# Calculating current -- Thevenin?

Can someone please remind me how to solve this very basic calculation?

I've got a Wye configuration of resistors and am giving DC values 1V, 2V, and 4V to points A, B, and C respectively. I want to calculate the current seen in each resistor. Is this solved with Thevenin equivalents? College physics was 20+ years ago and I'm quite rusty -- thank you :)

• This looks very much like a homework question, even if it has been 20 years since you took a physics course. If not, please give us the larger context of where you encountered this circuit. If it is homework we expect you to show us a substantial of your own effort and then ask a specific question. Sep 18, 2021 at 21:06
• Use nodal analysis.
– Chu
Sep 19, 2021 at 0:26
• Any circuit analysis techinique like nodal, mesh, superposition, etc will do the job. Nodal analysis is perhaps the easiest here. Sep 19, 2021 at 5:18
• @ElliotAlderson Hah, I wish! I'm trying to time the A2D readings on a three-phase motor so that I'm reading the current sense resistors while the low-side FET is closed, since the current sensing is on the low-side. I wanted to figure out what the current sense resistors should be reading. I know what voltages I'm putting on them, and if I can compare what they should be reading to what they are reading I can determine the optimal point to be sampling it. Still sound like homework? Sep 19, 2021 at 9:05
• @anhnha Thank you, those pointers were helpful. Sep 19, 2021 at 18:03

Assuming all voltage sources are referenced to the same point, voltage at the star point is

Vs*(1/R1 + 1/R2 + 1/R3) = V1/R1 + V2/R2 + V3/R3

Vs/Rp = V1/R1 + V2/R2 + V3/R3

This is Millman theorem. Once Vs is known, finding current is trivial:

I1 = (V1-Vs)/R1

For your specific example:

Vs = 7/3 V I1 = -4/3 A I2 = -1/3 A I3 = +5/3 A

Needless to say, sum of all currents at the star point (node) is zero. This is Kirchhoff’s current law.

• Agreed. All that results in $\frac{V_1 R_2 R_3+V_2 R_1 R_3+V_3 R_1 R_2}{R_1 R_2+R_1 R_3+R_2 R_3}$. (I mentally think: "voltage 1 times the product of all the opposing resistors + voltage 2 times the product of all the opposing resistors + ..." divided by the sum of the permutations nPr where r=n-1.) Anyway, nicely written approach. +1 (The down vote was not warranted and its author will hopefully decide to remove it.)
– jonk
Sep 18, 2021 at 21:26
• Thanks everyone. Seems like the summary is that Kirchoff's rule is sufficient, and it can be applied via mesh analysis, nodal analysis, etc. Millman's theorem is a specific shortcut that gives a quick answer to this specific configuration. Sep 19, 2021 at 18:02