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 R₃ connected in series with voltage source v_s₂.

In Circuit A, the voltage across the current source i_s is connected in parallel with the series combination of R₃ and v_s₂. You can find the voltage across R₃ and v_s₂, 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 R₃. 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 R₃ due to each independent source. Here’s how it works:

Source 1: Circuit B, first voltage source: You calculate the voltage across R₃ due to v_s₁ by first removing the voltage source v_s₂ 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_s₁. Consequently, the voltage divider technique applies, yielding a voltage v₃₁ across resistor R₃due to v_s₁:
Source 2: Circuit C, current source: You calculate the voltage across R₃due to i_s by first removing the voltage sources v_s₁and v_s₂ 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 i₃₂ through resistor R₃, resulting from current source i_s. Also not that the voltage polarity of V_s₂ is opposite that of v₃₃. Using the current divider for Circuit C yields the following:

Next, use Ohm’s law to find the voltage across R₃ due to current source i_s:
Source 3: Circuit D, second voltage source: You calculate the voltage across R₃due to v_s₂ by first removing the voltage source v_s₁and 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_s₂. Because this is a series circuit, the voltage divider technique applies, producing a voltage v₃₃ across resistor R₃due to v_s₂. Also note that the voltage polarity of v_s₂is opposite that of v₃₃. Using the voltage divider technique produces the following output: