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schematic

simulate this circuit – Schematic created using CircuitLab

The problem asks for finding the voltage across resistor R7 using KVL analysis. When I did the work, I identified 3 mesh currents. Ix, Iy, Iz.

  • Ix flows out of the positive terminal of the independent voltage source counter clockwise through R3, R6, R1.

  • Iy flows clockwise from the positive terminal of the voltage sources through R4, R5, R1.

  • The third mesh current in the out loop, which flows counter clockwise from R4, R5, R8, R10, R9, R7, R6, R3.

I set up my systems of equations and performed my ohm's law substitutions and got Ix = 0.340909 A, Iy = 0.340909 A and Iz = 0 A.

On the answer key, the professor had both Ix and Iy and Iz flowing counter clockwise and Iz = 2E-14 A.

I checked and I didn't make any arithmetical errors and the coefficients were the same, just with different signs depending on the direction I drew the currents in.

What would be the correct value of Iz according to this diagram? (The professor has been known to leave errors on the keys)

Even if the coefficients had different signs depending on which terminal the current is coming out of the source, it should all equal out. We were told in class currents only go in the same direction, when two meshes share a current source and this rule is not applicable to voltage sources.

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    \$\begingroup\$ Why would Iz not be zero? There's no voltage source. \$\endgroup\$ Sep 10 at 16:46
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    \$\begingroup\$ Your diagram is a bit ambiguous -- you probably mean for \$I_x\$ to be the current loop to the immediate left of the voltage source, but as drawn it looks more like it's the current through that voltage source. Just a nit... \$\endgroup\$
    – TimWescott
    Sep 10 at 18:31

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The Photon has the right of it. You can prove to yourself (and to anyone else, including your teacher) that \$I_z=0\:\text{A}\$. Just redraw the schematic.

You are allowed to set exactly one node to zero (ground reference.) Selecting your bottom node as ground and re-ordering \$V_1\$ and \$R_1\$ (which is fine since they are in series), find:

schematic

simulate this circuit – Schematic created using CircuitLab

From no more than a cursory glance you can see that \$I_z=0\:\text{A}\$.

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Iz is zero. You could redraw it so the whole Iz loop sits on one side or the other of the rest of the circuit and easily see that there is no source driving this loop.

Your professor likely used a simulator to solve the circuit, and a non-zero result came out due to rounding errors. Why a professor would blindly rely on a simulation result like this is an interesting question, but not one we can answer here.

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  • \$\begingroup\$ Sorry I meant to edit my answer but it ended up being your answer. \$\endgroup\$
    – DKNguyen
    Sep 10 at 17:09
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For checking my simulator ...I simulated it ... It gives really 0 A. Mine is ok.

By the way, are you sure of your currents? As pointed in comment by @TimWescott.

Ok for the mesh currents. ix = iy = 340.909 mA.

Have you checked also the voltages ?

enter image description here

Because this configuration can also appear ... If 80 Ohm and V1 are swapped.

enter image description here

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Current can only flow one way through a current source.

Current can flow in either direction through a voltage source. This only really happens if there is more than one voltage source and the circuit is such that a voltage source is being overpowered by another current source.

Your Iz loop contains no current because if you take the entire branch containing R7, R8, R9, and R10 and flip it vertically so it sits on the lower half of the page, you can see that the node 8 shorts that entire loop thus preventing the voltage source from affecting it.

This is engineering so you have to get used to working in the real world where numbers are not precise and information is imperfect and fuzzy. Recognize that \$I_z=2\times 10^{−14}\$ is a tiny number that is practically insignificant relative to the magnitudes of the other calculated currents.

If you built this circuit and your current probe was measuring \$I_z=2\times 10^{−14}\$, would you wondering if your calculated current of \$I_z=0A\$ was wrong? Or if your friend calculated an answer of \$I_z=2.1\times 10^{−14}\$? Simulators also have rounding error and finite precision.

Sometimes you need to take a step back and look at things.

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    \$\begingroup\$ Ok, great that makes sense. Since the mesh that Iz flows through is a short circuit that means no voltage is going across it, but there is a possibility of a non-zero current source going through it. \$\endgroup\$
    – Ned
    Sep 10 at 17:19
  • \$\begingroup\$ @Ned A calculated loop current is only a current that sticks with you all the way around the loop. Other currents can dip in and out of the loop but as long as they don't follow you all the way around they won't appear in the calculated current for that loop. \$\endgroup\$
    – DKNguyen
    Sep 10 at 17:23
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    \$\begingroup\$ @Ned An idealized circuit that you are studying, the Node 8 connection is zero resistance between all devices connected to Node 8 which means you'll get zero current for Iz. In a real world circuit, if the connections to Node 8 are spread out, there will be some finite amount of resistance between the components connected to Node 8 which could cause a minuscule value for Iz. This leads us to why we use star-grounds, something you will study in the future. \$\endgroup\$
    – qrk
    Sep 10 at 18:12

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