I am trying to gain an intuition with circuit analysis by going through MIT's free course on intro to electronics. Question S2E3 presents the circuit below. I solved the problem (find all currents at nodes and voltages across elements) by applying KCL and KVL and doing some algebra, but I had no real intuition that my answer was right until I submitted my answer and it told me I was correct. Which took several attempts. When experienced electronics gurus sees this circuit, what do they see? I see a voltage source and a current source fighting each other to decide who's in charge, but I'm pretty sure that's not the right way to see it. simulate this circuit – Schematic created using CircuitLab

• I see an application for superposition. Dec 30, 2014 at 4:47
• I mentally change the current source + 5$\Omega$ resistor into the Thevenin equivalent of a 15V source with 5 ohms in series, so you have 13V across 9 ohms, and the currents and voltages follow readily. Dec 30, 2014 at 4:55

Other people have given good quantitative methods. If you want a more qualitative method, one place you can start is to try to figure out which way the currents are flowing in each branch. That will usually help give you some intuition as to what's happening with the voltages, too.

There are three branches -- The V1/R1 branch, the R2 branch, and the I1 branch. The I1 branch has a constant current source, so you know which way the current is flowing. Next, look at the R2 branch. All of your sources are pointing towards the top of R2, which tells us that the current will be flowing from top to bottom.

That leaves the V1/R1 branch. The question here is, under what conditions will current flow into the branch (from the top node towards V1) and out of the branch (from V1 towards the top node)? Well, it depends on the voltage at the top node, which I'll call Vtop. When Vtop < V1, V1 sources current. When Vtop > V1, V1 sinks current. When V1 = Vtop, no current flows through V1.

Let's imagine that V1 = Vtop, so there's no V1 current. All of the I1 current goes through R2. What would Vtop be, then? $3A * 5\Omega = 15V$. Do we have anything close to 15V in V1's branch? Nope! Therefore, V1 must be sinking some of I1's current -- probably a fair bit since it's a small voltage source.

In this circuit, the resistors were all roughly the same size. That's not always the case. If one resistor is much larger than the others, anything in series with it will have a small effect on the rest of the circuit. Likewise, a small resistance will tend to short out whatever's in parallel with it. (You'll see this principle at work in passive filters once you get to AC circuit analysis.)

Lastly, there are two special cases that are very rare but may help with your intuition. The first is an ideal voltage source in parallel with an ideal current source. In this case, the current source has no effect at all. Likewise, when an ideal voltage source is in series with an ideal current source, the voltage source has no effect at all. You should make sure you understand why. Feel free to post a comment if you want to confirm anything.

• This is exactly the type of answer I was hoping for. Thanks. Dec 30, 2014 at 14:17

There are several ways to intuitively understand the circuit operation. I will list two most fundamental concepts that can be used to analyze this circuit: 1. Superposition principle 2. Thevenin/Norton equivalents

In both cases, analyses are simplified because they allow us to have a circuit with resistors and onnly voltage or current sources but not both.

1. Superposition principle: That basically means you can consider voltage/current sources one at at time in the circuit and the resultant circuit parameters would be just a direct sum of the individual results. This is a bit longer because the circuit has to be analyzed twice.

2. Thevenin/Norton equivalent: This allows us to convert a voltage source to an equivalent current source or vice versa. It is very easy, for example, to analyse the circuit if there were just two current sources in parallel, because they can be directly summed to give a single equivalent current source. the circuit would then have a current source of 2V/4 ohms = 0.5 A and a parallel resistance of 4 ohms. Therefore, you just have a net current of 3.5 A flowing through a resistor of (4 || 5) ohms.

All of these methods can be applied. The determining factor remains which quantity for the circuit is required to be known at a given instant. Some practice would get you to intuitively choose the least computationall intensive and the most intuitive method of analyzing the circuit.

I'll pass on the guru part, but at least in this case the intuition is pretty clear.

Start with the current source, and pretend that the 2v/4ohms doesn't exist. Then you'll get 15 volts across the 5 ohms, with a 3 volt/ohm drop. So if you now connect the voltage source and its resistor, the voltage will be roughly halved (since there's now a nearly 5 ohm resistor in parallel with the original. So figure about 7.5 volts at the 5 ohm resistor, with a drop of 1.5 volts per ohm. In the other branch, if the 2 volt source were replaced with a 1-ohm resistor, that resistor would have 1.5 volts, instead of 2.

So slightly less current will flow through the 4 ohm when you restore the 2 volt source, and the voltage across the 5 ohm will rise slightly. Say, about 0.1 volts, giving about 7.6 or 7.7 volts.

Does that make sense?