Library / Electrical / ELE-30
Electrical
Load current from a Thevenin source
Current drawn by a resistive load from any linear DC network, once the network has been replaced by its Thevenin equivalent: an ideal voltage source in series with a resistance.
- Ref
- ELE-30
- Rev
- v1
- Field
- Electrical
- Result
- IL [A]
Any linear network seen from two terminals behaves like a voltage source Vth in series with a resistance Rth. A load RL connected to these terminals draws the current IL = Vth/(Rth + RL). Vth is the open-circuit voltage and the short-circuit current is Vth/Rth.
Current drawn by a resistive load from any linear DC network, once the network has been replaced by its Thevenin equivalent: an ideal voltage source in series with a resistance.
Calculator
- Short-circuit current, Vth/Rth3 A
- Load voltage, ILRL6 V
- Voltage lost inside the source, ILRth6 V
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 12 V source with 4 Ω of internal resistance drives 1.5 A into a 4 Ω load: 50 % of the short-circuit current of 3 A (exactly half when the load equals the source resistance).
- A load much larger than Rth draws almost no current (an open circuit); a load much smaller draws nearly the short-circuit current.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (A) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Vth Thevenin (open-circuit) voltage | 1 to 100 V | 0.125 | 12.5 | 12.38 | +10% | +10 % |
| RL Load resistance | 0.1 to 1,000 Ω | 0.01195 | 2.927 | 2.915 | +10% | −4.76 % |
| Rth Thevenin resistance | 0.1 to 100 Ω | 0.1154 | 2.927 | 2.811 | +10% | −4.76 % |
Sweep of Load resistance
| Load resistance (Ω) | Load current (A) |
|---|---|
| 0.1 | 2.927 |
| 0.3162 | 2.78 |
| 1 | 2.4 |
| 3.162 | 1.675 |
| 10 | 0.8571 |
| 31.62 | 0.3369 |
| 100 | 0.1154 |
| 316.2 | 0.03747 |
| 1,000 | 0.01195 |
Sweep one variable
Variables and how each one contributes
Load current
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 current in direct proportion. Raising it by 1 V (to 13 V) moves the result by +8.33 %, from 1.5 to 1.625 A.
Thevenin resistance
The Thevenin resistance seen from the load terminals, in ohms: the source’s internal resistance. Lowers the load current: more resistance in the source limits the current. Raising it by 0.5 Ω (to 4.5 Ω) moves the result by −5.88 %, from 1.5 to 1.412 A.
Load resistance
The load resistance, in ohms. Lowers the load current: a larger load resistance draws less current. Raising it by 0.5 Ω (to 4.5 Ω) moves the result by −5.88 %, from 1.5 to 1.412 A.
Worked example with the values above
- Total series resistance: Rth + RL = 4 + 4 = 8 Ω.
- Apply Ohm’s law: 12 V over 8 Ω.
- IL = 1.5 A.
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.
Statistics and uncertainty
The current depends on the sum Rth + RL: a 5 % tolerance on each resistor gives about a 3.5 % error in the sum when they are equal, and so in the current. Measuring Vth and Rth by an open-circuit and a loaded test carries the meter’s errors as well.
Where it is used
- Finding the current drawn from a battery, supply or sensor with internal resistance.
- Reducing a complicated network to one equivalent loop.
Maths you need
- Thevenin equivalent: a source and one resistance that replace the rest of the circuit
- Ohm’s law: V = IR