Library / Electrical / ELE-32
Electrical
Power delivered to a load by a Thevenin source
Power dissipated in a resistive load fed by a Thevenin source.
- Ref
- ELE-32
- Rev
- v1
- Field
- Electrical
- Result
- PL [W]
The power delivered to the load is PL = IL²RL = Vth²RL/(Rth + RL)². It is zero for a short circuit (RL = 0) and for an open circuit, and has a maximum in between, at RL = Rth, where it equals Vth²/(4Rth). Half of the total power is then dissipated in the source, so the efficiency is 50 %.
Power dissipated in a resistive load fed by a Thevenin source. It rises to a maximum at a particular load resistance, then falls.
Calculator
- Maximum possible load power, Vth²/(4Rth)9 W
- Fraction of the maximum power delivered1
- Efficiency, RL/(Rth + RL)0.5
- Power lost inside the source9 W
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A 4 Ω load on a 12 V, 4 Ω source takes 9 W, which is 100 % of the maximum of 9 W; the source itself wastes 9 W.
- Matching gives the most power but only 50 % efficiency: power systems use loads far larger than Rth for efficiency, while signal and radio circuits match to get the most power from a weak source.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (W) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Vth Thevenin (open-circuit) voltage | 1 to 100 V | 0.0625 | 625 | 624.9 | +10% | +21 % |
| Rth Thevenin resistance | 0.1 to 100 Ω | 0.05325 | 34.27 | 34.21 | +10% | −9.3 % |
| RL Load resistance | 0.1 to 1,000 Ω | 0.1429 | 9 | 8.857 | +10% | −0.227 % |
Sweep of Load resistance
| Load resistance (Ω) | Load power (W) |
|---|---|
| 0.1 | 0.8566 |
| 0.3162 | 2.444 |
| 1 | 5.76 |
| 3.162 | 8.877 |
| 10 | 7.347 |
| 31.62 | 3.588 |
| 100 | 1.331 |
| 316.2 | 0.4441 |
| 1,000 | 0.1429 |
Sweep one variable
Variables and how each one contributes
Load power
What the equation gives you. Everything else feeds it.
Thevenin (open-circuit) voltage
The open-circuit (Thevenin) voltage of the source, in volts. Raises the load power with the square of the voltage. Raising it by 1 V (to 13 V) moves the result by +17.4 %, from 9 to 10.56 W.
Thevenin resistance
The Thevenin resistance seen from the load terminals, in ohms: the source’s internal resistance. Lowers the load power here, because the load is already matched to the source: any change in Rth moves away from the maximum. Raising it by 0.5 Ω (to 4.5 Ω) moves the result by −11.4 %, from 9 to 7.972 W.
Load resistance
The load resistance, in ohms. Lowers the load power slightly here, because the load is already matched to the source: a change in either direction moves away from the maximum. Raising it by 0.5 Ω (to 4.5 Ω) moves the result by −0.346 %, from 9 to 8.969 W.
Worked example with the values above
- Current: IL = Vth/(Rth + RL) = 12 / 8 = 1.5 A.
- Square it and multiply by the load: 1.5² × 4.
- PL = 9 W.
Assumptions and limits
- A linear circuit that can be replaced by a Thevenin equivalent: a voltage source Vth in series with a resistance Rth.
- Steady DC conditions, or AC with purely resistive impedances.
Common mistakes
- Using the load’s resistance as Rth: Rth belongs to the source side.
- Forgetting that the Thevenin voltage is the open-circuit voltage, measured with the load removed.
- Assuming matching also gives the best efficiency: it gives only 50 %.
Statistics and uncertainty
Near the match the power varies very little with the load resistance: a 20 % mismatch still delivers 99 % of the maximum, so the exact value of RL is not critical, but it is when the load is far from matched.
Where it is used
- Choosing a load to extract the most power from a battery, solar cell, antenna or sensor.
- Estimating how much power is wasted in the source.
Maths you need
- Thevenin equivalent: a source and one resistance that replace the rest of the circuit
- Ohm’s law: V = IR