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I know how to calculate resistance in parallel and series connected resistors, and how to reduce most circuits to these, but I missed my lecture on cases where you can't just reduce the circuit to series and parallel and now I'd like to catch up...

schematic

simulate this circuit – Schematic created using CircuitLab

How to calculate total resistance between A and B in such case? R5 rains over my parade because I don't know a'priori which direction current will travel through it, and I don't know how to transform this into something I'd know - I could try to calculate potentials at points p1 and p2, to get the current that flows through R5, except that current modifies these potentials...

How do I solve a circuit like this?

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I think you can try Y-Δ transform, transform half of the upside or downside 3 resistors to "Y", then it'll be easier to analyze.

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  • \$\begingroup\$ this is a version of a Wheatstone bridge. If the ratio of R1:R2 = R3:R4 then P1 and P2 have the same voltage, and R5 has no effect on overall impedance. Doesn't help solve the problem, BUT this is a useful circuit. \$\endgroup\$ – Alan Campbell Nov 10 '14 at 22:40
  • \$\begingroup\$ Yes, you are right. \$\endgroup\$ – diverger Nov 11 '14 at 1:19
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One way to look at this conceptually is to imagine a voltage applied accross A-B, then calculate the current drawn. From the voltage and the current you can figure the equivalent resistance.

With that mental model, you can replace R1 and R2 with a Thevenin source, and R3 and R4 with another Thevenin source. Now you can compute the current thru R5 because it is simply connected between two Thevenin sources, which themselves can be reduced to a single Thevenin source. Once you have the current thru R5, and therefore the voltage accross it, the current thru the remaining resistors should be easy to find.

There are lots of ways of attacking this problem. The above is actually somewhat roundabout, but it provides a good conceptual understanding of what is going on each step. If you're clever, you can use that method to derive a more direct method.

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  • \$\begingroup\$ How would I go about that without knowing the source? Do I just assume one? And do you mean the source goes between A and B, and I calculate the equivalent Thevenin source between the terminals of R1 and R2 at p1, as if I disconnected the circuit there? \$\endgroup\$ – SF. Nov 10 '14 at 16:04
  • \$\begingroup\$ @SF.: This is just a mental exercise. You can assume any value, like 1 V. In the end the voltage of the imagined voltage source will cancel out. \$\endgroup\$ – Olin Lathrop Nov 10 '14 at 17:21
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I'm probably in the minority here but I don't agree that you should do the Y-\$\Delta\$ transform. That is sort of a formal way of solving these problems. Since you are asking this question I think you'll learn more by working through this in gory detail.

Write down the loop laws and enforce that the sum of the currents into each node is zero. Set the potential between \$A\$ and \$B\$ to \$V_{AB}\$. Once you calculate the effective resistances you will find that \$V_{AB}\$ drops out of that calculation. The currents should all be proportional to \$V_{AB}\$. When you apply the current law you have to assume a direction for the currents. It doesn't matter what directions you assume. You just have to apply the laws consistently. For example if you assume that the currents at \$p1\$ all flow into \$p1\$ you will find $$i_1+i_2+i_5=0$$

When you apply the current law at \$p2\$ assuming \$i_3\$ and \$i_4\$ flow into \$p2\$ you will have $$i_3 + i_4 - i_5 = 0$$

The drop through \$R_4\$ then is $$V_4 = i_4 R_4$$ The drop across \$R_3\$ from \$p2\$ to the top of the diagram is $$V_3 = -i_3 R_3$$ (because you wrote the current laws with \$i_3\$ pointing into \$p2\$), etc.

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