7
\$\begingroup\$

I'm missing something here. The maximum power transfer theorem states that max power transfer from source to load happens when source impedance matches load impedance.

In the case of a 10V source with an impedance of 5 ohms driving a load impedance of 5 ohms, power transfer to the load would be 5 watts, for an efficiency of 50%.

In the case of a 10V source with an impedance of 1 ohm driving a load impedance of 9 ohms, power transfer to the load would be 9 watts, with an efficiency of 90%.

This seems to contradict the theorem, what am I missing?

\$\endgroup\$
1
  • 3
    \$\begingroup\$ The maximum power transfer theorem actually states that max power transfer from source to load happens when load impedance matches source impedance (and not the other way round). If you thought about it this way (the correct and proper way), your follow-up analysis would be seen to be irrelevant. \$\endgroup\$
    – Andy aka
    Commented Jun 14 at 17:33

3 Answers 3

10
\$\begingroup\$

What you are missing is, that if you reduce your load to 1 ohm, it gets even more power:

  • Total resistance: \$R_\text{tot}=R+R=1+1=2\text{ ohms}\$
  • Current: \$U/R_\text{tot} = 10\text{ V}/2\text{ ohms} = 5\text{ A}\$
  • Power for the load \$P=RI^2=1 \times 5^2=25\text{ W}\$

So if you reduce your load to 1 Ω (i.e. same as your source impedance), you get far more power for the load than with a 9 Ω load (at the cost of a lower efficiency).

Basically:

  • if your load resistance is greater than the source resistance: you get a lot of voltage, but not much current. Energy efficiency is "rather good" (at least >50%)
  • if load resistance = source resistance: it's the best trade off between current and voltage (you get maximum power for your load)
  • if the load resistance is smaller than the source resistance: even if you get more current, the voltage at the load is so small that the power starts decreasing. Efficiency is very bad (<50%)

If you take it to the limits:

  • if your load has infinite resistance: you get all your voltage, but 0 A current, so no power.
  • if your load has zero resistance (so 0 V), i.e. a short circuit: you get no power in your load (because no voltage), but your source wastes a lot of power in its internal resistance.
  • the balance to get the maximum power is in between: it happens that it is when both resistances are equal (I don't know any "logical" explanation excepting doing the math).

Note that this "equal resistance to get maximal power" is quite useful for signals where you get a fixed power (like an radio receiver), or when you need very long cables but not a lot of power. On the contrary, it's a very bad idea if you care about efficiency (you waste 50% of your energy).

\$\endgroup\$
3
  • \$\begingroup\$ Great explanation, thank you. \$\endgroup\$ Commented Jun 14 at 18:04
  • \$\begingroup\$ Note that if the source impedance is reactive, things can get more interesting. I've often heard the MPT described in terms of impedance, but if e.g. the source is capacitive and the load is reactive, the magnitude of the combined impedance may be smaller than the source impedance. \$\endgroup\$
    – supercat
    Commented Jun 15 at 9:40
  • \$\begingroup\$ @supercat I believe the MPT extended for complex source impedance states that the load should be the complex conjugate of the source impedance; i.e. resistive impedance is the same, while reactive impedance should have equal magnitude and opposite sign. Of course, in the case of fully real impedances or at DC, this reduces to the familiar equality. \$\endgroup\$
    – Hearth
    Commented Jun 15 at 13:30
3
\$\begingroup\$

What you are missing is that you have completely different amount of power available from 10V if you have 5 ohms or 1 ohm as source impedance.

If you have 5 ohm source impedance, you only have 5W max available, and it happens with 5 ohm load which gets 5W.

If you have 1 ohm source impedance, 9 ohms does get 9W but you have 25W available and you will get 25W with 1 ohm load resistor, so if you want to transfer the maximum amount of power then clearly 9 ohms is a poor choise and 1 ohm is the best choise.

\$\endgroup\$
0
\$\begingroup\$

Electricity distribution authorities around the world disregard the Jacobi theorem because they seek maximum efficiency, not maximum power. The maximum efficiency if you use the Jacobi theorem is 50%; electricity distribution authorities achieve closer to 95% transmission efficiency.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.