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 09-30-2019, 11:25 PM #1 Topper Addon Developer Calculate simple circuit? Hello, is it possible to calculate I1, I2, I3 and I4 assumed that the resistance of the wires is negligible small (let's say they are superconductors )? Let's say voltage is 12V if required. Sorry but I don't get it at the moment... Last edited by Topper; 09-30-2019 at 11:28 PM.
 10-01-2019, 10:41 AM #2 GLS Addon Developer (I'm half asleep, so this might not be correct) Try solving this system: Code: ```i4 + i3 = 4 i1 - i2 = 8 i2 - i3 = 1 i4 + i1 = 15``` ---------- Post added at 11:41 AM ---------- Previous post was at 01:57 AM ---------- Quote: Originally Posted by GLS  (I'm half asleep, so this might not be correct) Try solving this system: Code: ```i4 + i3 = 6 i1 - i2 = 8 i2 - i3 = 1 i4 + i1 = 15``` *now awake* I corrected the system.
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 10-01-2019, 07:07 PM #3 Topper Addon Developer by solving this system I get: I1= -I4 + 15 I2= -I4 + 7 I3= -I4 + 6 I4= I4 but the question is waht's the value of I4? And can we assume that I1 = -I3 and I4 = I2 if all conductors have no resistance ? Last edited by Topper; 10-01-2019 at 07:13 PM.
 10-01-2019, 07:16 PM #4 GLS Addon Developer Quote: Originally Posted by Topper  but the question is waht's the value of I4? I'd guess so, as those 4 nodes, assuming no wire resistance, behave like just one.
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 10-01-2019, 07:26 PM #5 Topper Addon Developer Quote: Originally Posted by GLS  I'd guess so, as those 4 nodes, assuming no wire resistance, behave like just one. That were also my thoughts. So if R=0, then U=R*I=0, but I=U/R, so it's a division by zero, thats not allowed, even 0/0 is not allowed! This is why I got confused assuming that these are superconductors. And this brought me to this Topic: https://en.wikipedia.org/wiki/London_equations, I don't want to say I can understand that ;-) Quote: While it is important to note that the above equations cannot be formally derived,[10] the Londons did follow a certain intuitive logic in the formulation of their theory. Substances across a stunningly wide range of composition behave roughly according to Ohm's law, which states that current is proportional to electric field. However, such a linear relationship is impossible in a superconductor for, almost by definition, the electrons in a superconductor flow with no resistance whatsoever. Last edited by Topper; 10-01-2019 at 07:33 PM.
 10-01-2019, 07:59 PM #6 kuddel Donator I am assuming you should apply Kirchhoff's laws, right? Last edited by kuddel; 10-01-2019 at 08:05 PM.
 10-01-2019, 09:23 PM #7 Topper Addon Developer Quote: Originally Posted by kuddel  I am assuming you should apply Kirchhoff's laws, right? Yes of course we have
 10-01-2019, 11:19 PM #8 kuddel Donator Numerically those "wire-loops" are a not a thing easy to calculate... If you try the circuit below at http://www.falstad.com/circuit/ for example, you immediately get an error message once you close both switches at the same time (thus creating the wire-loop) Direct link: http://tinyurl.com/y4wupqqm
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 10-02-2019, 01:48 PM #9 Miner1 Orbinaut I think the problem here is that Kirchhoff's current law produces a set on linearly dependent equations for the loop. You need another independent condition. Kirchhoff's voltage law would, assuming no voltage drop between the nodes, cause the loop to collapse into just a single node. Another possible path is to assume that the red triangles are diodes that only allow current in the direction indicated. With this assumption, I've found two possible solutions. i4 = 0 => i1 = 15, i2 = 7, and i3 = 6 and i3 = 0 => i1 = 9, i2 = 1, and i4 = 6
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