# Resistors in Series vs Parallel

## Main Question or Discussion Point

I am confused how I would add certain resistors, like the ones in the attached picture (sorry it's a bad picture). If I add the first two resistors as if they were in parallel (as they are), couldn't I then add that equivalent resistance to resistor 3 as if they were in series, since the equivalent resistor would be drawn as if it were in series with resistor 3? Any insight would be greatly appreciated.

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Simon Bridge
Homework Helper
The circuit diagram and the description of the problem is incomplete.
If the idea is to find the total resistance between two points in the network, you need to specify which two points.
When you do that, you will see how to complete the calculation.

Drakkith
Staff Emeritus
From your drawing, which as Simon said is incomplete, it appears that all three resistors are in parallel. As such you cannot add one of the resistors resistance as if it were in series with the others, as it is not.

Simon, I see your point, I would find it easier to calculate the resistance if given two points within the network, but the question I came upon in my studying asked for the total resistance in a circuit like this one (with no battery). I couldn't decide between calculations.

Thanks Drakkith, I think that makes sense to me. I just always thought one could "replace" two resistors in a circuit with one with equivalent resistance. Doing that to two of these would just make this circuit a circle with the two resistors. But that would be incorrect?

The more I think about it, I think the problem is too ambiguous. Whether or not resistors are in parallel of series depends on whether they split the current or all experience the same current, right? If these were in parallel, then the current would have to enter by the middle resistor to create a junction splitting the current to each resistor. However I took the problem to mean a voltage source could be added in on the side somewhere, which would force one of the resistors to experience the total current at some point.

Drakkith
Staff Emeritus
Thanks Drakkith, I think that makes sense to me. I just always thought one could "replace" two resistors in a circuit with one with equivalent resistance. Doing that to two of these would just make this circuit a circle with the two resistors. But that would be incorrect?
You can. But the issue is whether or not resistors 1 and 2 are in parallel with each other AND in series with resistor 3. Without knowing where a voltage source would be located it isn't possible to know.

The more I think about it, I think the problem is too ambiguous. Whether or not resistors are in parallel of series depends on whether they split the current or all experience the same current, right? If these were in parallel, then the current would have to enter by the middle resistor to create a junction splitting the current to each resistor. However I took the problem to mean a voltage source could be added in on the side somewhere, which would force one of the resistors to experience the total current at some point.
At first I assumed that all 3 were in parallel with the voltage source 'off screen' and connected to each side of the circuit where the circuit splits into 3 paths and then again where they converge on the other side.

Simon Bridge
Homework Helper
When you reduce the first pair of resistors with just one - you effectively calculated the resistance between two points. Draw those points in.

In the picture of the two resistors basically making a "circle" by themselves - where are the two points which would allow you to add the resistors in series?

Equivalent resistance for three resistors in parallel

I believe the problem is to find the equivalent (total) resistance, call it Rt.

If it is then you can do this two ways. (1) use product over sum twice or (2) use reciprocal formula for resistors in parallel.

(1) find R2 in parallel with R3: Req = (R2)(R3)/(R2+R3). Then find Rt = (R1 in parallel with Req): Rt = (R1)(Req)/(R1+Req)

(2) 1/Rt = (1/R1) + (1/R2) + (1/R3). So first find (1/Rt) then take the reciprocal of the result to get Rt

Simon Bridge
Homework Helper
@kevlat: You realize, of course, that "product over sum" is the "reciprocal formula".

$$\frac{1}{R_t}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\ \Rightarrow R_t = \frac{R_1 R_2 R_3}{R_2 R_3 + R_1 R_3 + R_1 R_2}$$ ... you can also add the conductances: $$Y_x=\frac{1}{R_x}\Rightarrow Y_t=Y_1+Y_2+Y_3$$

However, this does not answer OPs question: how did you choose to use the parallel resistor formula and not the series resistor formula - in the case where there is only Req and R1 (your example #1)? i.e. can you use this to clear up the "circle of resistors" confusion?

I'm hoping OP will play around with inserting the missing reference points into the diagram (as it stands it is not a "circuit") and comparing the total resistances in each case ... there is only one configuration that makes sense in terms of the question.

The situation would be more ambiguous if the diagram showed a loop of three resistors in series and no reference points...

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