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_{o} across the current source. The voltage across the current source is equivalent to the voltage across resistor 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_{o}, with the preceding equation.
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_{s} by replacing it with an open circuit.
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_{s} by first removing the voltage sources 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_{s}. As a result, the current divider technique applies. This produces a current i_{32} through resistor R_{3}, resulting from current source i_{s}. Also not that the voltage polarity of 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_{s} by replacing it with an open circuit.

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_{AB}) across the current source (or voltage v_{AB} across Terminals A and B):