A Thevenin simulator: a network beside its equivalent
Mokxi is a free online Thevenin theorem simulator that runs in your browser, with nothing to install and no account needed. A three-resistor network on 12 V, and beside it the one source and one resistor Thevenin’s theorem says it reduces to. Each drives the same 10 k load with a meter across it, and the two meters agree. Change the loads and they keep agreeing.
- 455parts on the bench
- 43boards running now
- 1.00xreal time, on every board
What you can do here
In the circuit above, or in the editor it opens.
- Work out Vth and Rth and check them against a running circuit
- Measure the open-circuit voltage and the short-circuit current
- Change the load and watch the network and its equivalent agree
- Find the load that takes the most power from the network
- Swap a resistor for a potentiometer and find Rth by experiment
- Give every student a different network to reduce
What the running circuit shows
Two circuits, one answer.
The top half is 12 V into R1 (4.7 k), with R2 (10 k) from the middle to ground and R3 (2.2 k) out to the load. Thevenin’s theorem says that, seen from the load, any network of sources and resistors is one voltage behind one resistance. Take the load off and R1 and R2 are a divider: Vth = 12 × 10 / 14.7 = 8.16 V. Replace the source with a short and look back in: R1 and R2 in parallel are 3.20 k, and with R3 that is Rth = 5.40 k.
The bottom half is that equivalent, 8.16 V behind 5.40 k. With a 10 k load on each, the top meter reads about 5.30 V and so does the bottom one. Make both loads 1 k and both read within a millivolt of 1.275 V. Delete both loads and both read the open-circuit voltage, 8.16 V. Make both loads 1 ohm and each carries about 1.51 mA, the short-circuit current, and 8.16 V / 1.51 mA gives Rth back, 5.40 k. That is how you find Rth on a real bench without taking the network apart.
The equivalent also says which load gets the most power: one equal to Rth. With both loads at 5.4 k each reads about 4.08 V, about 3.1 mW. At 10 k the load gets about 2.8 mW, and at 2.2 k about 2.5 mW. That is the maximum power transfer theorem, measured rather than derived.
How the circuit is modeled
One exact solve, and meters that load it a little.
Every resistor and source is in the same nodal solve, so the voltages are the circuit’s own, to the millivolt, not a formula the page is quoting. The bottom source is rounded to 8.16 V and its resistor to 5.40 k, the way you would write them down, so the bottom meter reads about 3 mV under the top one: 8.156 V against 8.159 V open. The volts meters have a 10 megohm input, which loads a 5.4 k source by about half a part in a thousand, as a real meter does.
What it leaves out: nothing that matters for a network of resistors and sources, which is what Thevenin’s theorem covers. Real resistors have a tolerance, 5 percent for the common ones, so on a bench the two halves agree only as well as the parts match. Change any one value here and the equivalent stops matching until you work it out again, which makes a good exercise.
Questions
Is it free?
Yes. The circuit on this page, every part in it and the editor it opens in are on the free plan, with no card.
Do I need an account?
No. The circuit runs the moment the page does, and the editor opens without one. A free account is only for saving a project to the cloud and sharing it by link.
How do I find the Thevenin equivalent of a circuit?
Remove the load and find the voltage across its terminals: that is Vth. Then replace every voltage source with a short and every current source with an open, and find the resistance looking back in: that is Rth. Or measure both, since Rth is the open-circuit voltage divided by the short-circuit current.
What is the maximum power transfer theorem?
A load takes the most power from a source when its resistance equals the source’s Thevenin resistance. Here that is 5.40 k, which gets about 3.1 mW, more than a 10 k or a 2.2 k load does.
Can I do a Thevenin lab without hardware?
Yes. Build the network on the breadboard, measure the open-circuit voltage and the short-circuit current with the meter, build the equivalent beside it, and check that the two agree on a few loads. Every value is editable.
More simulators
Each one opens on a circuit that is already running.
Reduce a network
Open the circuit, change the loads and watch both halves agree. Nothing to install, and no account needed.