Verification of the superposition theorem
Superposition theorem experiment: replacing each source in turn by its internal resistance, measuring the individual and combined currents, why power does not superpose, tabular column, and viva questions with answers.
Aim
To verify the superposition theorem in a two-source DC network by measuring the branch current produced by each source acting alone and confirming that their algebraic sum equals the current measured with both sources acting together.
Apparatus required
| Apparatus | Specification | Qty |
|---|---|---|
| Regulated DC power supplies | 0–30 V, 1 A, two independent outputs | 2 |
| Resistors | 220 Ω, 470 Ω, 1 kΩ, ½ W | 1 each |
| Digital multimeter | DC milliamps and DC volts | 1 |
| Shorting link or patch cord | To replace a source by its internal resistance | 1 |
| Breadboard and patch cords | — | 1 set |
Theory
The superposition theorem states that in any linear bilateral network containing more than one independent source, the current through or voltage across any element equals the algebraic sum of the currents or voltages produced by each source acting alone, with all other independent sources replaced by their internal resistances.
Replacing a source by its internal resistance is the part most often got wrong. An ideal voltage source has zero internal resistance, so it is replaced by a short circuit. An ideal current source has infinite internal resistance, so it is replaced by an open circuit. A practical source is replaced by its internal resistance alone, not removed. Dependent sources are never suppressed — they are controlled by the circuit and must stay in place throughout.
The theorem rests entirely on linearity. Superposition follows from the fact that in a linear network every branch current is a linear combination of the source values, so the response to a sum of inputs is the sum of the responses. Any element whose characteristic is not a straight line — a diode, a transistor, an iron-cored inductor driven into saturation — breaks that premise, and the theorem does not apply.
Power does not superpose, and this experiment shows why directly. Power is proportional to the square of the current, and squaring is not a linear operation: if I1 and I2 are the currents from each source alone, then (I1 + I2)²R is not I1²R + I2²R, the difference being the cross term 2·I1·I2·R. Computing the power from the individually measured currents and comparing it with the power from the combined current makes the discrepancy plain.
The practical value of the theorem is that it turns one difficult multi-source problem into several easy single-source ones, each solvable by simple series-parallel reduction. It also underlies the way a small AC signal is analysed separately from the DC bias in an amplifier.
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.
- A two-mesh network: source V1 with its series resistor R1 in the left branch, source V2 with its series resistor R2 in the right branch, and the common resistor R3 forming the shared middle branch.
- The ammeter in series with the middle branch R3, which is the element whose current is being verified; keep it in the same place and the same polarity for all three readings.
- Both supplies referenced to a single common node, so the two sources share a ground and the mesh is genuinely closed.
- A shorting link ready at each source's terminals, so that source can be replaced by a short circuit while the other acts alone.
- Both supplies set to minimum before switching on, and the ammeter polarity noted so that a reversal of current direction is recorded as a sign rather than ignored.
Procedure
- 1Build the two-mesh network and have the connections checked, with both supplies at minimum.
- 2Set V1 and V2 to their chosen values, note them, and record the current through R3 with both sources acting. This is the combined reading.
- 3Switch off, disconnect V2 and replace it with a shorting link across its terminals, leaving R2 in circuit. Switch on and record the current through R3 due to V1 alone, with its sign.
- 4Switch off, restore V2, then remove V1 and replace it with a shorting link across its terminals. Switch on and record the current through R3 due to V2 alone, again with its sign.
- 5Add the two individual currents algebraically and compare the sum with the combined reading, computing the percentage difference.
- 6Repeat the whole set for at least two more pairs of source voltages, and if time allows for a different branch of the network.
- 7Compute the power in R3 from each individual current and from the combined current, and compare the sum of the first two with the last to show that power does not superpose.
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
- Replace a suppressed voltage source with a short circuit across its terminals, not by removing it and leaving the branch open — an open branch changes the network topology and gives a wrong result.
- Leave the series resistance of a suppressed source in circuit; only the EMF is removed, never the resistance.
- Record the direction of every current. Superposition is an algebraic sum, and treating an opposing current as positive is the commonest source of a large error.
- Switch off the supplies before making or breaking any connection, particularly when inserting the shorting links.
- Keep the ammeter in the same branch and the same orientation for all three readings so the sign convention stays consistent.
- Check that no resistor exceeds its power rating at the highest source voltages used.
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.
- The burden voltage of the ammeter: its own resistance sits in the branch being measured and reduces the current slightly, and the effect differs between readings if the range is changed.
- Resistor tolerance — ordinary ±5 % resistors mean the network is not quite the one on paper, though the theorem is still verified within that band.
- The finite internal resistance of the supplies, so a real source shorted by a link is not a perfect short.
- Self-heating of the resistors during a long run, which drifts their values between the first and last readings.
- Rounding the individual currents before adding them, which can make a genuinely small discrepancy look larger than it is.
Viva questions with answers
State the superposition theorem and the conditions under which it applies.
In a linear bilateral network with several independent sources, the response in any branch is the algebraic sum of the responses caused by each source acting alone with the others replaced by their internal resistances. It requires linearity and bilaterality, so it fails for circuits containing diodes, transistors operated over a non-linear range, or saturating magnetic elements.
How is an ideal voltage source replaced when it is suppressed, and how is a current source?
An ideal voltage source is replaced by a short circuit, because its internal resistance is zero. An ideal current source is replaced by an open circuit, because its internal resistance is infinite. A practical source is replaced by its internal resistance alone; the EMF is removed but the resistance stays in the network.
Why can the superposition theorem not be used to calculate power directly?
Power depends on the square of the current, and squaring is non-linear. The power due to the combined current is (I1 + I2)²R, which exceeds I1²R + I2²R by the cross term 2·I1·I2·R. Superposition must therefore be applied to currents or voltages first, and power computed afterwards from the resultant.
What happens to dependent sources when superposition is applied?
They are left in the circuit for every step. A dependent source is not an independent input but a response controlled by a voltage or current elsewhere in the network, so suppressing it would change the network itself. Only independent sources are taken one at a time.
Why must the direction of each current be recorded rather than just its magnitude?
Superposition is an algebraic sum, so a current produced by one source may oppose that produced by another. If the opposing current is added as a positive quantity the sum will be badly wrong, often larger than the measured combined value. The sign carries the physics and must be preserved.
What is the practical advantage of the theorem?
It reduces one multi-source network, which would otherwise need simultaneous mesh or nodal equations, into several single-source networks that can each be solved by simple series-parallel reduction. It is also what justifies analysing the DC bias and the AC signal of an amplifier separately.
Is superposition applicable to an AC network?
Yes, provided the network is linear. With sources at the same frequency the responses are added as phasors. With sources at different frequencies each response must be found separately and the results combined in the time domain, since phasors of different frequencies cannot be added.