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.

Wye configuration

Is this solved with Thevenin equivalents? College physics was 20+ years ago and I'm quite rusty -- thank you :)

  • \$\begingroup\$ 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. \$\endgroup\$ Sep 18, 2021 at 21:06
  • \$\begingroup\$ Use nodal analysis. \$\endgroup\$
    – Chu
    Sep 19, 2021 at 0:26
  • \$\begingroup\$ Any circuit analysis techinique like nodal, mesh, superposition, etc will do the job. Nodal analysis is perhaps the easiest here. \$\endgroup\$
    – emnha
    Sep 19, 2021 at 5:18
  • \$\begingroup\$ @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? \$\endgroup\$ Sep 19, 2021 at 9:05
  • \$\begingroup\$ @anhnha Thank you, those pointers were helpful. \$\endgroup\$ Sep 19, 2021 at 18:03

1 Answer 1


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.

  • 2
    \$\begingroup\$ 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.) \$\endgroup\$
    – jonk
    Sep 18, 2021 at 21:26
  • \$\begingroup\$ 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. \$\endgroup\$ Sep 19, 2021 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.