Solving Circuits with Superposition Theorem
Even with KCL and KVL, as circuits get more complicated, sometimes the setup and the math can become quite complicated. However, using the superposition theorem, in certain situations, you can simplify the circuit by turning off or “suppressing” all the independent power sources except one and solving the circuit, and doing that with all of the power sources, adding up the end result into a single outcome.
That, I imagine, didn’t make much sense. But come back after reading the rest of the tutorial and I bet it’ll be a lot clearer. Let’s try it again. The summation of all of the current produced by the power sources individually is the same as the summation of all of the current produced by the power sources together. It’s linear - each power source linearly affects all other power sources. To simplify things, you can remove all other voltage sources and replace them with a short circuit and remove all other current sources and replace them with an open circuit and solve the circuit. Repeat this with all the power sources, and then sum them up to figure out what the total current will be.
Now that I’ve said the same thing in two different ways, I think an example would be the best way to truly understand it in practice.
Let’s use this simple circuit as an example:
As you can see, there is a voltage source and a current source and a single resistor that we want to know the current through. If you need the superposition theorem to simplify this circuit, I highly recommend you check out our tutorials on KVL and on KCL and do some practice circuits! But, with the voltage source and current source seemingly in contention, I can see how this might cause some confusion. With the superposition theorem, this will be simplified.
First, choose a power source and suppress all other power sources. Again, a suppressed voltage source will become a short circuit and a suppressed current source will become an open circuit. Let’s suppress the current source first.
With the current source now open, we can make two statements based on inspection. With the open circuit in series with the resistor, there will be no current flow through the resistor. Thus, the amount of current through the resistor because of the voltage source is 0. Second, as there’s no current through the resistor, we know that the voltage on BOTH sides of the resistor is 10V.
Now, let’s choose the current source and suppress the voltage source.
As the voltage source becomes a short circuit, we can see that we simply have the current source and the 100 ohm resistor. By inspection, we know that the current, flowing from right to left, is 5A. However, let’s take a moment. To create that 5A, the current source will need to generate 500V (V=IR -> 500V = 5A * 100Ω). So on the right side of the resistor, we’re seeing 500V, and on the left side of the resistor, it is grounded, so it’s at 0V.
Now, the final step of doing the superposition theorem is putting it together.
From the first scenario, we had zero amps going through the resistor and ten volts on both sides of the resistor. From the second, we had 5 amps going through the resistor (right to left), with 500 volts on the right side and 0 volts on the left side. Combining this all together, we get that there are 5 amps going through the resistor (right to left), 510 volts on the right side of the resistor and 10 volts on the left side of the resistor. The common ground between the voltage source and the current source is, as expected, at 0V.
Now, does this make sense intuitively? We know that a current source will create any voltage necessary to output the expected current and we know that a voltage source will create or absorb any current necessary to output the expected voltage. Since the voltage source is raising the voltage on the left hand side, the current source has to generate 510V, so that the voltage across the resistor is 500V, to give the desired 5A. The voltage source on the left, to maintain that 10V, will absorb the 5A. Both power sources have their functional requirements satisfied and the summation of the two sources working separately is the same as the two working at the same time.
Let’s do one more example to see if we can cement your understanding of this concept.
Another circuit that we could almost certainly figure out without using the superposition theorem but since we’re here already, let’s work through it!
First, suppress the 10V voltage source. When it’s suppressed, what do we do with it - open or short the circuit?
Yep, we short it. So, we get the following circuit:
Now we solve this like normal. Do whatever is most comfortable for you, I think I’m going to do this an odd way - not using KCL or KVL but just Ohm’s Law and what we know of series and parallel resistors.
We have 200Ω in series with 300Ω||500Ω (an equivalent resistance of 187.5Ω), so the total current through the series circuits will be:
Which is also the current through the 200Ω resistor. Now that we know the current through that resistor, we can find the voltage across it.
Now we know we have 5 - 2.58V = 2.42V across the other two resistors and can solve for their currents independently.
And, summing 8.1mA and 4.8mA gives us 12.9mA so everything makes sense and is internally consistent! I love it when the math works out, even if it’s simple.
We’ve solved for this source, let’s solve with the 10V source.
I won’t go through the steps this time, but here are the currents through the different resistors - feel free to do it yourself for practice.
200Ω resistor: 16.2mA
300Ω resistor: 22.6mA
500Ω resistor: 6.5mA
Notice that the current through the 200Ω and 500Ω resistors don’t actually add up to the same current as in the 300Ω resistor? That’s just due to rounding errors - don’t worry about it.
BUT! Now we have an interesting dilemma. I didn’t identify the direction the current was flowing in these resistors, but in my calculations, with the two scenarios, the 200Ω and the 500Ω resistors were labelled with the current flowing the opposite way. So we need to decide which direction we will assume the current will flow so we know which of these values are positive and which are negative. In this case, we’ll assume that the current is flowing left to right in the 200Ω resistor, and right to left in the 300Ω resistor.
With that assumption, we get the following values:
200Ω resistor: 12.9mA
300Ω resistor: -8.1mA
500Ω resistor: 4.8mA
200Ω resistor: -16.2mA
300Ω resistor: 22.6mA
500Ω resistor: 6.5mA
200Ω resistor: -3.3mA
300Ω resistor: 14.5mA
500Ω resistor: 11.3mA
Surprisingly, the 5V power supply is actually absorbing 3.3mA — not something I was anticipating when randomly putting these values together. There’s not a lot we can do to make a sanity check here, unfortunately, other than it makes sense that current is flowing away from the 10V supply more than the 5V supply and that the currents going into the center node equal zero (again, disregarding rounding issues).
A couple of notes as we wrap things up.
- Again, this concept only works with linear circuits. Op-amps, microcontrollers, 555 timers, or other non-linear elements will render this invalid.
- This works with AC circuits as well as DC circuits, as long as the circuit is linear.
- Any dependent power source must remain in the circuit at all times. You cannot take it out (or suppress it) or leave it in as the sole power source in a circuit.
And that’s it for the superposition theorem. As stated at the beginning of this tutorial: Using the superposition theorem, in certain situations, you can simplify the circuit by suppressing all the independent power sources except one and solving the circuit. Then you do that with all of the power sources, summing the end result into a single outcome. Now that we’ve been through it, that hopefully makes a lot more sense! If you have any questions or comments, leave them here on CircuitBread.com and we’ll try and clarify the issue for you.
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