Superposition:
Analyzing circuits with multiple sources.

Caveat:  Superposition only works for linear circuits. Circuits with R, L, C.

The current or voltage at any point in a circuit containing multiple sources (current and/or voltage) is the superposition (sum) of the currents or voltages imposed separately by each source.

Procedure for analyzing a circuit using superposition.

• Turn on the sources one at a time.
• For each source that is “off”, replace it with its characteristic resistance. (Ideal voltage sources R=0, a short circuit. Ideal current sources R=∞, an open circuit.)
• Analyze for the desired quantity due to each source.
• The full solution for all sources active is the sum of the contributions due to each source individually.
• Carefully observe the signs of the individual voltages and currents. The polarity of the voltages and the direction of the currents may not be the same for all sources.

Example:  Find the Thevenin equivalent voltage, VTH and the source resistance RS of the following network as seen at the terminals. Because VTH is the open circuit voltage, we simply need the voltage at the terminals, V, which is the same as the voltage across R3.

Because the network has 3 internal sources we will use superposition to solve the problem. Superposition says that V is the sum of the voltages contributed from each of the 3 sources. V = V1 + V2 + V3. We will determine the voltage contribution from each voltage separately by turning on only one source at a time. Sources that are “off” are replaced with their characteristic resistance.

Find VTH and RS using superposition.

Solving for V1 due to IS1.

The terminal voltage is the voltage drop across R3. We also note that R1 and R2+R3 form a current divider. To find V1 we need i3 which we can find from the current divider equation.

Solving for V2 due to IS2.

We see that circuit reduces to almost the same problem as for IS1, except that the direction of the current is opposite. We have chosen to keep the same direction of i3 to remind us that we need a negative sign. The lower terminal will obviously be more positive that the upper terminal. We must be very careful of the signs!

Solving for V3 due to VS3.

Again, the terminal voltage is the voltage drop across R3. However we also recognize that R1+R2 and R3 form a voltage divider. To find V3 we simply use the voltage divider equation.

The VTH of this network as seen at the terminals is the sum of each of these component voltages. Be careful of the signs!!

To solve for the source resistance, RS, we simply turn off ALL of the sources and solve for the equivalent resistance of the resulting resistor network.