Superposition theorem

A DC network with two or three independent sources feels harder than one with a single battery — every branch current seems to depend on everything at once. The superposition theorem cuts that tangle apart: switch off all but one source, solve the leftover single-source circuit with plain Ohm's law, repeat for each source in turn, then add the results. Because resistive circuits are linear, this shortcut is mathematically exact, not an approximation. It is the same DC-networks unit that also covers node-voltage, mesh-current and star-delta transformation, and it is the direct stepping stone to Thevenin's and Norton's theorems, which lean on the identical linearity fact. Get comfortable killing sources correctly and adding contributions with the right sign, and superposition becomes the fastest hand-tool in the whole unit for a quick branch current or voltage.

The core insight

In a linear resistive network, the current or voltage produced by several independent sources acting together equals the algebraic sum of what each source produces acting alone, with every other independent source deactivated (voltage sources shorted, current sources opened).

The core idea

What linearity buys you. Ohm's law, $V = IR$ , is a straight-line relationship — double the voltage and the current doubles too, with no squared or cross terms. A network built only from resistors and independent sources inherits this property: the response (any branch current or voltage) to a sum of source excitations equals the sum of the responses to each excitation taken separately.

Deactivating a source. To isolate one source you must silence every other independent source without disturbing the circuit's topology:

An independent voltage source is deactivated by replacing it with a short circuit ( $V = 0$ ) — remove the source symbol, join its two terminals with a wire.
An independent current source is deactivated by replacing it with an open circuit ( $I = 0$ ) — remove the source symbol, leave a gap.

Any series resistance that sat alongside a deactivated source stays exactly where it was; only the source itself is replaced.

The four-step procedure. Pick the target branch quantity, then for each independent source in turn: activate only that source (deactivate all the rest as above), solve the resulting single-source circuit by Ohm's law and series-parallel reduction (or KCL/KVL if the circuit is not a simple ladder), and record its contribution to the target. Once every source has had its turn, add the contributions algebraically, respecting a single fixed reference direction:

X_{\text{total}} = X^{(1)} + X^{(2)} + \cdots + X^{(n)}

where: $X$ stands for the branch current or voltage being found, and $X^{(k)}$ is the contribution of the $k$ -th independent source acting alone with all others deactivated; a contribution that points opposite to the chosen reference direction is added as a negative number. Superposition applies to currents and voltages only. Power cannot be superposed, because power depends on the square of current or voltage, $P = I^2R$ , and $(a+b)^2 \ne a^2+b^2$ in general — the cross term $2ab$ is lost if you naively add individual-source powers. The correct route is always: superpose the currents or voltages first, then compute power once from the true total.

Figure

A resistor fed by two sources shown split into two single source sub circuits that are then added

Derivation ASuperposed voltage across a resistor fed by two independent sources

The goal is to derive, from Ohm's law and series-parallel reduction alone, the exact expression for the voltage $V_R$ across a resistor $R$ that is fed simultaneously by an independent voltage source $E$ (in series with resistance $R_1$ ) and an independent current source $I_s$ , both connected in parallel across $R$ , so that the final numeric answer can be checked term by term against each single-source sub-circuit.

Step 1. Deactivate the current source first, replacing it with an open circuit, since an open branch carries no current and simply drops out of the circuit. What remains is a single loop: $E$ , $R_1$ and $R$ in series. By Ohm's law the loop current is $I' = E/(R_1+R)$ , and the resulting voltage across $R$ from this source alone is

V_R' = I' R = \frac{E}{R_1+R}\,R.

Step 2. Restore the current source and instead deactivate the voltage source, replacing it with a short circuit but leaving its series resistance $R_1$ in the circuit. Now $I_s$ feeds a parallel combination of $R_1$ (the shorted branch) and $R$ . The current-divider rule gives the share of $I_s$ that flows through $R$ :

I_R'' = I_s\,\frac{R_1}{R_1+R}, \qquad V_R'' = I_R''\,R = I_s\,\frac{R_1}{R_1+R}\,R.

Step 3. Add the two contributions algebraically, using one fixed reference polarity for $V_R$ (say, the top of $R$ positive in both sub-circuits):

V_R = V_R' + V_R'' = \frac{E R}{R_1+R} + \frac{I_s R_1 R}{R_1+R}.

Step 4. Sanity-check the structure: both terms share the same denominator $R_1+R$ , confirming that the same series-parallel path carries both contributions, and each term individually reduces to the correct single-source answer when the other source is set to zero ( $I_s=0$ recovers Step 1; $E=0$ recovers Step 2). This is the general template behind the worked numbers below.

Worked example

A $12\,\text{V}$ source in series with $2\,\Omega$ , and a $2\,\text{A}$ current source, are both connected in parallel across a resistor $R = 4\,\Omega$ . Find the voltage across $R$ using superposition, and check what happens if you (wrongly) try to superpose power instead.

Step 1. Deactivate the $2\,\text{A}$ source (open it); it drops out entirely, leaving $12\,\text{V}$ in series with $2\,\Omega$ feeding $R = 4\,\Omega$ as a simple loop:

I' = \frac{12}{2+4} = 2\ \text{A}, \qquad V_R' = I' \times R = 2 \times 4 = 8\ \text{V}.

Step 2. Deactivate the $12\,\text{V}$ source (short it, keeping its series $2\,\Omega$ in place). The $2\,\text{A}$ source now sees $2\,\Omega$ and $R = 4\,\Omega$ in parallel; by current-divider,

I_R'' = 2 \times \frac{2}{2+4} = 0.667\ \text{A}, \qquad V_R'' = I_R'' \times R = 0.667 \times 4 = 2.67\ \text{V}.

Step 3. Superpose: both contributions act with the same reference polarity across $R$ , so

V_R = V_R' + V_R'' = 8 + 2.67 = \mathbf{10.67\ \text{V}}.

Competitive stretch

If you tried to superpose power instead, you would compute $P' = (I')^2R = 2^2\times4 = 16\,\text{W}$ and $P'' = (I_R'')^2R = 0.667^2\times4 = 1.78\,\text{W}$ , and be tempted to add them to $17.78\,\text{W}$ . But the true total current through $R$ is not $I'+I_R''$ in general (here it happens to be, since both currents flow through the same $R$ in the same direction: $2+0.667 = 2.667\,\text{A}$ ), so the correct power is $P = I^2R = 2.667^2\times4 = 28.4\,\text{W}$ — far more than $17.78\,\text{W}$ , because $(a+b)^2$ includes the cross term $2ab$ that adding $a^2$ and $b^2$ misses. Always superpose currents or voltages first, and take power only from the final total.

Common traps

Superposing power directly — power is nonlinear ( $P=I^2R$ ), so add currents or voltages first and compute power only once, from the total.
Leaving another source active while solving one sub-circuit — every other independent source must be deactivated simultaneously at each step.
Deactivating the wrong way — a voltage source becomes a short, a current source becomes an open; swapping these destroys the topology and the answer.
Dropping a series resistance that sat next to a deactivated source — only the source itself disappears (as a short or open); its companion resistor stays in the circuit.
Losing sign/reference direction when adding contributions — each term must be added relative to one single, fixed reference, not whichever direction felt natural in that sub-circuit.

Ask the tutor: Try asking the AI tutor to walk through superposition on a three-source circuit mixing voltage and current sources, or to check your own worked answer step by step.

Why does superposition work for currents and voltages but not for power?

Superposition relies on linearity — Ohm's law relates current and voltage by simple proportion, so the response to a sum of sources equals the sum of the responses to each source alone. Power, however, is $P=I^2R$ , a squared quantity, and $(a+b)^2 = a^2+b^2+2ab$ has a cross term $2ab$ that vanishes only if one of the currents is zero. Adding individual-source powers throws away that cross term, giving a wrong (too small) answer. Always add currents or voltages first, then compute power once from the genuine total.

Do I need to redraw the whole circuit for every source, or is there a shortcut?

For a small circuit like the worked example, redrawing each single-source sub-circuit is fast and safest, especially early on. As you get comfortable, you can often track deactivated sources mentally, directly writing the reduced series-parallel resistance seen by the active source. But until source-deactivation is second nature, sketch each sub-circuit explicitly — it is the step most students skip and then get wrong, especially forgetting to keep a deactivated source's series resistor in place.

How is superposition different from just solving the whole circuit at once with mesh or node analysis?

Mesh and node analysis solve the full multi-source circuit in one pass using simultaneous equations, and always work, even for nonlinear-looking sign patterns. Superposition instead breaks the same circuit into several single-source problems, each solvable with basic Ohm's law and series-parallel reduction, then adds the results — often faster by hand for two or three sources, and it directly demonstrates the linearity property that Thevenin's and Norton's theorems build on next. For circuits with many sources or complex topology, mesh or node analysis usually wins on effort.

Superposition theorem

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads