Maximum Efficiency and Power Transfer Theorem

Maximum Power Transfer Theorem

The condition of maximum power transfer does not result in maximum Efficiency, they both are totally different theorems with different conditions.

If we define the efficiency η as the ratio of power dissipated by the load to power developed by the source, then it is straightforward to calculate from the above circuit diagram that

${\displaystyle \eta ={\frac {R_{\mathrm {load} }}{R_{\mathrm {load} }+R_{\mathrm {source} }}}={\frac {1}{1+R_{\mathrm {source} }/R_{\mathrm {load} }}}.}$

Consider three particular cases:

• If ${\displaystyle R_{\mathrm {load} }=R_{\mathrm {source} }}$, then ${\displaystyle \eta =0.5,}$
• If ${\displaystyle R_{\mathrm {load} }=\infty }$ or ${\displaystyle R_{\mathrm {source} }=0,}$ then ${\displaystyle \eta =1,}$
• If ${\displaystyle R_{\mathrm {load} }=0}$, then ${\displaystyle \eta =0.}$
• 
  

The efficiency is equal or less than 50% when maximum power transfer is achieved, but approaches 100% as the load resistance approaches infinity, though the total power level tends towards zero.

Efficiency also approaches 100% if the source resistance approaches zero, and 0% if the load resistance approaches zero. In the latter case, all the power is consumed inside the source (unless the source also has no resistance), so the power dissipated in a short circuit is zero.

The mathematical proof explains the theorem clearly.

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:

${\displaystyle I={\frac {V}{R_{\mathrm {S} }+R_{\mathrm {L} }}}.}$

The power P_L dissipated in the load is the square of the current multiplied by the resistance:

${\displaystyle P_{\mathrm {L} }=I^{2}R_{\mathrm {L} }=\left({\frac {V}{R_{\mathrm {S} }+R_{\mathrm {L} }}}\right)^{2}R_{\mathrm {L} }={\frac {V^{2}}{R_{\mathrm {S} }^{2}/R_{\mathrm {L} }+2R_{\mathrm {S} }+R_{\mathrm {L} }}}.}$

The value of R_L for which this expression is a maximum could be calculated by differentiating it, 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.

${\displaystyle R_{\mathrm {S} }^{2}/R_{\mathrm {L} }+2R_{\mathrm {S} }+R_{\mathrm {L} }}$

 Differentiating the denominator with respect to R_L:

${\displaystyle {\frac {d}{dR_{\mathrm {L} }}}\left(R_{\mathrm {S} }^{2}/R_{\mathrm {L} }+2R_{\mathrm {S} }+R_{\mathrm {L} }\right)=-R_{\mathrm {S} }^{2}/R_{\mathrm {L} }^{2}+1.}$

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

${\displaystyle R_{\mathrm {S} }^{2}/R_{\mathrm {L} }^{2}=1}$

or

${\displaystyle R_{\mathrm {L} }=\pm R_{\mathrm {S} }.}$

In practical resistive circuits, R_S and R_L are both positive, so the positive sign in the above is the correct solution.

To find out whether this solution is a minimum or a maximum, the denominator expression is differentiated again:

${\displaystyle {{d^{2}} \over {dR_{\mathrm {L} }^{2}}}\left({R_{\mathrm {S} }^{2}/R_{\mathrm {L} }+2R_{\mathrm {S} }+R_{\mathrm {L} }}\right)={2R_{\mathrm {S} }^{2}}/{R_{\mathrm {L} }^{3}}.\,\!}$

This is always positive for positive values of ${\displaystyle R_{\mathrm {S} }\,\!}$ and ${\displaystyle R_{\mathrm {L} }\,\!}$, showing that the denominator is a minimum, and the power is therefore a maximum, when

${\displaystyle R_{\mathrm {S} }=R_{\mathrm {L} }.\,\!}$