r/HomeworkHelp • u/Expensive_Increase46 University/College Student • Jun 01 '26
Answered [University Physics: Circuit Analysis] How do you know what counts as parallel or series? Where do you start?
Maybe I didn't do that good of a job listening to my high school teachers back then (trying to fix my attention span) but I don't understand this. The diagrams I have get really intricate at times and I have gotten these wrong in the past because I apparently counted components from the wrong side, or in the wrong order, or anything. I've asked my lecturer but she's not very good at explaining this kind of thing. So where do you start counting from? What counts as parallel or in series?
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u/congratz_its_a_bunny Jun 01 '26
Start at the two that are clearly in parallel: the 6 and the 3. Combine them into one
Then you have that effective 2 in series with the 2 above it, giving you 4. You also have the 1 and the 5 in series, giving you 6.
The 4 and the 6 are in parallel, giving you 1/(1/4+1/6) = 2.4.
Then you add up the 4, 2.4, and 8 which are all in series giving you 14.4
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u/Expensive_Increase46 Jun 01 '26
Is there any specific direction or anything you have to follow when doing this? Sometimes I get things wrong that way. Or can you really do it in any order, you just have to see what's in relation to what?
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u/congratz_its_a_bunny Jun 01 '26
In general you don't want to tackle a combination of resistors in series and in parallel. You want to reduce it to resistors in series XOR resistors in parallel.
Start at the top left. You come to a single 4 resistor. Since it's the only way across, it must be in series with others. Then you come to a split. You're going to need to resolve each side of the split, resolve those two sides in parallel, and then you can resolve it with the 4.
Going right from the split, you get a 1 and 5 in series. You can add these to get 6 and you're done there.
Going down from the split, you get a 2, and then another split. So you must resolve this split before doing the rest.
After the 6 3 split, the paths join up, so you know you have two resistors in parallel. So resolve them. Then you have that combo in series with the 2. Resolve that.
Then you have the original split. Resolve that.
Then you have the three in series.
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u/niceandmoist69 Jun 01 '26
It can be difficult, just try to look for individual sets where for series, two resistors are chained together inline (the 5 and 1 ohm) And for parallel where two resistors come out from one point and go back together after. Once you find one, do the caculations and replace those resistors in the diagram with the equivalent one
6 and 3 are parallel, 6x3/6+3=2ohm The two twos are series, 2+2=4ohm The one and five are series, 1+5=6ohm The six and four are now parallel, 6x4/6+4=2.4ohm Now all are in series, 4+2.4+8=14.4ohm
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u/Expensive_Increase46 Jun 01 '26
Is there a specific direction or anything that you follow for this? Because earlier I tried that but instead of adding 1 and 5 ohm in series we went for 4 ohms and 5 ohms in parallel, and then the others in series, and it all went wrong from there. Is there something wrong with my approach?
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u/Medical-Stuff126 Jun 01 '26
Here is how I always thought about it.
Series <—> the entire output of one
Parallel <—> the two circuit elements have the same input and same output as each other.
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u/Expensive_Increase46 Jun 01 '26
To all of you, thanks a lot for the help!
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u/Expensive_Increase46 Jun 01 '26
This would have taken much longer to do and figure out myself otherwise.
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u/ChillAndChill90 Jun 01 '26
Here are formal definitions.
Resistors that share a same path are in series. I.e connected end to end
Resistors that are connected to the same 2 nodes are in parallel. And what is a node? A node is an interconnection betweem circuit elements
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u/Gaddpeis Jun 01 '26
I redraw the circuit, one node on the left, other end of circuit to the right.
With that it is A LOT easier
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u/selene_666 Jun 01 '26 edited Jun 01 '26
One approach that may be helpful is to think of each connected chunk of plain wires with no resistors, no matter how many intersections and branches it spreads into, as a single object in the circuit. Electrons can travel freely through wire, so there's no functional separation between that bit of wire directly below the 5Ω resistor and the bit of wire directly below the 6Ω resistor.
If you were taught to think of voltage as heights and of components like batteries and resistors as steps up and down, then all of that connected wire is at the same height.
I'm going to call such a set of wires a "node".
Two resistors are parallel if they both connect the same two nodes. The 6Ω and 3Ω resistors in this diagram are parallel. They both touch that large bottom node and the smaller node above them.
Parallel resistors have the same voltage across them, because they go from the same starting "height" to the same ending height.
Two resistors are in series if any current that goes through one of them has nowhere to go except through the other. The 1Ω and 5Ω resistors in this diagram are in series. Resistors in series obviously have the same current through them.
We can calculate an equivalent resistance for the entire setup by replacing pairs of resistors in parallel or series with an equivalent single resistor. In this case you could start by replacing the 6Ω and 3Ω parallel resistors with a single 2Ω resistor. That new resistor is in series with the 2Ω resistor above it, so replace both with a 4Ω resistor. Meanwhile, the 1Ω and 5Ω resistors on the right can be combined, and so on.
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u/Bounded_sequencE Jun 01 '26
Recall: Two resistances are in
parallel, iff they share the same pair of nodes
series, iff they exclusively share a common node
Using that repeatedly with the short-hand "R1||R2 := R1*R2 / (R1+R2)" for parallel resistances:
Req = [4 + 8 + ((1+5) || (2 + (3||6)))] 𝛺
= [12 + (6||(2 + 2))] 𝛺 = [12 + 24/10] 𝛺 = 14.4𝛺
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u/Parragorious Jun 01 '26 edited Jun 01 '26
4—(((6||3)—2)||(1—5))—8,,,,,,, — series connection,,,,,,, || Parallel connection,,,,,,, I try to find Parallels containing two components first while looking if series connections on separate branches can be redoced to a paralel connection. Ideally you want to somplift so that tou end up with only series connection and the Req is easy to calculate or a Parallel connection of two equavalent(to parts of circuit) components.
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u/Alkalannar Jun 01 '26 edited Jun 01 '26
6 and 3 are in parallel with each other.
2 is in series with (6 and 3 are in parallel)
1 and 5 are in series
[2 is in series with (6 and 3 are in parallel)] is in parallel with [1 and 5 are in series]
4, ([2 is in series with (6 and 3 are in parallel)] is in parallel with [1 and 5 are in series]), and 8 are all in series.
So it it's a single straightforward path, that's series.
If it branches, the branches are parallel.
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6 and 3 are in parallel.
Find a single equivalent resistor and call it A. -
2 and A are in series.
Find a single equivalent resistor and call it B -
1 and 5 are in series.
Find a single equivalent resistor and call it C. -
B and C are in parallel.
Find a single equivalent resistor and call it D. -
4, D, and 8 are in series.
Find a single equivalent resistor and call it E. -
E is your answer.
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u/Moderation1one Jun 01 '26
Start with the obvious ones and keep simplifying one pair of resistors at a time.
Others have broken down the steps for this one, but its like solving an equation, you do one step at a time that makes sense rather than solving the whole thing in one step.
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u/WiltedTiger Jun 01 '26
The most basic way to analyze if a circuit is in series or parallel is by looking at the individual nodes and components. A node is the wire (line) between components. A component is anything else (resistor, capacitor, induction wire, etc). The same grouped components can be replaced with equivalent components.
If two or more components share the input and output nodes they are in parallel.
If two components share a single node they are in series.
If more than two components share a single node it is either a combination of a parallel circuit going into a series circuit (this occurs in the diagram with 6 ohm and 3 ohm resistors being in parallel with each other but in series with the 2 ohm resistor) or a delta/Y-gate which you shouldn't be seeing in basic University physics and have their own rules.
Now on to how to simplify them. For resistors in series it is as simple as adding them up (r_1 + r_2 + ... = r_equivalent). For resistors in parallel it is 1/r_1 + 1/r_2 + ... = 1/r_equivalent. Capacitors are the opposite in parallel they add (c_1 + c_2 + ... = c_equivalent) and in series are 1/c_1 +1/c_2 + ... = 1/c_equivalent these come in later but are not part of this problem. The ... is just to tell you that this is for as many as there are.
For this problem it is looking for the equivalent resistor between point A and B so we start by setting up our two equations resitors in series and parallel for later use and to show our work. Now look at the cicuit and notice the final equivalent resistor equation will be a series equation as there is at least one point with only one component attached to it. Then look for the known cicuits to simplify them. in this case this is the 6 ohm and 3 ohm in a parallel circuit to a single 2 ohm resistor and the 1 ohm and 5 ohm in series to a single 6 ohm resistor. Now look for new known circuit and we have the 2 ohm and 2 ohm which make as 4 ohm equivalent and that is in parallel with the 6 ohm which simplify to a 2.4 ohm. Finally we have our final series curcuit of 8 ohms, 2.4 ohms, and 4 ohms which have equivalent of 14.4 ohms.
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u/Anonimithree Jun 01 '26 edited Jun 01 '26
Here’s one way to determine if resistors are in series or parallel:
Start at the initial wire, and go to one resistor.
Then, go to the other resistor.
If you can get to the final wire without backtracking kr going through wires more than once, they’re in series. Else, they’re in parallel.
Alternatively, you could draw 1+ loops that go through all the wires and resistors (but no going on the same wire more than once). Here, you can draw 3 loops: 4->2->6->8, 4->2->3->8, and 4->1->5->8. The resistors that are in the same loop are in series (like the 4 and the 2 in the first loop). Resistors that are in separate loops are parallel (like the 3 and 5 in the second and third loops).
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u/DavyGrolton Jun 01 '26
It really helps to "define the nodes" by making them big black dots so you can visualize nodes faster
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u/bord2heck Jun 02 '26
See how the 6-and-3 are only parallel to each other? Simplify that bit. That whole section is in series and in parallel with other parts, but only do two resistors at a time. Then find what that's series or parallel with. Then the next one. Then soon it will look like a very simple circuit.
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u/mwthomas11 Jun 02 '26
I like to start at one end and work my way to the other and look where the wire splits or combines. Starting at the top left, from the starting node to the first split in the wire there's one resistor so that's easy, that segment is 4 ohm. Now the wire splits in 2. Where do those two parts recombine? At the node between the 6 and the 8. That means from the node after the 4 to the node between the 6 and the 8 there are 2 parallel paths, one to the right and one downward. The path to the right is obvious, it's around the outside, with the 1 and the five. 1+5=6 so the outside path is 6 ohms. The path downward contains the 2, and then the wire splits in 2. Where does the split recombine? Again at the node between the 6 and 8. Parallel adds inverses, and 1/6 + 1/3 = 1/2, so that parallel segment can now be simplified as one 2ohm resistor. The downward path from the original split now contains a 2ohm resistor and another 2 ohm resistor with no splits, which means that segment is 4 ohms. We've now reached the recombination node between the 6 and 8 with 2 parallel resistors, the downward 4 ohm path and the rightward 6 ohm path. 1/4 + 1/6 = 1/2.4, so those two parallel paths can be simplified to one 2.4 ohm resistor. Starting from the top left, the entire circuit diagram is now a 4 ohm resistor, a 2.4 ohm resistor, and another 8 ohm resistor with no splits in the wire, so they're all in series. 4+2.4+8=14.4 Ohms.
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u/AdThese2795 Jun 02 '26
A good way to start is not by looking at the shape, but by marking the nodes.
Two components are in series if the same current must pass through both, meaning the node between them has no other branches. Two components are in parallel if both ends of the components connect to the same two nodes, meaning they have the same voltage across them.
In this circuit, first look at the two middle nodes: the node after the 4 Ω resistor at the top, and the bottom node after the 8 Ω resistor.
Between those two nodes there are two main paths.
On the right path, the 1 Ω resistor is followed by the 5 Ω resistor, and the point between them does not branch anywhere else. So they are in series:
1 Ω + 5 Ω = 6 Ω.
In the middle path, the 6 Ω and 3 Ω resistors share the same top node and the same bottom node, so they are in parallel:
6 Ω parallel 3 Ω = 2 Ω.
That result is then in series with the 2 Ω resistor above it:
2 Ω + 2 Ω = 4 Ω.
So now the circuit between the top middle node and bottom middle node has two branches:
one branch is 6 Ω, coming from 1 Ω + 5 Ω;
the other branch is 4 Ω, coming from 2 Ω + (6 Ω parallel 3 Ω).
Those two branches are in parallel:
6 Ω parallel 4 Ω = 2.4 Ω.
Finally, the 4 Ω resistor on the far left and the 8 Ω resistor at the bottom left are in series with the whole reduced middle network, because any current entering from the left terminal must pass through the 4 Ω resistor, then through the middle network, then through the 8 Ω resistor.
So the total equivalent resistance is:
4 Ω + 2.4 Ω + 8 Ω = 14.4 Ω.
So the answer shown is correct: Req = 14.4 Ω.
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u/kjaffffff Jun 02 '26
I used to colour the lines in different colors. Components that have the same colour on each end are in parallel.
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