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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 :)

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  • \$\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\$
    – Elliot Alderson
    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\$
    – Dan Sandberg
    Sep 19, 2021 at 9:05
  • \$\begingroup\$ @anhnha Thank you, those pointers were helpful. \$\endgroup\$
    – Dan Sandberg
    Sep 19, 2021 at 18:03

1 Answer 1

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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.

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    \$\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\$
    – Dan Sandberg
    Sep 19, 2021 at 18:02

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