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Electricity, resistance

  1. Apr 11, 2017 #1

    Kajan thana

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    Gold Member

    1. The problem statement, all variables and given/known data
    Find the effective resistance between X and Y. I have attached the circuit to this post.

    2. Relevant equations
    V=IR

    3. The attempt at a solution
    I did 0.25+0.5+0.5=1.25 then flipped the fraction to get 0.8ohm. I considered as if two resistors are connected in series and rest are connected in parallel .
     

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  2. jcsd
  3. Apr 11, 2017 #2

    gneill

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    Staff: Mentor

    It's hard to tell from what you've posted which resistors you've considered to be in series and which in parallel. Here's your diagram with the resistors labeled:

    upload_2017-4-11_18-58-48.png

    Take us through your calculation step by step, identifying which resistors are involved.
     
  4. Apr 11, 2017 #3

    Kajan thana

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    Thank you for labelling it for me.

    Basically I combined R1 and R2 to give me 4 ohms. Then treated (R1&R2), R3 and R4 as if they are connected in parallel to get the effective resistance 0.8 ohms.
     
  5. Apr 11, 2017 #4

    gneill

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    Staff: Mentor

    Okay, well since X is the junction between R1 and R2 they can't be in series for any path that starts at X and ends elsewhere (at Y for example). Automatically there are two paths leaving point X, one through R1 and one through R2. The best you could hope for is that they are in parallel, but a quick look at the circuit tells you that they are not (they only share one node in common: X).

    Take a look for other series or parallel opportunities to start the simplification. What can you find?
     
  6. Apr 11, 2017 #5

    Kajan thana

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    I got the answer as 1.2 ohms which matches the book's answer as well, But I don't understand it properly. Can you explain to me please? I basically took R3 and R4 as parallel and got the overall resistance 1 ohm, then combined that with R2 to give me 3ohms. Then I took the combined 3ohm and R2 to be parallel to give me the answer 1.2. The part that does not make sense to me is why are we taking R3 and R4 parallel, because they are coming from the same node isn't?
     
  7. Apr 11, 2017 #6

    Kajan thana

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    I don't understand when you said X is the junction...., can you please expand on it?
    Thank you so much.
     
  8. Apr 11, 2017 #7

    gneill

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    Your analysis is correct.

    R3 and R4 are in parallel because their connections share two nodes. Anther way to tell that they are in parallel is that you can follow a closed path that passes through only through those two components:

    upload_2017-4-11_20-25-39.png

    The node X is where R1 and R2 are connected. Thus it is the junction between those two components.

    You're looking for the resistance between nodes X and Y. So you need to consider them as starting and ending points, or the terminals of the network. Node X happens lie at the junction of two components, R1 and R2, so there are two paths leading from X into the network.
     
  9. Apr 11, 2017 #8

    Kajan thana

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    Thank you so much.
    Make sense now
     
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