# Resistor simplification

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.

Thanks

• Resistance requires two points. So do you mean the resistance between a and b, a and c, and b and c? – jonk Oct 16 '17 at 4:15
• a to b and b to c, I clarified in the edit. – thePinochleKid Oct 16 '17 at 4:41
• 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. – jonk Oct 16 '17 at 4:46
• Just to complete my understanding, can you explain a-c should it come up? Thanks for answering my question up to this point – thePinochleKid Oct 16 '17 at 4:49
• That case is just $R_1\mid\mid R_2+R_3\mid\mid R_4$. – 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 $\frac{R_1×R_2}{R_1+R_2}$+$\frac{R_3×R_4}{R_3+R_4}$