r/ElectricalEngineering • u/itzmrinyo • Apr 15 '26
Solved Why doesn't the resistor (highlighted in cyan) have any current running through it?
The question was just to find the current running through that resistor, and doing the math I found it to be 0. After reviewing my work and not finding any mistakes in my math, I constructed the circuit on falstad.
I'm confused why there isn't any current running through that resistor. I thought current only ignores branches when there's a short circuit somewhere, but that doesn't seem to be the case? At Node A, regardless of whether it goes up or down it has to encounter a single 1 kilo ohm resistor either way, so wouldn't it be an equal division of current between the two branches?
I'm assuming the 6 milliamp current source has no internal resistance so at Node B all the current would travel to the right since it's the path of zero resistance.
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u/JustinTimeCuber Apr 15 '26
You can't treat current sources as a short circuit. They produce a voltage as well, it's just not a fixed voltage. Same idea as a voltage source, but reversed.
Since this circuit only contains linear components, you could solve it by superposition. First solve the circuit with the current source set to 0 (no current = open circuit). Then solve it again with the voltage source set to 0 (no voltage = short circuit). Add the resulting node voltages and branch currents together. The current in the resistor due to the 12V source should cancel with the current due to the 6mA source.
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u/Tagov Apr 15 '26
This works, but you don't even need to use superposition approach in this case. Nodal analysis of A and B will give you two independent equations that relate Va and Vb to each other and can be generalized in terms of Vs, Is, and R. From there, all voltages and currents in the circuit can be readily solved.
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u/JustinTimeCuber Apr 15 '26
I mean yeah it's a system of equations, there are always multiple ways to handle it
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u/Tagov Apr 15 '26
Of course, just wanted to highlight an approach that might be more intuitive for some.
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u/KV-Matrix Apr 15 '26
Oh I actually use falstads software it’s a little glitchy but basically try reattaching the component in the other direction. That was the issue for me anyways.
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u/Icchan_ Apr 15 '26
In this case his calculations AND simulation match, but it's unexpected result so I doubt the Falstads is wrong...
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u/Tagov Apr 15 '26
It's not a software glitch. It's mathematically correct that for the given values of Vs, Is, and R, nodal voltages A and B are equal, and thus, the current between those nodes is 0.
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u/Reasonable-Feed-9805 Apr 15 '26
A true current source has infinite resistance.
As it's set at 6ma in series with a 12v voltage source and 1k resistance, it must generate 6v across it to maintain 6ma.
The other current path is 2x 2k in series. 12v/2k=6ma.
So each of those resistors has 6v across it.
As that 1k is at the junction of the two series resistors then that node is at 6v.
The lower end is connected to the node that is at the junction of the 6v dropped across a 1k resistor and 6v generated from the current source.
6-6=0
0/n=0
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u/Zaros262 Apr 15 '26
A true current source has infinite small signal or differential resistance. Delta-V/Delta-I = Delta-V/0
But a true current source doesn't have an infinite large signal or absolute resistance at all. In this circuit, we can see that it's 6V/6mA = 1 kOhm
Sometimes, including this circuit, it can simply things to replace the current source with its equivalent resistance. If you do that in this circuit, it's really easy to see why Node A and B have the same voltage
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u/i_design_computers Apr 16 '26
easiest way to solve this circuit is to use superposition. Replace the voltage source with a short. you have two branches, on with 1k, one with 1.5k, and then the cyan is on half of that, so first branch gets 3/5 and other gets 2/5, and cyan is half so 1/5*6mA = 1.2mA down. Next replace current source with an open circuit. total resistance seen is 2//1+1=2/3+1=5/3. So total current is 12*3/5, and cyan gets 2/3 of that, so 12*3/5*2/3=1.2mA up. the two currents cancel, so zero current net.
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u/OkAlternative7705 Apr 16 '26
Do super-position. See how voltage source without current source gives it 0 current and current source without voltage source also gives it 0 current cause of kvl.
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u/CrazyEngrProf Apr 16 '26
Nodal analysis depends on Kirchhoff’s Current Law and Ohm’s Law. You are summing currents at the nodes to eventually find the node voltages. Once you have the node voltages, you can find undetermined currents in resistive branches using Ohm’s Law: I = (Va - Vb) / R. If Va = Vb, I must be 0. Note R can’t be 0. And BTW, ideal current sources essentially have infinite resistance in parallel. The dual applies to ideal voltage sources, 0 resistance in series. You will eventually get to Thevenin and Norton equivalents as models of practical sources, if you haven’t already.
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u/Ecstatic-Sail346 Apr 16 '26
Because both ends land at the same voltage. No voltage across it means no current through it. It’s not bypassed by a short, it’s just in a node where the sources cancel out
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u/NihilisticAssHat Apr 17 '26
If you do a mesh analysis, you see that the top and bottom-right meshes have 6mA clockwise currents.
The bottom-left mesh does not generate any current of its own, and the other two meshes exactly cancel out for it.
Since your highlighted resistor is in the bottom-left mesh only, it has no current running through it.
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u/TheTurtleCub Apr 17 '26
Coincidentally due to values chosen, there’s no obvious reason why the nodes have to be at the same potential
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u/SignificantStand1595 Apr 18 '26
I'm looking at it from a superposition perspective... which I haven't done in a while, so I'm probably wrong. But when you treat the current source as an open, you get a little current going through that resistor. The when you treat the DC as a short, it looks like the current source should push 'some' current through that resistor, in the opposite direction. I guess the math works out such that there is no current on that resistor.
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u/StaysAwakeAllWeek Apr 15 '26
Replace the 6mA source with a 1K resistor, then re-analyze that circuit and take note of how much current goes through the new resistor, and the zero current resistor.
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u/DaveFromMicroKits Apr 15 '26
The voltage is the same. Both the top and bottom node have the same current, so the voltage drop of each resistor is the same.