# Analyze Circuits with Three Independent Sources Using Superposition

Use superposition to analyze circuits that have lots of voltage and current sources. Superposition helps you to break down complex linear circuits composed of multiple independent sources into simpler circuits that have just one independent source. The total output, then, is the algebraic sum of individual outputs from each independent source.

You can use superposition when faced with a circuit that has three (or more) independent sources. With three independent sources, you find the output voltage of three simplified circuits, where each circuit has one source working and the others turned off. Then add the outputs due to the three power sources.

Circuit A in the sample circuit shown here has two voltage sources and one current source. Suppose you want to find the output voltage across the current source *i** _{s}*.

To help you follow the analysis, the voltage *v** _{AB}* is identified by the labels Terminals A and B. This voltage is equal to the output voltage

*v*

*across the current source. The voltage across the current source is equivalent to the voltage across resistor*

_{o}*R*

_{3}connected in series with voltage source

*v*

_{s}_{2}.

In Circuit A, the voltage across the current source *i** _{s}* is connected in parallel with the series combination of

*R*

_{3}and

*v*

_{s}_{2}. You can find the voltage across

*R*

_{3}and

*v*

_{s}_{2}, which is equal to the output voltage

*v*

_{o}.

Applying Kirchhoff’s voltage law (KVL) to describe this situation, you wind up with

Essentially, finding *v** _{o}* involves finding the voltage across resistor

*R*

_{3}. When you know this voltage, you can easily calculate the output voltage,

*v*

*, with the preceding equation.*

_{o}You can break down Circuit A, with three independent sources, into simpler Circuits B, C, and D, each having a single independent source with the other sources removed or turned off. To analyze the simpler circuits with one source, you apply voltage and current divider techniques.

You need to first find the voltage across *R*_{3} due to each independent source. Here’s how it works:

**Source 1: Circuit B, first voltage source**: You calculate the voltage across*R*_{3}due to*v*_{s}_{1}by first removing the voltage source*v*_{s}_{2}and replacing it with a short. You also remove the current source*i*by replacing it with an open circuit._{s}After removing two independent sources, you have Circuit B, a series circuit driven by a single voltage source,

*v*_{s}_{1}. Consequently, the voltage divider technique applies, yielding a voltage*v*_{31}across resistor*R*_{3}_{ }due to*v*_{s}_{1}:**Source 2: Circuit C, current source**: You calculate the voltage across*R*_{3}_{ }due to*i*by first removing the voltage sources_{s}*v*_{s}_{1}and*v*_{s}_{2}and replacing them with shorts.After removing two independent voltage sources, you have Circuit C, a parallel circuit driven by a single current source

*i*. As a result, the current divider technique applies. This produces a current_{s}*i*_{32}through resistor*R*_{3}, resulting from current source*i*. Also not that the voltage polarity of_{s}*V*_{s}_{2}is opposite that of*v*_{33}. Using the current divider for Circuit C yields the following:Next, use Ohm’s law to find the voltage across

*R*_{3}due to current source*i*:_{s}**Source 3: Circuit D, second voltage source**: You calculate the voltage across*R*_{3}_{ }due to*v*_{s}_{2}by first removing the voltage source*v*_{s}_{1}_{ }and replacing it with a short circuit. Also remove the current source*i*by replacing it with an open circuit._{s}After removing two independent sources, you have Circuit D, a series circuit driven by a single voltage source,

*v*_{s}_{2}. Because this is a series circuit, the voltage divider technique applies, producing a voltage*v*_{33}across resistor*R*_{3}_{ }due to*v*_{s}_{2}. Also note that the voltage polarity of*v*_{s}_{2}_{ }is opposite that of*v*_{33}. Using the voltage divider technique produces the following output:

To find *v*_{R}_{3}, add up the voltages across resistor *R*_{3} due to each independent source:

Here’s the total output voltage (*v** _{o}* +

*v*

*) across the current source (or voltage*

_{AB}*v*

*across Terminals A and B):*

_{AB}