Calculations of Power Delivered to a Load

Background


We use the Thévenin-equivalent circuit to determine an appropriate load resistance to achieve maximum power transfer.  However, there are some potential pitfalls associated with this technique which we explore below and which are summarized at the end.
 

Analysis


In this circuit, what load resistance RL should be selected to maximize the amount of power transferred to the load?  To answer this question requires finding the Thévenin-equivalent resistance and making the load resistance equal it.  But when we want to find out how much power actually gets transferred to the load and how much is actually generated by the voltage source, we cannot rely on the Thévenin-equivalent circuit:  we must revert to the original circuit.  Let's see why.

Finding the Thévenin-equivalent resistance is easy:  remove the load, short the voltage source, and consider the resistance from node A around to node B.  The simplified circuit looks like this:

The resistance from A to B consists of R1 in parallel with R2 or 5 W.

Finding the Thévenin-equivalent voltage also is easy:  it's the voltage appearing across R2 with the load removed and can easily be computed using the voltage divider law.

The Thévenin-equivalent circuit therefore is

The combined series resistance of RTH and RL is 10 W so the 5 V source produces 500 mA of current.  The voltage across RL is half that of the Thévenin-equivalent source or 2.5 V.  Therefore the power dissipated by the load is (500 mA)(2.5 V)=1.25 W.

How much power is produced by the source?  This is deceptive:  one might suppose it is the power produced by the Thévenin-equivalent voltage source but this would be false.  That is a fictitious source and should only be used as an analytical tool to determine the appropriate load resistance RL.  The real source is VS = 10 V.  Let's do the power calculations both ways and compare the results.

First let's consider what power we would believe is generated if we treated the Thévenin-equivalent source as a real source.

Now let's consider what power would be generated by the real source, VS .  We need to find the total current through this source.  We can get that by regarding R2 and RL in parallel and R1 in series with that combination.

Obviously these are significantly different values.  The second of these is the real power generated by the source.  The first is a fictitious value and should not be regarded as real in any sense.

To compute the efficiency of power delivery, use PS , then, and the power dissipated by the load.  (This latter value is correctly predicted from the Thévenin-equivalent model.)

Had we instead used the incorrect power PTH = 2.5 W, we would have arrived at the following incorrect value:


 

Summary


Use the Thévenin-equivalent model to determine the load resistance to maximize transfer of power from the source to the load.  However, do not determine the power generated by the source by considering the Thévenin-equivalent voltage source:  use the actual voltage source.

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*Last updated 25 September 1998