Superposition Theorem: Solve Linear Circuits One Source at a Time

The superposition theorem is a method for analyzing linear electrical circuits that contain two or more independent sources. Instead of solving the complete circuit with every source active at once, you calculate the contribution from each independent source separately and then add the contributions algebraically. It is especially useful for circuits containing multiple voltage sources, current sources, and resistors.

The theorem applies to linear circuits: the response must be proportional to the applied input and satisfy additivity. Ideal resistors and linear dependent-source models fit this framework. Components such as diodes and transistors can operate nonlinearly, so a simple superposition calculation is not generally valid across arbitrary operating regions. Small-signal linear models may be used only within their assumptions.

A crucial rule is that superposition applies to voltage and current, not directly to power. Because power depends quadratically on voltage or current, calculating power from each source contribution separately and adding those powers can give the wrong answer. This guide develops the procedure, examples, dependent-source handling, and common mistakes.

1. The theorem in plain language

In a linear circuit with multiple independent sources, the voltage or current in any selected branch equals the algebraic sum of the contributions produced by each independent source acting alone, with all other independent sources set to zero. The passive components remain in the circuit, and dependent sources remain active.

“Set a source to zero” means replace it with its ideal zero-value equivalent. An ideal voltage source set to 0 V becomes a short circuit. An ideal current source set to 0 A becomes an open circuit. It does not mean removing every component connected to that source, nor does it mean shorting a current source or opening a voltage source.

The theorem is valuable because it separates a complicated problem into simpler subproblems. For example, a branch current may be driven partly by a 12 V source and partly by a 2 A current source. Find each signed contribution using a consistent reference direction, then add the results. If one contribution flows opposite to the reference direction, it is negative and reduces the total.

2. Conditions and limits

Superposition requires a linear circuit model. For resistive networks, this usually means ideal voltage sources, ideal current sources, and linear resistors with fixed values. Linear inductors and capacitors can also be included in appropriate linear time-domain or frequency-domain analysis. The theorem is not a shortcut for adding nonlinear device behavior.

If a diode changes between conducting and nonconducting states as sources are considered separately, the overall circuit may not obey a single linear model. Similarly, a transistor's operating point can shift substantially when a source is removed. A small-signal equivalent circuit can be analyzed by superposition if the device is linearized around a defined operating point, but the result is valid only within that model's range.

Superposition can be used with ideal independent sources and with dependent sources, but dependent sources are not switched off. A dependent source is controlled by a voltage or current elsewhere in the circuit; its value changes according to the controlling variable. Turning it off would change the circuit model and generally produce an incorrect result.

3. The source-suppression rules

For each subproblem, choose one independent source to remain active and suppress the other independent sources:

• An ideal independent voltage source of value (V_s) is replaced by a short circuit when suppressed, because its voltage becomes zero.
• An ideal independent current source of value (I_s) is replaced by an open circuit when suppressed, because its current becomes zero.
• A resistor or other passive component remains in place.
• A dependent source remains active because its output depends on another circuit variable.

Real sources often have internal resistance. If a practical source is modeled as an ideal source in series or parallel with a resistance, the resistance stays in the circuit when the ideal source is suppressed. Replacing the entire real-source model with a wire or open circuit can discard a meaningful part of the circuit.

Before calculating, redraw the circuit for each source case. Mark the selected branch voltage polarity or current reference direction and keep it consistent across every case. This prevents sign errors when the partial contributions are added.

4. Worked example with two voltage sources

Consider a single loop containing a 6 Ω resistor and a 3 Ω resistor in series, driven by two ideal voltage sources whose polarities oppose each other: a 12 V source and a 3 V source. Define the clockwise loop current as positive, and orient the sources so the 12 V source drives clockwise current while the 3 V source drives counterclockwise current.

With both sources active, the net driving voltage is (12-3=9) V and total resistance is (6+3=9) Ω. The loop current is (I=9/9=1) A clockwise.

Now use superposition. First keep the 12 V source and replace the 3 V source with a short circuit. The current contribution is (I_1=12/9=1.333) A clockwise. Next keep the 3 V source and short the 12 V source. The 3 V source drives counterclockwise, so relative to the chosen clockwise reference its contribution is (I_2=-3/9=-0.333) A. Add the contributions: (I=I_1+I_2=1.333-0.333=1) A clockwise, matching the direct solution.

This simple example illustrates why reference directions matter. The contribution from the opposing source is not “negative current” in a physical sense; the negative sign simply indicates that it flows opposite to the chosen reference direction.

5. Worked example with a current source and resistor network

Suppose a node is connected to ground through two resistors, 6 Ω and 3 Ω, and an ideal 2 A current source injects current into the node. The voltage across both resistors is the same node voltage. Their parallel equivalent resistance is (R_{eq}=(6\times3)/(6+3)=2) Ω, so the node voltage is (V=IR=2\times2=4) V. The 6 Ω branch current is (4/6=0.667) A, and the 3 Ω branch current is (4/3=1.333) A; together they sum to 2 A.

Now add an independent 6 V source connected to the same node through a 3 Ω resistor, while retaining the two shunt resistors. Superposition can be applied to find the node voltage contribution from each source. With the current source suppressed, open its branch; the voltage source drives current through its series resistor into the parallel load. With the voltage source suppressed, short its ideal voltage and retain the series resistor, while the current source drives the resulting network.

The exact circuit must be redrawn carefully for each case. This example emphasizes a general point: suppressing a source changes its ideal value, not the surrounding resistors. If you remove a branch resistor along with a source, the equivalent resistance and answer can change.

6. Calculate voltage and current contributions

Superposition can be used to calculate a chosen branch current or a chosen node voltage. For each source case, solve the reduced circuit with Ohm's law, series/parallel reduction, nodal analysis, mesh analysis, or another valid linear-circuit technique. You do not need to use the same solving method for every subproblem, as long as the circuit model and sign convention remain consistent.

If the target is a branch voltage, define a polarity, such as positive at the upper terminal relative to the lower terminal. Each partial voltage is measured using that same polarity. If the target is current, define a reference arrow. Contributions that oppose the arrow are negative. Algebraic addition then yields the total.

When a circuit is more complex, use a table to organize results: source case, active source, suppressed sources, target voltage/current, and signed contribution. This makes it easier to catch a missed source or an incorrect sign. After summing the contributions, verify the result with Kirchhoff's current and voltage laws or a direct circuit solution when practical.

7. What happens to dependent sources?

A dependent source is controlled by another circuit quantity. For example, a dependent voltage source might be (2v_x), or a dependent current source might be (0.5i_x). When using superposition, independent sources are suppressed one at a time, but the dependent source remains in the circuit. Its value is determined by the controlling voltage or current in each subproblem.

If the dependent source is present, simple resistor reduction may not be enough. Nodal or mesh equations are often clearer. Write the controlling relationship and include it in the equations for each source case. The resulting contribution can then be added like any other linear response.

When finding the equivalent resistance seen by a circuit containing dependent sources, do not simply suppress all sources and combine resistors. A test source may be needed to determine the port voltage-current relationship. This is a related but distinct application of linear-circuit analysis. The main superposition rule remains: dependent sources stay active because their behavior is part of the circuit's model.

8. Why you cannot directly superimpose power

Power in a resistor is (P=I^2R) or (P=V^2/R). Because the current or voltage is squared, the power due to the total response is generally not the sum of the powers calculated from each source contribution separately.

For example, if two sources independently produce 1 A each through the same 1 Ω resistor in the same direction, the total current is 2 A and total resistor power is (P=2^2\times1=4) W. Calculating each contribution's power gives 1 W and 1 W, whose sum is 2 W—not the correct 4 W. The missing term comes from the interaction when the contributions combine.

If the source contributions oppose each other, the discrepancy can go the other way. First use superposition to find the total branch voltage or current, then calculate power from that total value. This rule applies to average power in the usual resistive example and should be handled carefully in AC circuits with phase relationships.

9. Common mistakes and debugging

Shorting a current source. A suppressed ideal current source becomes an open circuit, not a short.

Opening a voltage source. A suppressed ideal voltage source becomes a short circuit, not an open.

Removing source resistance. Keep any explicit series or parallel resistance that belongs to the source model.

Turning off dependent sources. Keep dependent sources active and maintain their controlling relationships.

Changing the reference direction. Measure every partial contribution using the same target polarity or current arrow.

Adding powers instead of voltages or currents. Add the signed voltage or current contributions first, then calculate power from the total.

Using the theorem on a nonlinear circuit without justification. Confirm that the model is linear over the operating range, or use an appropriate nonlinear analysis instead.

A good check is to solve the complete circuit directly if it is small enough, then compare the result with the superposition sum. Agreement is a strong check against arithmetic and sign mistakes, though it does not replace checking whether the original circuit model is valid.

10. A repeatable solution workflow

1. Identify the target branch voltage or current and choose a reference polarity or direction.
2. Confirm that the circuit model is linear for the analysis being performed.
3. List all independent sources and any dependent sources.
4. Keep one independent source active and suppress every other independent source using the correct zero-value equivalent.
5. Retain all passive components and all dependent sources.
6. Solve for the target response in that source case and record the signed contribution.
7. Repeat for every independent source.
8. Add the contributions algebraically using the same reference convention.
9. Verify the result using KCL, KVL, or a direct solution if practical.
10. Calculate power only after obtaining the total voltage or current.

Superposition is not always the fastest method for every circuit, but it is a valuable way to understand how multiple sources influence one response. It also provides a useful check on nodal and mesh analysis. Careful source suppression, consistent signs, and a clear distinction between linear responses and nonlinear power calculations make the method reliable.

11. Using superposition with AC sources

In a linear AC circuit, superposition can be applied to sinusoidal signals, but each source contribution must be represented consistently. If all sources operate at the same frequency, use phasors with a common frequency reference and add the complex voltage or current phasors. Adding only magnitudes discards phase information and can produce the wrong result. For sources at different frequencies, analyze each frequency separately using the circuit impedance at that frequency, then combine the time-domain responses when appropriate.

For example, two same-frequency sources may contribute currents with different phase angles. Their total RMS current is not generally the arithmetic sum of the two RMS magnitudes; it is the magnitude of the vector sum of their phasors. Once the total waveform or phasor response is known, calculate power using the correct AC power relationships. Do not mix a DC result, a peak value, and an RMS phasor without converting them to a consistent representation. This extends the same core lesson: superposition adds linear responses with signs and phase preserved, not arbitrary numbers detached from their definitions.