CR · DC analysis

Superposition

This unit shows how to split a linear circuit with several sources acting together into easy problems with one source switched on at a time, and then add the results. You will also see for yourself that the partial responses add up to exactly the full response.

Turn on the sources one at a time and add

V₁ and R₁ sit on the left, V₂ and R₂ on the right, and the current I is measured in the middle branch through R₃. Turning a source off with its button redraws it in the figure as a shorted wire. Note the current with only one source on, then the other, and compare their sum with the current when both are on.

Current I in the middle branch

I = 1.5 A

I = 1.0 + 0.5 = 1.5 A

Both · sum = total ·

Linearity is what lets you add

In a linear circuit, doubling the input exactly doubles the output, and feeding in the sum of two inputs produces the sum of their outputs. Thanks to this proportionality and additivity, even a circuit with several sources can be solved one source at a time and the results simply added. Once a nonlinear element such as a diode enters the circuit, the theorem no longer holds.

Keep one source and set the rest to 0

To see the contribution of one source alone, set the other sources to 0. Setting a voltage source to 0 V means joining its ends with a wire, a short circuit; setting a current source to 0 A means cutting it, an open circuit. Any internal resistance a source has stays in the circuit. In this figure, turning off V₂ turns its place into a shorting wire, leaving the current that V₁ produces alone.

Add the partial responses, signs and all

Find the current each source produces alone in the same branch and with the same reference direction, then add them with their signs matched to direction, and the result is the total current. If a source pushes current the opposite way through that branch, its share is subtracted. Here both sources push current the same way through the R₃ branch, so 1.0 A and 0.5 A simply add.

ObserveI = I₁ + I₂

The total response is the sum of the partial responses.

ChooseI₁ = ?

The current with V₁ alone.

Fill inI₂ = ?

The current with V₂ alone.

On your ownI = 1.0 + 0.5 = ?

Adding the two partial currents gives the total current.

Back to the first screen

With one source on, the middle branch carried 1.0 A or 0.5 A, and with both on it carried exactly their sum, 1.5 A. There was no need to solve both sources at once; solving the easy one-source circuit twice and adding was enough. As long as the circuit is linear, a complicated multi-source problem can always be split up this way.

The superposition theorem states that any current or voltage in a linear circuit equals the sum of the responses each independent source produces alone. All but one source are set to 0: voltage sources are shorted and current sources opened, while internal resistances stay. It applies only to linear circuits, where proportionality and additivity hold.