Resistors in series/parallel

  • Thread starter monahanj09
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  • #1
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Homework Statement



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.


Homework Equations



Ohm's law: V=IR


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.
 
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Answers and Replies

  • #2
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I actually already solved the first one, but I'm not certain I broke the circuit down correctly. I had 8k and 10k in series, which was then parallel to 6k. Upon doing some research I think that the 8k may be in series with the 6k, which would then be in parallel to the 10k.
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
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We must be looking at different circuits becuase I see no 8k and 10k resistors in the first circuit.
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
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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.
That is correct.
 
  • #5
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That is correct.
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
.Scott
Homework Helper
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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
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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.
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
STEMucator
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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?
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
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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.
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
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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?
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:
attachment.php?attachmentid=62227&stc=1&d=1380246047.png
 

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  • #11
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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?
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
 
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  • #12
STEMucator
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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?
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
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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
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
gneill
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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
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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.
They can also be nodes, but not always either.
 
  • #16
gneill
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They can also be nodes, but not always either.
Essential nodes are where multiple components share a connection, so you'll probably find dots there :smile:
 

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