Verification of Thevenin's theorem
Verifying Thevenin's theorem experimentally: measuring V_th and R_th, predicting load current, comparing with the measured value, maximum power transfer, and viva questions with answers.
Aim
To verify Thevenin's theorem experimentally by determining the Thevenin equivalent of a linear network and comparing the load current it predicts with the measured value.
Apparatus required
| Apparatus | Specification | Qty |
|---|---|---|
| Regulated DC power supply | 0–30 V, 2 A | 1 |
| Resistors | As per the circuit, ¼ W, ±5 % | As required |
| Decade resistance box | For the variable load | 1 |
| Digital multimeter | For voltage and resistance | 1 |
| Milliammeter (MC) | 0–50 mA | 1 |
| Breadboard and patch cords | — | 1 set |
Theory
Thevenin's theorem says that any linear, bilateral, two-terminal network of sources and resistances can be replaced, as far as anything connected to those two terminals is concerned, by a single voltage source V_th in series with a single resistance R_th.
V_th is the voltage measured across the two terminals with the load removed — the open-circuit voltage. R_th is the resistance looking back into those terminals with every independent source replaced by its internal resistance: voltage sources shorted, current sources opened.
R_th can also be found without disturbing the sources at all, by measuring the short-circuit current at the terminals: R_th = V_oc / I_sc. That is the route this experiment uses, because it needs no rewiring of the network, and comparing it with the looking-back value is a check on both.
Once the equivalent is known, the current through any load is simply I_L = V_th/(R_th + R_L) — a one-line calculation instead of a full network solution, and that is the whole value of the theorem. It also makes maximum power transfer obvious: the load draws the most power when R_L = R_th, and that maximum is V_th²/4R_th.
The theorem holds only for linear, bilateral elements. A diode, a transistor or any element whose resistance depends on the current through it puts the network outside its scope.
Circuit connections
Check every point below against your board before switching on. There is no diagram here on purpose — a wrong diagram is worse than none, and this is the list a demonstrator actually walks through with you.
- Build the given network on the breadboard, leaving the two load terminals A and B free.
- For V_oc: connect the voltmeter across A–B with nothing else attached. A digital voltmeter's input resistance is high enough not to load the network appreciably.
- For I_sc: connect the milliammeter directly across A–B. Choose a range above the expected current before connecting.
- For the load reading: connect R_L across A–B with the milliammeter in series.
Procedure
- 1Assemble the network and set the supply to the specified voltage. Note the actual value, not the dial marking.
- 2With the load disconnected, measure the open-circuit voltage across A–B. This is V_th.
- 3Replace the open circuit with the milliammeter alone and measure the short-circuit current I_sc. Compute R_th = V_oc/I_sc.
- 4As a cross-check, switch the supply off, replace it by a short, and measure the resistance across A–B with the multimeter. It should agree with the value above.
- 5Connect the load resistance R_L with the milliammeter in series and measure the load current.
- 6Compute the predicted current from the Thevenin equivalent and find the percentage error against the measured value.
- 7Repeat for two or three values of R_L, including one equal to R_th, to show maximum power transfer.
Work out your readings
Type in the numbers off the meters. This fills the tabular column, works the calculation through step by step, plots the characteristic — and tells you when a reading cannot physically be right, which is the part a manual can't do. Everything stays on this device, and it works with the network off.
Nameplate and machine data
Fill in the machine data to see the results, the worked calculation and the curve.
Precautions
- Switch the supply off before shorting it out for the looking-back resistance measurement — measuring resistance in a live circuit damages the meter and gives a meaningless reading.
- Never connect the milliammeter across a supply directly; it must always be across the network's output terminals, whose current is limited by the network.
- Choose meter ranges before connecting, and start on the highest.
- Check the actual resistor values with the multimeter rather than trusting the colour code — a ±5 % band is a real ±5 %.
Sources of error
Every record asks for these, and every record gets the same three lines copied from the one before. These are the errors this particular experiment actually has.
- Ammeter resistance: the milliammeter's own resistance is in series with the load, so the measured current is always slightly less than the true one. In a high-resistance network this is the dominant error.
- Voltmeter loading: the voltmeter draws a small current when measuring V_oc, so the reading is slightly below the true open-circuit value.
- Resistor tolerance — several ±5 % resistors in one network can easily give a 5–8 % discrepancy on their own.
- Contact resistance at breadboard connections, which matters when the network resistances are small.
Viva questions with answers
State Thevenin's theorem.
Any linear, bilateral, two-terminal network can be replaced by an equivalent circuit consisting of a single voltage source V_th, equal to the open-circuit voltage at the terminals, in series with a single resistance R_th, equal to the resistance looking back into the terminals with all independent sources replaced by their internal resistances.
How do you handle sources when finding R_th?
Independent voltage sources are replaced by a short circuit, independent current sources by an open circuit. Dependent sources are not removed — they are kept, and R_th is found by applying a test source at the terminals and taking the ratio of test voltage to test current.
What is the relationship between Thevenin's and Norton's theorems?
They are duals. The Norton equivalent is a current source I_N = V_th/R_th in parallel with the same resistance R_N = R_th. Either can be converted into the other by a source transformation.
When does Thevenin's theorem not apply?
When the network contains non-linear elements such as diodes or transistors, or unilateral elements that behave differently in the two directions, or when the load is magnetically or otherwise coupled back into the network.
State the maximum power transfer condition and the efficiency at that point.
Maximum power is transferred to the load when R_L = R_th, and the maximum is V_th²/4R_th. The efficiency at that point is only 50 %, because an equal amount is dissipated in R_th. That is why power systems are never operated at maximum power transfer — they are operated for maximum efficiency instead.
Why does the measured current come out slightly lower than the predicted one?
Chiefly because the ammeter has its own resistance, which adds to R_L and reduces the current. Resistor tolerance and voltmeter loading account for the rest.
What does bilateral mean?
That the element behaves identically for current in either direction — a resistor is bilateral, a diode is not.