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### Network # Energy transfer problem

#### jacklam2048

Feb 22, 2021
28
Hi,
I have a fundamental problem about energy transfer:
According to the Maximum Power Transfer Theorem, maximum power is transferred from source to load when the load resistance is equal to the source's internal resistance. In this case, the efficiency is 50%.

Now the question comes:
suppose I have a lamp which is 10W at 12V, and I have a battery pack at 12V, 2000mAH (assuming the output voltage is stable). When I calculate how long the battery pack can power the lamp. I use 10W / (12V x 2AH) = 0.42 hours, but this is assuming 100% efficiency of energy transfer.
So, if according to the Maximum Power Transfer Theorem, should I multiply this 0.42 hours by 50% ?

Thank you.

#### Bluejets

Oct 5, 2014
5,881
Visit battery university as batteries lose their capacity at differing charge and discharge rates and it also depends on the battery rating itself.

#### jacklam2048

Feb 22, 2021
28
Hi, my point is whether the Maximum Power Transfer Theorem should apply ?
If not, why ?

#### Nanren888

Nov 8, 2015
623
The battery has an internal resistance which will limit the power that it can supply according to the theorem you talk about.
This is not energy, but energy per second, power.
Generally the internal resistance is low, so the maximum power (energy per second) it can supply is large. The power transfer for maximum power transfer is a lot greater than the lamp wants or can handle, so the resistance of the lamp is much higher than the battery internal resistance.
.
If you'd like to look up a realistic internal resistance for a battery you can calulate the maximum power and also the amount of power lost in the internal resistance when powering that lamp. Integrating this will give you energy.

• Harald Kapp

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,614
The maximum power transfer theorem states that maximum power transfer is achieved when the load impedance matches the source impedance. Source impedance here is the internal resistance of the battery, load resistance is the resistance of e.g. your lamp.
So, if according to the Maximum Power Transfer Theorem, should I multiply this 0.42 hours by 50% ?
No, makes no sense. Why would you multiply by 50 %? Your load impedance (lamp) and source impedance (battery) are not equal, so why should the power you calculated be divided equally between the two?

Besides, your calculation is wrong:
10W / (12V x 2AH) = 0.42 hours
10W / (12V × 2Ah) = 0.42 1/h which is a useless unit in this case. The correct calculation goes:
12 V × 2 Ah = 24 Wh
24 Wh / 10 W = 2.4 h
The battery will last ~ 2.4 hours.

#### AnalogKid

Jun 10, 2015
2,665
Hi, my point is whether the Maximum Power Transfer Theorem should apply?
If not, why ?
Not. Because the theorem applies to a constant-voltage source with infinite current capability. The effects of the theorem appear when the load impedance is nearly equal to the source impedance. If we assume that the battery output impedance (ESR) is 0.1 ohm, and the load is 14.4 ohms, then clearly the majority of the circuit energy (10 W) will be dissipated in the load rather than the ESR. But that is not the maximum power available.

To see the MPTP in action, let's change to a larger battery, with a 12 V output at up to 500 amps, and an internal impedance of 0.1 ohm.

Now let's decrease the load to 0.2 ohm. The circuit current is 40 A, the power dissipated in the load is 320 W, and the power dissipated in the battery's internal resistance is 160 W.

Now let's decrease the load to 0.1 ohm. The circuit current increases to 60 A, the power dissipated in the load increases to 360 W, and the power dissipated in the battery's internal resistance increases to 360 W.

Keep going, and decrease the load impedance even more, to 0.05 ohm. The circuit current is 80 A, the power dissipated in the load is *decreases* to 320 W, and the power dissipated in the battery is 640 W Decreasing the load impedance more causes the load power to decrease more. With a theoretically perfect zero ohm load, the circuit current is 120 A, but the power in the load is zero watts.

And, your math is off.10 W at 12 V >> i =0.833 A
2 Ah / 0.833 A = 2.4 hours

ak

Last edited:

#### jacklam2048

Feb 22, 2021
28
Thank you very much ! your explanation makes me understood !

#### jacklam2048

Feb 22, 2021
28
The maximum power transfer theorem states that maximum power transfer is achieved when the load impedance matches the source impedance. Source impedance here is the internal resistance of the battery, load resistance is the resistance of e.g. your lamp.

No, makes no sense. Why would you multiply by 50 %? Your load impedance (lamp) and source impedance (battery) are not equal, so why should the power you calculated be divided equally between the two?

Besides, your calculation is wrong:

10W / (12V × 2Ah) = 0.42 1/h which is a useless unit in this case. The correct calculation goes:
12 V × 2 Ah = 24 Wh
24 Wh / 10 W = 2.4 h
The battery will last ~ 2.4 hours.

#### jacklam2048

Feb 22, 2021
28
Thank you very much for pointing out my mistake as well as to your explanation !

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