Superposition is a technique used to analyze linear electric circuits with multiple independent sources. It is based on the principle that the voltages and currents in such a circuit can be found by algebraically summing the voltages and currents produced by each independent source acting alone, with all other independent sources disabled (or set to zero).

Example 1

For instance, consider the circuit shown in Fig. 1(a), in which the sought variable is $I_0$. Using the principle of superposition we can write $I_0$ as a superposition of $I_{01}$ and $I_{02}$, where the latter can be computed using the circuits in Fig. 1(b) and Fig. 1(c). Using KVL and current division we have $$\begin{equation}I_0=I_{01}+I_{02}=-\frac{10}{2+3}+4\cdot\frac{2}{2+5}=-2+1.6=-0.4\ {\class{mjunit}A}\end{equation}$$

Example 2

Let us now consider the circuit shown in Fig. 2(a), which contains both dependent and independent sources and let us compute $I_0$. Using the principle of superposition we can write $I_0=I_{01}+I_{02}$, where $I_{01}$ and $I_{02}$ can be computed using the circuits in Fig. 2(b) and Fig. 2(c). Since these circuits contain both dependent and independent sources, we should normally solve them using nodal or mesh analysis. For instance, applying KVL in Fig. 2(b) we get $10+3I_{01}+3I_{01}+2I_{x1}=0$, which gives $I_{01}=-1.25\ {\class{mjunit}A}$. Applying mesh analysis for the circuit in Fig. 2(c) we have $I_{02}=4+I_{x2}$ and $3I_{02}+2I_{x2}+3I_{x2}=0$, which can be solved for $I_{02}=2.5\ {\class{mjunit}A}$. Hence: $$\begin{equation}I_0=I_{01}+I_{02}=-1.25+2.5=1.25\ {\class{mjunit}A}\end{equation}$$ and current $I_x$ from Fig. 2(a) becomes $$\begin{equation}I_x=I_0-4\ {\class{mjunit}A}=-2.75\ {\class{mjunit}A}\end{equation}$$

Assume we have a circuit that contains $n \ge 2$ independent sources and $m$ dependent sources and the sought variable is a current or voltage in the circuit. To compute the sought variable we split the circuit into a superposition of $n$ circuits, in which we activate only one independent source at a time.

1. Keep only one independent source and deactivate all other independent sources. Notice that if the circuit contains dependent sources you need to keep those in the circuit (in a way, they are treated like resistors).
2. Calculate the potentials and currents in the circuit.
3. Repeat the last two steps for each source in the circuit.
4. Superimpose the values of potentials and currents obtained at Step 2; pay special attention to the direction of the voltage drops and current flows.

Miscellaneous

• To deactivate a current source we remove it from the circuit (replace it with open-circuit or break); to deactivate a voltage source we replace it with a short-circuit (or wire).
• Unlike mesh and nodal analysis that can be applied to both nonlinear and linear circuits, superposition can be applied only to linear circuits. Therefore, in general, we cannot use superposition to compute the potential and currents in a circuit containing diodes, transistors, or other nonlinear elements.
• Special attention needs to be paid when computing the power using the method of superposition. If we need to compute the power generated or dissipated by a component, we need to compute the voltage across and the current flowing through that component, and then multiply the two to obtain the power.
• Superposition can be applied to circuits containing both dependent and independent sources. In this case superposition is applied only to the independent sources, while the dependent sources are always kept in the circuit.

It is clear from the two examples above that superposition can be easier to apply than nodal or mesh analysis when the circuit has no dependent sources (as in Fig. 1). In such cases, simple methods like resistor simplification and current or voltage division can be used to find the desired quantities in each superimposed circuit. However, if dependent sources are present (as in Fig. 2), we may still need to use nodal or mesh analysis to solve the superimposed circuits. In this situation, it is often simpler to solve the original circuit directly using nodal or mesh analysis.

In the previous section we saw that superposition can be applied to circuits containing both independent and dependent sources (see Example 2 and Fig. 2). When we apply superposition to circuits containing dependent sources, we normally keep the dependent sources in the circuit at all times, while we deactivate only the independent sources one at a time. Therefore, if the circuit contains $n$ independent sources and $m$ dependent sources, we end up with $n$ circuits, each containing all $m$ dependent sources and only one independent source activated.

The question now is, can we apply superposition to dependent sources as well (like in Fig. 3 - notice that Fig. 3(b) and (c) do not contain the dependent source, while Fig. 3(d) contains only the dependent source)? In other words, can we write the original circuit as a superposition of $n+m=2+1=3$ circuits? Despite what is often suggested in the literature, the answer is YES, however, you need to do it carefully. For more information about this topic please refer to this paper by W. Marshall Leach from Georgia Tech.

Example 3

Consider again the circuit from Example 2, but now solve it using the superposition of dependent sources (see figure below).

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Let us compute current $I_x$ using the superposition of dependent sources. Using Ohm's law, current division, and KVL, it is easy to show that

• $I_{x1} = -\frac{10}{2+3}=-2\ {\class{mjunit}A}$
• $I_{x2} = -4\cdot\frac{3}{2+3}=-2.4\ {\class{mjunit}A}$
• $I_{x3} = -\frac{3I_{x3}}{2+3}$

Notice that the solution of the last equation is $I_{x3}=0$, however, we do not want so solve for $I_{x3}$. Instead, we replace $I_{x3}$ with $I_x$ in the right-hand-side of the last equation and appply superposition as follows $$\begin{equation}I_x=I_{x1}+I_{x2}+I_{x3}=-2-2.4-\frac{3I_{x}}{2+3}=-4.4-0.6 I_x\end{equation}$$ which can be solved for $I_x$ $$\begin{equation}I_x=\frac{-4.4}{1+0.6} = -2.75\ {\class{mjunit}A}\end{equation}$$ Notice that we obtained the same answer as when we used the superposition of independent sources (in Example 2). To compute $I_0$ we use superposition again $$\begin{equation}I_0=I_{01}+I_{02}+I_{03}=-\frac{10}{2+3} + 4\cdot\frac{2}{2+3} - \frac{3Ix}{2+3} = -2 + 1.6 + \frac{3\cdot 2.75}{5}=1.25\ {\class{mjunit}A} \end{equation}$$ which is again the same as when use used the superposition of independent sources.