# Resistors in series/parallel

1. Sep 26, 2013

### monahanj09

1. The problem statement, all variables and given/known data

Here are the two problems I'm working with.

We have to figure out the current and voltage for all components in the circuit. Sorry for the image, I wasn't sure how else to present the circuit on here.

2. Relevant equations

Ohm's law: V=IR

3. The attempt at a solution

I actually already solved the first one, but I'm not certain I broke the circuit down correctly. I had 1k and 200 in series, 350 and 50 in series, and then those two resultant resistors in parallel.

The second one 2k and 500 are in parallel, the 3k and 5k are in parallel, and the 7k and the 7.5k are in parallel, and then the three resultant components are all in series with each other. I'm honestly not too sure though, I'm pretty bad at telling series vs. parallel if it's not blatantly obvious.

I know exactly what to do once the circuit is broken down, I just don't know how to break it down.

Last edited: Sep 26, 2013
2. Sep 26, 2013

### jegues

We must be looking at different circuits becuase I see no 8k and 10k resistors in the first circuit.

Also you are interpreting the second circuit incorrectly.

Start by making all the series simplifications you can, followed by parallel simplifications.

3. Sep 26, 2013

### monahanj09

Whoops, was looking at a different problem and got my numbers crossed up. On the first one I had 1k and 200 in series, 350 and 50 in series, and then those two resultant resistors in parallel.

4. Sep 26, 2013

### jegues

That is correct.

5. Sep 26, 2013

### monahanj09

Okay, cool.

As for the second one, I don't see anything that's in series. My professor has never explained to us what those heavy black dots on a schematic mean, and I think that's what is throwing me off.

6. Sep 26, 2013

### .Scott

Okay. You second one is the one labelled "f".
You have a 3K, 7K, and 5K in series - so that a 15K.
In parallel with that 15K, you have a 7.5K, so 1((1/15K)+(1/7.5K)) = 1((1/15K)+(2/15K)) = 5K
Then that 5K is in series with the 2K and 0.5K for a total of 7.5K.
12V across a 7.5K give you the current for the circuit.

I think you can work it from there.

7. Sep 26, 2013

### monahanj09

Why are the 3K, 7K, and 5K in series? I think the 7.5K in the middle is really throwing me off. I don't know how to interpret the heavy black dots on the schematic at all. Do those cut off the nodes like any other component would?

8. Sep 26, 2013

### Zondrina

Notice that the 3k, 7k and 5k are all in series because the current across each one will be the same. So you could just use one resistance instead of doing 3 different calculations.

9. Sep 26, 2013

### monahanj09

I know that series means all components have the same current, so does that mean that the heavy black dots on the schematic divide/distribute the current?

10. Sep 26, 2013

### iRaid

No the dots are just there connecting the wires (nodes), you can ignore them if you're just doing this kind of analysis for now.

Also for me it's good practice re-drawing the circuits after each step. Let me show you what I mean:

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Last edited: Sep 26, 2013
11. Sep 26, 2013

### B4ssHunter

once the current takes the path which has 3 , 7 and 5 . it cannot go anywhere else
i mean , there is only one way between 3 and 7 and only one way between 7 and 5 , thus they are all series , it would be better if you start off collecting the series ones

12. Sep 26, 2013

### Zondrina

The dots are known as nodes. They have important usage when performing nodal analysis.

Current travels along the path of least resistance, so you could say that nodes are important points in the circuit in determining which way current goes.

You should worry more about identifying which elements are in series and parallel first though.

13. Sep 26, 2013

### monahanj09

Ah, okay. The way my professor explained series to us was "same current, only 2 connected components", so the fact that they were connected to other components threw me off. I see what you mean by the current can't change as it travels between 3k, 7k, and 5k, though. Makes a lot more sense.

14. Sep 26, 2013

### Staff: Mentor

The "heavy black dots" are just indicating where wires connect. They are not components and have no function other than to emphasize that the wires at that point are connected. They serve no purpose other than to assure you that the wires that pass through that point are indeed connected and don't just pass by each other.

In some schematic diagrams, wires that cross without a dot are to be considered as not connected -- only junctions with dots are to be considered connected. NOTE that this is not a universal standard! You will find schematics that assume crossing wires are connected unless a semi-circular "overpass" indicates that wires are not joined. No dots in that case, but there will be obvious "jumps" over the intersections.

In the schematics you have shown, dots indicate junctions of wires -- connection points.

15. Sep 26, 2013

### iRaid

They can also be nodes, but not always either.

16. Sep 26, 2013

### Staff: Mentor

Essential nodes are where multiple components share a connection, so you'll probably find dots there