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Finding the current through the battery

  1. Mar 25, 2017 #1
    1. The problem statement, all variables and given/known data
    I am confused what to do but I have to find the current through the battery:
    gFFgItf.jpg

    2. Relevant equations
    V=IR

    3. The attempt at a solution
    I decided to find the equivalent resistance and do V/R = I using the 14V, but I believe my Req is wrong. What is throwing me off is the horizontal 1 ohm resistor.
     
  2. jcsd
  3. Mar 25, 2017 #2

    gneill

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    Staff: Mentor

    Show the details of your attempt. How did you try to reduce the network to a single equivalent resistance?

    What's your backup plan? :smile:
     
  4. Mar 25, 2017 #3

    berkeman

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    There may be a trick to simplify it because of the symmetry of the resistance values, but it's probably just easiest to write the KCL equations and solve them. Can you show us that work?

    EDIT -- Oops, beaten out by gneill again! :smile:
     
  5. Mar 25, 2017 #4
    Would this be right?

    14V - I1(1Ω) - I1(2Ω) = 0
    So that would give me I1 = 4.67 A
     
  6. Mar 25, 2017 #5

    gneill

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    You'll have to show us what ##i_1## is on your diagram, but I would say that it's very unlikely. Any current through R3 would "break" any attempt to consider the vertical paths as isolated series connected resistors.
     
  7. Mar 25, 2017 #6
    My bad. Here's what I was trying to do:
    e8LKYVG.jpg I realized that the I1 splits as soon as it reaches the junction and i can either keep going to the right or down. So that means my equation for my loop has to involve I2 right?
     
  8. Mar 25, 2017 #7

    berkeman

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    You can either write the multiple KVL equations, or the multiple KCL equations. I personally prefer KCL, but either will work. Are you familiar with how to write them?
     
  9. Mar 25, 2017 #8
    Yes. How my professor taught it to us is that we have to set directions for the current and create a clockwise or a counter-clockwise loop. My biggest problem with it is keeping it to only 3 unknown currents (I1, I2, and I3) as he advised. I get confused when im past assigning where I3 is.
     
  10. Mar 25, 2017 #9
    I think how i labeled the diagram was wrong. Would the current going through R3 be (I2 + I3) ?
    EA5xuJ0.jpg
     
  11. Mar 25, 2017 #10

    gneill

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    One way to minimize the number of currents you "invent" while labeling the circuit is to only ever add the minimum required number of currents at any junction you come across while proceeding systematically through the circuit. If you can, label new paths using mathematical combinations of existing currents.

    For example, if you go into a junction with i1 and two unlabeled paths diverge from there, make one current i2 and the other i1-i2. Thus only one "new" current is introduced at that junction rather than two.
     
  12. Mar 25, 2017 #11
    Okay. I relabeled everything. Can you check to see if i did it right? u4GDwaL.jpg
     
  13. Mar 25, 2017 #12

    gneill

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    Check the KCL sum at the center right node. Everything else looks okay.
     
  14. Mar 25, 2017 #13
    would it be -I3-I2?
     
  15. Mar 25, 2017 #14

    gneill

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    No. remember, whatever is flowing out of the node must also be flowing into the node. You have I2 and I3 both flowing out, so what must flow in on the remaining path?
     
  16. Mar 25, 2017 #15
    I2 + I3 ?

    Sorry, I'm a little confused about this stuff :sorry:
     
  17. Mar 25, 2017 #16

    gneill

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    Yup.
    No worries. After a bit of practice it'll suddenly "cick" and you'll wonder what all the fuss was about :smile:
     
  18. Mar 25, 2017 #17
    Oh okay! Thanks :smile:

    So if I were to assign three loops like this a8MvWOt.jpg
    would these be the right eqns?

    Eqn #1: (loop with 14V, starting after the 14V)
    14V - (I2+I1)*(1Ω) - (I1+I2+I3)*(2Ω) = 0

    Eqn #2: (upper loop, starting above I2+I1)
    (I2+I1)*(1Ω)+(I2)*(2Ω) - (I3)*(1Ω) = 0

    Eqn #3: (lower loop, starting above I1+I2+I3)
    (I1+I2+I3)*(2Ω) + I3*(1Ω) + (I3+I2)*(1Ω) = 0
     
  19. Mar 25, 2017 #18

    gneill

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    I'd say that they look good. I don't spot any problems.
     
  20. Mar 25, 2017 #19
    :) Awesome! So finding the current through the battery, i have to find I1, I2 and I3 and sum them together to use in the V=IR equation, where V = 14V?
     
  21. Mar 25, 2017 #20

    gneill

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    Your I1 flows through the battery. Find I1 and you're done (of course you need to mostly solve everything to get there... unless you use something like Cramer's Rule to find just the one current).
     
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