity

### Asked by | 4th Jun, 2008, 12:31: PM

power transfer is maximum when r = R which means internal resistance of source equals the load resistance. It is called the maximum power theorem.

Note that we are talking about maximum power and not maximum efficiency. If resistance of the load is greater than the resistance of the source, efficiency is more but overall powr is low.

If internal impedance is made larger than the load the most of the power is dissipated in the source itself, but although the total power dissipated is higher, due to a lower circuit resistance, it turns out that the amount dissipated in the load is reduced.

This can be proved using calculus. I have copied and pasted the proof from wikipedia, url of website is also pasted for further reading.

http://en.wikipedia.org/wiki/Maximum_power_theorem

In the diagram opposite, power is being transferred from the source, with voltage *V* and fixed source resistance *R*_{S}, to a load with resistance *R*_{L}, resulting in a current *I*. By Ohm's law, *I* is simply the source voltage divided by the total circuit resistance:

The power *P*_{L} dissipated in the load is the square of the current multiplied by the resistance:

We could calculate the value of *R*_{L} for which this expression is a maximum, but it is easier to calculate the value of *R*_{L} for which the denominator

is a minimum. The result will be the same in either case. Differentiating with respect to *R*_{L}:

For a maximum or minimum, the first derivative is zero, so

or

In practical resistive circuits, *R*_{S} and *R*_{L} are both positive. To find out whether this solution is a minimum or a maximum, we must differentiate again:

This is positive for positive values of *R*_{S} and *R*_{L}, showing that the denominator is a minimum, and the power is therefore a maximum, when

*R*_{S}=*R*_{L}.

### Answered by | 4th Jun, 2008, 12:49: PM

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