Verification of the maximum power transfer theorem
Maximum power transfer theorem experiment: varying the load resistance, plotting power against load, finding the peak at R(L) = R(th), the 50 % efficiency at match, and viva questions with answers.
Aim
To verify the maximum power transfer theorem by varying the load resistance across a source of known Thevenin equivalent, plotting the power delivered against load resistance, and confirming that the power is greatest when the load resistance equals the Thevenin resistance.
Apparatus required
| Apparatus | Specification | Qty |
|---|---|---|
| Regulated DC power supply | 0–30 V, 1 A | 1 |
| Decade resistance box | 0–10 kΩ in 1 Ω steps, as the load | 1 |
| Resistors for the source network | 470 Ω and 1 kΩ, ½ W | 1 each |
| Digital multimeters | DC milliamps and DC volts | 2 |
| Breadboard and patch cords | — | 1 set |
Theory
The maximum power transfer theorem states that a source delivers the greatest power to a load when the load resistance equals the internal, or Thevenin, resistance of the source network seen from the load terminals. For a source of open-circuit voltage V(th) and internal resistance R(th) feeding a load R(L), the load current is V(th)/(R(th) + R(L)) and the power in the load is P = V(th)²·R(L)/(R(th) + R(L))².
Differentiating that expression with respect to R(L) and setting the result to zero gives R(L) = R(th) as the condition for a maximum, and substituting it back gives the maximum power itself: P(max) = V(th)²/4R(th). The curve of power against load resistance rises from zero at short circuit, peaks at the match, and falls away slowly towards zero at open circuit — and the shape matters, because it is noticeably flat near the peak. A load within a factor of two of the match still delivers about 89 % of the maximum, which is why practical matching need not be exact.
At the matched condition exactly half the power drawn from the source is dissipated in its own internal resistance, so the efficiency is 50 %. That is the point students most often find surprising: maximum power transfer and maximum efficiency are different objectives and occur at different load resistances. Efficiency rises steadily towards 100 % as R(L) increases, but the power delivered falls once past the match.
The distinction decides where the theorem is used. In communications, instrumentation and audio, the available signal power is small and getting the most of it into the load is what matters, so circuits are matched. In power systems the opposite holds: a generator of a fraction of an ohm internal resistance is never matched to its load, because doing so would waste half the generated energy as heat in the machine. There the load is deliberately made much larger than the source resistance so the efficiency is high.
The theorem extends to AC networks, where maximum power is delivered when the load impedance is the complex conjugate of the source impedance — equal resistance and equal but opposite reactance, so the reactances cancel and the circuit is resistive at the operating frequency.
Circuit connections
Check every point below against your board before switching on. There is no diagram here on purpose — a wrong diagram is worse than none, and this is the list a demonstrator actually walks through with you.
- The source network built as a Thevenin equivalent: the supply in series with a resistor that represents R(th), brought out to a pair of load terminals.
- The decade box connected across those load terminals as R(L).
- The ammeter in series with the load to read the load current, and the voltmeter directly across the load to read the load voltage.
- Supply set to minimum before switching on, and the decade box set to a mid-range value rather than zero so the first switch-on is not into a short circuit.
- All connections short and tight; contact resistance in the load branch adds directly to R(L) and shifts the apparent match point.
Procedure
- 1Determine the Thevenin equivalent of the source network first: measure the open-circuit voltage at the load terminals, then either measure the short-circuit current and take R(th) = V(oc)/I(sc), or compute R(th) from the network with the source suppressed.
- 2Connect the decade box as the load and set it to a value well below the expected R(th).
- 3Switch on and record the load resistance, the load voltage and the load current together.
- 4Increase the load resistance in steps, taking readings at each, and cluster the steps closely around the expected match point so the peak is well defined. Take at least ten points spanning from well below to well above R(th).
- 5Compute the power delivered at each step as P = V·I, and also as I²R(L) as a check that the two agree.
- 6Plot power against load resistance and read the resistance at which the curve peaks. Compare it with the measured R(th).
- 7Compute the efficiency at each point as the ratio of load power to total power drawn from the source, and confirm that it passes through 50 % at the match.
Work out your readings
Type in the numbers off the meters. This fills the tabular column, works the calculation through step by step, plots the characteristic — and tells you when a reading cannot physically be right, which is the part a manual can't do. Everything stays on this device, and it works with the network off.
Nameplate and machine data
Load readings
Take at least ten points, clustered closely around the expected match so the peak is well defined.
| # | Load resistance R(L)(Ω) | Load voltage(V) | Load current(mA) |
|---|---|---|---|
| 1 | |||
| 2 | |||
| 3 | |||
| 4 | |||
| 5 | |||
| 6 |
Fill in the machine data and at least one complete row of readings to see the results, the worked calculation and the curve.
Precautions
- Do not set the decade box to zero while the supply is on; that short-circuits the source through only R(th) and may exceed the resistor's rating.
- Check the power rating of the resistor representing R(th) — at the match it dissipates as much as the load does.
- Take voltage and current readings simultaneously at each step, since a drifting supply or a warming resistor makes readings taken at different times inconsistent.
- Keep the decade box within its own current rating, which is usually lowest on the smallest decade.
- Take closely spaced readings near the expected peak; a coarse sweep can miss the maximum entirely and make the curve look flat-topped.
- Switch off before changing the source network, and re-measure V(oc) and R(th) if any part of it is altered.
Sources of error
Every record asks for these, and every record gets the same three lines copied from the one before. These are the errors this particular experiment actually has.
- Ammeter burden: the meter's own resistance adds to the load branch, so the effective load is slightly larger than the decade box setting and the peak appears at a lower dial reading.
- Voltmeter loading, which draws a small current in parallel with the load and matters most when R(L) is large.
- Contact and lead resistance in series with the load, which shifts the apparent match point downward by that amount.
- The flatness of the power curve near its peak, which makes the exact maximum hard to locate from measurements alone — the reason for taking many closely spaced readings there.
- Self-heating of the source resistor, which raises R(th) during the run so that the match point measured at the end differs slightly from the one at the start.
Viva questions with answers
State the maximum power transfer theorem and give the condition for maximum power.
A source delivers maximum power to a load when the load resistance equals the Thevenin resistance of the network seen from the load terminals. The condition is R(L) = R(th), obtained by differentiating P = V(th)²R(L)/(R(th) + R(L))² with respect to R(L) and equating to zero, and the maximum power is then V(th)²/4R(th).
What is the efficiency at maximum power transfer, and why is it not higher?
It is exactly 50 %. At the match the load resistance equals the internal resistance, so the same current flows through both and each dissipates the same power. Half the power drawn from the source is therefore lost inside it, and only half reaches the load.
Why is the maximum power transfer condition not used in power systems?
Because it would waste half the generated energy in the generator and transmission network. A power system is designed for efficiency, so the load resistance is made far larger than the source resistance; efficiency then approaches 100 % even though the power delivered is well below the theoretical maximum. Matching is used only where the available power is small and efficiency is secondary.
How does the condition change for an AC circuit?
Maximum power is transferred when the load impedance is the complex conjugate of the source impedance: the resistances are equal and the reactances are equal in magnitude but opposite in sign. The reactances then cancel, leaving a purely resistive circuit at that frequency, and the resistive matching condition applies to what remains.
Why is the power curve flat near its maximum, and what does that mean in practice?
Because the maximum is a stationary point, the power varies only to second order in the departure from the match. A load anywhere within a factor of two of R(th) still delivers roughly 89 % of the maximum, so practical matching is forgiving and an approximate match is usually good enough.
How would you find the Thevenin resistance experimentally before the main measurement?
Measure the open-circuit voltage at the load terminals, then measure the short-circuit current at the same terminals; R(th) is their ratio. Alternatively suppress the independent sources — voltage sources shorted, current sources opened — and measure the resistance looking back into the terminals with a multimeter.
What happens to the load power when the load resistance is very small or very large?
In both cases it tends to zero. With a very small load the current is large but almost all the voltage is dropped internally, so the load sees almost none of it. With a very large load the voltage across it approaches the full open-circuit value but the current becomes negligible. The product peaks between the two extremes, at the match.