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I know resistors in series have the same current, and resistors in parallel have the same voltage drop, but in practice, I can't get the equivalent resistance right. Wondering if anyone has tips/advice/etc

I have the below circuit: enter image description here

First I label the nodes (did it with color so it's easier to see) and the direction of passive current.

enter image description here

Apparently R2 and R6 are in series, and that equivalent resistance is in parallel with R4. In total, that makes up the resistance between nodes A and B.

But how? It doesn't look like R2 and R6 have the same current, because current splits into two direction after R1, and only part of it goes to R2, whereas R6 gets more current.

In any case, doesn't current take the path of least resistance, so it shouldn't even go through R2?

I know if might seem really simple, but I'm pretty lost.

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  • \$\begingroup\$ "take the path of least resistance", that's not how electricity works, no. It takes all paths, with different currents. Be more precise than saying "the current splits": how so? Evenly? 1/3 in one path, 2/3 in the other? \$\endgroup\$ – Marcus Müller Dec 17 '19 at 1:18
  • \$\begingroup\$ Unless you set V1==0, you can#t use the same color for the nodes left and right of it. R2 and R6 are not really in series, since there's I1 adding up to the current through R2 to make the current through R6, so that is wrong, too. You might really want to go back a bit and read on the basics of network analysis, you've got something wrong there :) \$\endgroup\$ – Marcus Müller Dec 17 '19 at 1:20
  • \$\begingroup\$ R3 and R7 can be combined, however, and actually you can also exchange the order fo V2 and R1 and then combine R1, R2 and R7, for example. \$\endgroup\$ – Marcus Müller Dec 17 '19 at 1:21
  • \$\begingroup\$ Do you mean the Thevenin resistance looking into terminals AB? \$\endgroup\$ – Chu Dec 17 '19 at 8:32
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Apparently R2 and R6 are in series, and that equivalent resistance is in parallel with R4.

R2 and R6 are not in series.

However, you might want to solve this circuit using the superposition method. Then you would zero out all sources except one, solve the circuit, then choose another source, zero all the other sources, solve again, etc.

One of these steps will be to zero \$I_1\$ and \$V_1\$, and solving for the circuit response to \$V_2\$. Stop now and re-draw the circuit with these sources zero'd.

You can see that when you zero \$I_1\$, then it's effect is like an open circuit. In this step, you can now treat \$R_2\$ and \$R_6\$ as being in parallel. And when you zero \$V_1\$, its effect is like a short circuit, so now your \$(R_1+R_6)\$ is effectively in parallel with \$R_4\$.

So probably when your text suggested these combinations, they were talking just about this particular step of a solution by superposition.

In the steps where you consider the effect of \$V_1\$ and zero other sources, or the step where you consider the effect of \$I_1\$ and zero the other sources, you won't be able to make these substitutions.

It's also possible your book was just working out the problem of finding the Thevenin or Norton resistance of the circuit. To find that, you can zero out all the sources, and again the suggested combinations apply. But that is only for one step of finding the Thevenin or Norton equivalent. When you're working on finding the Thevenin or Norton source values, you won't be able to make those substitutions (except in certain steps if you solve using superposition, as discussed above).

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You can combine R1, 3 and 7 in series. Then you have 4 voltage nodes and 4 current loops. You'll therefore have 4 simultaneous equations to solve. There are no numbers given so don't expect numeric answers...

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