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The superposition theorem says that in a linear circuit with multiple independent sources, the response (a voltage or current) at any point equals the sum of the responses caused by each source acting alone, with all other independent sources deactivated — voltage sources replaced by a short circuit, current sources replaced by an open circuit. This calculator applies it to the classic case of two voltage sources (V1 through R1, V2 through R2) feeding a common node that also has R3 to ground.
| Step | Formula | Meaning |
|---|---|---|
| 1. V2 deactivated (shorted) | Vnode1 = V1×(R2∥R3)/(R1+R2∥R3) | Divider of V1 into R1 and (R2∥R3) |
| 2. V1 deactivated (shorted) | Vnode2 = V2×(R1∥R3)/(R2+R1∥R3) | Divider of V2 into R2 and (R1∥R3) |
| 3. Superpose | Vnode = Vnode1+Vnode2 | Total node voltage with both sources active |
| R3 current | I3 = Vnode/R3 | Sum of each source's individual R3 contribution |
Superposition only works for linear elements (resistors, linear sources) — it cannot be applied directly to power, since power is a nonlinear (squared) function of voltage/current. Always sum voltages or currents first, then compute power from the totals.
In a linear circuit with multiple independent sources, the total response at any point is the sum of the responses each source produces acting alone, with all other independent sources deactivated.
Replace it with a short circuit (0V, a plain wire) — its internal resistance, if any, stays in the circuit.
Replace it with an open circuit (remove it, leaving a break) — the branch simply carries no current from that source.
No. Power is proportional to voltage or current squared, which is nonlinear, so you cannot superpose power contributions directly. Sum the voltages or currents first, then compute power from the totals.
No, it only applies to linear circuits (resistors and linear dependent/independent sources). Diodes, transistors in nonlinear regions, etc. break the principle.
It isolates each source's individual effect, which is valuable for troubleshooting, sensitivity analysis, and understanding how much each supply or signal contributes to a shared node.
The same method extends: deactivate all sources except one, find that source's contribution, repeat for each source, then sum all contributions.
Yes, and it is essential there — each source's contribution is found using phasors/impedances at its own frequency, then the individual time-domain waveforms are summed (they cannot be combined as phasors across different frequencies).
It means R2 and R3 in parallel: (R2×R3)/(R2+R3), the equivalent resistance the V1 source \"sees\" beyond R1 once V2 is shorted.
Because each partial voltage is computed with the other source's resistor still loading the node, so the two dividers interact; the calculator accounts for this automatically.
By symmetry, each source contributes equally to the node voltage, and Vnode ends up equal to that common source value scaled by the loading of R3.
Yes, if that source's polarity is negative (e.g. a −9V rail) its contribution term is negative and reduces the total node voltage.
I3 = Vnode/R3 = (Vnode1+Vnode2)/R3 = I3 due to V1 alone + I3 due to V2 alone — currents superpose just like voltages.
For two or three sources it is usually quicker and more intuitive; for many sources or complex topologies, matrix-based nodal/mesh analysis is typically faster.
Yes — the resistive network in front of an op-amp summing amplifier follows the same math; just note the amplifier's virtual-ground node changes R3 to the feedback path's effective input impedance.
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