# How do I use superposition to solve a circuit?

Yes, this is a pedagogical question. While answering another recent question, I wanted to refer the OP to concise instructions for using superposition to solve circuits. I found that all the easily found resources online were somewhat deficient. Typically they were unclear about what kinds of circuits superposition applies to, or about the actual method to apply the superposition theorem to a circuit problem. So,

What kinds of circuits can be solved by superposition?

How are different kinds of sources treated when solving by superposition?

What are the steps to solve a circuit using the superposition theorem?

• Since this is to have a place to point to, how about a community wiki answer so it can be tweaked for this purpose? Commented Mar 30, 2015 at 21:55

Superposition theorem
"The superposition theorem for electrical circuits states that for a linear system the response (voltage or current) in any branch of a bilateral linear circuit having more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, where all the other independent sources are replaced by their internal impedances."

What kinds of circuits can be solved by superposition?

Circuits made of any of the following components can be solved using superposition theorem

• Independent sources
• Linear passive elements - Resistor, Capacitor and Inductor
• Transformer
• Linear dependent sources

What are the steps to solve a circuit using the superposition theorem?

1. Answer = 0;
2. Select the first independent source.
3. Replace all independent sources in original circuit except the selected source with its internal impedance.
4. Calculate the quantity (voltage or current) of interest and add to Answer.
5. Exit if this was the final independent source. Else Goto step 3 with selecting next source.

The internal impedance of a voltage source is zero and that of a current source is infinity. So replace voltage source with a short circuit and current source with open circuit while executing step 3 in the above algorithm.

How are different kinds of sources treated when solving by superposition?

The independent sources are to be treated as explained above.

In case of dependant sources, do not touch them.

Superposition only applies when you have a purely linear system, i.e.:

\begin{align*} F(x_1 + x_2) &= F(x_1) + F(x_2)\\ F(a x) &= a F(x) \end{align*}

In the context of circuit analysis, the circuit must be composed of linear elements (capacitors, inductors, linear transformers, and resistors) with N independent sources, and what you're solving for must be either voltages or currents. Note that you can take a super-imposed solution to voltage/current to find other quantities which are not linear (ex. power dissipated in a resistor), but you cannot superimpose (add) non-linear quantities to find the solution for a larger system.

For example, let's take a single resistor and look at Ohm's law (I'm using U and J for voltage/current respectively, no particular reason) and see how current contributed from source $i$ affects the voltage:

\begin{align*} U = J R = R \left(\sum_{i=1}^N J_i\right) = \sum_{i=1}^N R J_i = \sum_{i=1}^N U_i \end{align*}

So I can find the voltage across a resistor by summing up the current contribution from every source independent of any other source. Similarly, to find the current flowing through the resistor:

\begin{align*} J = \frac{U}{R} = \frac{1}{R} \sum_{i=1}^N U_i = \sum_{i=1}^N \frac{U_i}{R} = \sum_{i=1}^N J_i \end{align*}

However, if I start looking at power, superposition no longer applies:

\begin{align*} P = J U = \left(\sum_{i=1}^N J_i\right) \left(\sum_{j=1}^N U_j\right) \neq \sum_{i=1}^N J_i U_i = \sum_{i=1}^N P_i \end{align*}

The general process for solving a circuit using superposition is:

1. For each source $i$, replace all other sources with their equivalent null source, i.e. voltage sources become 0V (short circuits) and current sources become 0A (open circuits). Find the solution $F_i$, for whatever unknowns you are interested in.
2. The final solution is the sum of all solutions $F_i$.

# Example 1

Take this circuit with two sources:

simulate this circuit – Schematic created using CircuitLab

I want to solve for the current J flowing through R1.

Pick V1 as source 1, and I1 as source 2.

Solving for $J_1$, the circuit becomes:

simulate this circuit

So we know that $J_1 = 0$.

Now solving for $J_2$, the circuit becomes:

simulate this circuit

So we can find that $J_2 = I_1$.

Applying superposition, \begin{align*} J = J_1 + J_2 = 0 + I_1 = I_1 \end{align*}

# Example 2

simulate this circuit

Now I am interested in the current through R4 $J$. Following the general process outlined earlier, if I denote V1 as source 1, V2 as source 2, and I1 as source 3, I can find:

\begin{align*} J_1 &= -\frac{V_1}{R_1 + R_2 + R_5 + R_4}\\ J_2 &= \frac{V_2}{R_2 + R_1 + R_4 + R_5}\\ J_3 &= -I_1 \frac{R_2 + R_5}{R_1 + R_4 + R_2 + R_5} \end{align*}

Thus the final solution is: \begin{align*} J &= J_1 + J_2 + J_3 = \frac{V_2 - V_1}{R_1 + R_2 + R_4 + R_5} - I_1 \frac{R_2 + R_5}{R_1 + R_2 + R_4 + R_5} = \frac{(V_2 - V_1) - I_1 (R_2 + R_5)}{R_1 + R_2 + R_4 + R_5} \end{align*}

The power of superposition comes from asking the question "what if I want to add/remove a source?" Say, I want to add a current source I2:

simulate this circuit

Instead of starting over from the beginning, the only thing I need to do now is find the solution for my new source I2 and add it to my old solution: \begin{align*} J_4 &= I_2 \frac{R_1 + R_2 + R_5}{R_1 + R_2 + R_5 + R_4}\\ J &= \sum_{i=1}^4 J_i = \frac{(V_2 - V_1) - I_1 (R_2 + R_5) + I_2 (R_1 + R_2 + R_5)}{R_1 + R_2 + R_4 + R_5} \end{align*}

• I have a few comments that I hope will be useful: 1. I find using U and J is somewhat confusing, V and I are better; 2. The first equation for U should not be summation, as it's for the i'th source alone; 3. The other summations should, I believe, be taken from i=1 to N, not from i to N; 4. Superposition in circuit theory is only used for current and voltage, so I would move the discussion on power later in the text; 5. In the example following the simple one of I1 and R1, shouldn't J3 = -I1 (...), as I1 acts in the opposite direction to J3?
– Chu
Commented Apr 1, 2015 at 12:43
• 1. I chose to use U and J because I labeled my sources with V and I, and I didn't want confusion caused by $I_3 = I_1 \cdot (\textrm{blah})$. I clearly state what U and J are in the hopes to limit confusion. 2. Yes, I made the notation clearer for what the summation variable and starting index is. 4. My idea was to put all of the basic information on when superposition theory before the examples. I made the examples sections clearer to separate the two. 5. Yes, that was my mistake. Commented Apr 1, 2015 at 16:39