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Without giving resistor values, what are the steps to breaking down this circuit to find total resistance?

circuit  

Edit:
Like this?

enter image description here

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    \$\begingroup\$ We won't directly answer it for you, Jack, but will help you along. Add in another photo (or use the CircuitLab button on the editor toolbar) showing what series resistors can be combined and what parallel resistors can be combined. \$\endgroup\$ – Transistor Aug 15 '18 at 21:31
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    \$\begingroup\$ Redraw it so it is less confusing. Try to combine resistors and try to make the current all flow in one direction through the resistors (eg. to the right on top and a return wire on the bottom). \$\endgroup\$ – Spehro Pefhany Aug 15 '18 at 21:32
  • \$\begingroup\$ Are these two schematics supposed to be of the same circuit? If so, you've made an error in your second schematic -- R1 should not be in series with R2. \$\endgroup\$ – duskwuff -inactive- Aug 15 '18 at 21:50
  • \$\begingroup\$ couldn't see the forest for the tree \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 15 '18 at 21:51
  • \$\begingroup\$ Good. Now what pairs of resistors can be combined? This is much easier when you have actual values to calculate. You're going to end up with a big messy (but usable) formula by the time you're finished this. \$\endgroup\$ – Transistor Aug 15 '18 at 21:53
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First, identify the nodes that are have the same voltage on one side and the same drop voltage on the other end. Take the parallel of the resistors and substitute that value for the R-equivalent.

If you redraw the circuit, you will see that voltage drop is the same for

  • R6 and R1 path
  • R5 and R1 path
  • R4 and R2 path

Let's call the low side node, attached to the voltage source, X; and, the upper side, where R5, R6 and R4 are connected, Y. The voltage drop across the aforementioned resistors are Vy-Vx, corresponding to the nodes at location high side Y and low side X.

Focus on this portion of the circuit and simplify: Here we see R5 and R6 share the same node with R1. Call this node Z. These two are in Parallel, because R5 and R6 have the same voltage = Vy-Vz. Simplify these two resistors, and this becomes R56 equivalent.

Now R56 is in series with R1; Simplify and substitute this with R561 and attach it between node Y and node X.

R561 is parallel with R4 (and R2 along that path), b/c it is between nodes X and Y. Simplify this one and make the substitution.

The process is identifying the parallel combination of resistors and then resolving any series resistors attached. Make sure that when two or more resistors are in parallel, we can identify the total voltage across to be the same. In your example, it is easy to identify R5 and R6 in parallel, but to identify (R5||R6 + R1) || (R4+R2) is not so easily seen.

Redraw the circuit each time you simplify with an equivalent resistor.

The rest is left as an exercise. :)

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Try to break down the circuit into simpler series/parallel resistors and get the equivalent resistor, then repeat until you get only one resistor.

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  • \$\begingroup\$ For instance, R5 and R6 are in parallel and can be replaced by a resistor value of of Rt where 1/Rt = 1/R5 + 1/R6. \$\endgroup\$ – BobT Aug 15 '18 at 21:42
  • \$\begingroup\$ Yes @BobT and the same for R3 and R7. Add R2 and R4. But I think the op is getting the idea. He will see how what follows is just more of the same thing. \$\endgroup\$ – lakeweb Aug 15 '18 at 22:01
  • \$\begingroup\$ @lakeweb Yes, I didn't want to do it for him, but a gentle pointer seemed to be OK. \$\endgroup\$ – BobT Aug 16 '18 at 0:40
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Here is ... (+) -- (R3 || R7) -- ( (R5 || R6) -- R1) || (R4 - R2) ) -- (-)

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enter image description here

Figure 1. OP's answer.

Yes, you appear to have it.

You could improve the notation a bit using '+' for series and '||' for parallel. Your end result would be

Upper:    (R1 + (R5 || R6)) || (R2 + R4)
Lower:    R3 || R7
Total:    (R1 + (R5 || R6)) || (R2 + R4) + (R3 || R7)

That way it's clear how you reached your conclusion and you can generate the mathematical expression to calculate from the series and parallel resistor formulas.

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