I am trying to simplify the following portion of a circuit for analysis. I am not sure if it would simply be shown as 2 resistors in parallel or something else as the nodes are connected. I do not have the resistor values currently, so I need to derive an equation for the resistance at a,b, and c.

Edit: I am looking for the resistance between a-b and b-c.

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    \$\begingroup\$ Resistance requires two points. So do you mean the resistance between a and b, a and c, and b and c? \$\endgroup\$ – jonk Oct 16 '17 at 4:15
  • \$\begingroup\$ a to b and b to c, I clarified in the edit. \$\endgroup\$ – thePinochleKid Oct 16 '17 at 4:41
  • \$\begingroup\$ Well, resistance from b to c shorts out \$R_1\$ and \$R_2\$, so they don't matter. That just leaves two resistors to deal with. The same is true for the resistance from a to b, except now \$R_3\$ and \$R_4\$ are shorted out and don't matter. For the case of a to c, it's a little bit different. But you don't care about that case. \$\endgroup\$ – jonk Oct 16 '17 at 4:46
  • \$\begingroup\$ Just to complete my understanding, can you explain a-c should it come up? Thanks for answering my question up to this point \$\endgroup\$ – thePinochleKid Oct 16 '17 at 4:49
  • \$\begingroup\$ That case is just \$R_1\mid\mid R_2+R_3\mid\mid R_4\$. \$\endgroup\$ – jonk Oct 16 '17 at 5:03

The equivalent resistance (aka impedance) that can replace resistors in parallel can be written like this:

\$R_{parallel} = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+..}Ω\$

This means that if \$R_1 = 53Ω\$ and \$R_2= 205Ω\$ then you can replace both of those two with one resistor that has the value \$\frac{1}{\frac{1}{53}+\frac{1}{205}}=42.1Ω\$

Or it can be written like this if it's just two resistors in parallel.

\$R_{parallel} = \frac{R_1×R_2}{R_1+R_2}\$

With the same example numbers as above, \$\frac{53×205}{53+205}=42.1Ω\$

For 3 resistors in parallel it's not \$\frac{R_1×R_2×R_3}{R_1+R_2+R_3}\$, it's \$\frac{R_1×R_2×R_3}{R_1×R_2+R_2×R_3+R_1×R_3}\$. I derived this expression from the first equation above (\$\frac{1}{...}\$).

The equivalent resistance that can replace resistors in series can be written like this:

\$R_{series} = R_1+R_2+...\$

When I write "..." I mean that the terms can be repeated if you have more resistors.

With the information above together with the schematic we can arrive to the conclusion:

The resistance between a and rotated b is equal to \$\frac{R_1×R_2}{R_1+R_2}\$

The resistance between rotated b and c is equal to \$\frac{R_3×R_4}{R_3+R_4}\$

The resistance between a and c is equal to \$\frac{R_1×R_2}{R_1+R_2}\$+\$\frac{R_3×R_4}{R_3+R_4}\$


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