# understanding maximum power transfer theorem

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#### PG1995

##### Full Member level 5 Hi :smile:

I'm trying to understand 'background' working of the maximum power transfer which says that power transfer takes place across R_L when it is equal to R_th (R_L is load resistance and R_th is Thevenin resistance). It is easily proved by doing some math.

P=VI, P=I^2.R, P=V^2/R, V=IR

Let's focus on P=V^2/R_L. To have maximum power transfer to R_L, V should be as large as power (V is volts dropped across R_L). V across R_L could be found using voltage divider rule: (R_L x E)/(R_L + R_th). Further R_L should be as little as possible because it is denominator. But also note that making R_L smaller would reduce the volts dropped across R_L.

But now focus on P=I^2.R_L. Compare it with the previous analysis of P=V^2/R_L. In (I^2.R_L), R_L should be as large as possible which is in contrast with the previous analysis which required R_L to be minimum. So some compromise is needed. But there is another point to note. R_L and R_th are in series so if R_L is made too big then the value of I would drop.

I hope you could see where I'm coming from (or, rather trying to come from! ). Could you please help me to really understand the working of maximum power transfer theorem? Many thanks.

#### vinodstanur  [/url][/IMG]
Just look above figure. It is a practical power source which will have an internal resistance greater than 0.

Now if u want to take maximum voltage at O/P then by using voltage divider law, u will get RL=\infty
In this case, power=0. ({v}^{2 }/infinity )=0

Now if u want maximum current O/P then u can decrease ur RL. right?
So if u decrease it to 0, then u will ger max current as V/Rth. At this time maximum power is delivered across the Rth.
But still , power O/P(power transfer) is 0. ({I}^{ 2} * 0)=0
U could not use {V}^{ 2} /R equation because it is 0/0 form. (V out=0; RL=0)

So the power O/P is zero for RL=0 & RL=infinity.

Now, the power O/P is maximum at a particular value of RL . But it depends on the internal resistance Rth.

For finding the value of RL for maximum ,
u can simply differentiate the Power Output with RL.
Then at peak power value, the rate of change of power with respect to RL will be zero.
So to find it, just substitute
dP/dRL=0;
Now u will get Rth =RL

This means the maximum value of power output will be when Tth = RL.

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##### Super Moderator
Staff member As vinodstanur states, the V and A conditions for peak power output depends on the invisible (internal) resistance of the source.

I once made graphs of a few power supplies. I attached loads of various resistances (R_L).

The x axis of my graph was load resistance (R_L).

For the y axis I superimposed several plots, namely V, A, and W.

The V plot was a gentle curve. The A plot was a gentle curve.
Low V coincided with high A.
High V coincided with low A. (Not surprising.)

When I plotted W for all loads, the curve was shaped like a bell. The top was where peak power occurred.

That's when I realized I had been unable to predict ahead of time where the top would be, that is, which R_L would yield V and A which would draw the maximum W.

I had to find out by experimentation.

I'd seen the rule that says peak power transfer occurs when load resistance equals source resistance. So I guess I took it for granted that the peak power would be at the point R_L was identical to the internal R of my supply.

I also remember that I had to custom connect a series of various loads, to get fine enough resolution of points at the top of the bell curve. The peak could not have been determined without my doing so.

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#### Miguel Gaspar #### PG1995

##### Full Member level 5 Hi again,

1: Just curious to know if there is also a minimum power transfer theorem of some kind.

2: When R_L is not equal to R_Th, then the power dissipated by the R_L won't be maximum. Is this because of the voltage drop across R_Th?

#### enjunear 1: Just curious to know if there is also a minimum power transfer theorem of some kind.

There is no such theorem, but it would be when V on R_L = 0V or I through R_L = 0 amps. Then you have minimum power transfer (like vinodstanur said, when R_L = infinity or R_L = short-circuit).

2: When R_L is not equal to R_Th, then the power dissipated by the R_L won't be maximum. Is this because of the voltage drop across R_Th?

The reason is that you are not at the optimal point on the dP/dR_L curve (try solving your circuit equation for power in R_L for several values of R_L and plotting them, P_R_L vs. R_L). If R_L = R_Th is the maximum power condition, when R_L is smaller than R_Th, the same current passes through R_Th and R_L, but more voltage is dropped across R_Th, so R_L has less voltage across it, and less dissipated power. When R_L is larger than R_Th, more voltage is dropped across R_L, but less current flows through both parts, also causing less power to be dissipated by R_L.

Power is made up of both voltage and current, so when you decrease one, the total power dissipation value falls. R_L = R_Th is the sweet spot where the product of the voltage across R_L and current through it is maximized.

• PG1995

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