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Basic resister circuit with two batteries

  1. Jul 20, 2016 #1
    1. The problem statement, all variables and given/known data
    https://i.gyazo.com/ec36f9345b56810598a5b2cebc62aeb7.png
    ec36f9345b56810598a5b2cebc62aeb7.png
    Determine the current in the 7ohm, 8ohm, and 4ohm resister.

    2. Relevant equations
    The series/parallel equations for equivalent resisters. and V=RI.

    3. The attempt at a solution
    I'm normally pretty good with these circuit questions, but I'm really lost when it comes to using a second battery. When I look at the circuit, it looks like the 9V battery flows into a parallel system with the 12V battery and the resisters. At the same time I see spots where I would imagine the current opposing each other. Basically, I'm wondering how to approach a problem with two batteries.
     
    Last edited by a moderator: Jul 20, 2016
  2. jcsd
  3. Jul 20, 2016 #2

    CWatters

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    Have you studied Kirchoff's voltage and current laws?
     
  4. Jul 20, 2016 #3

    epenguin

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    For such problems you use the "superposition theorem":
    The total current in any part of a linear circuit equals the algebraic sum of the currents produced by each source separately.

    This is explained in many texts, for instance http://hyperphysics.phy-astr.gsu.edu/hbase/electric/suppos.html#c2 with application examples following. But before or instead of wading through texts, you would do yourself more good by first trying to see that it makes sense, and possibly even proving it by yourself - you almost got the essence of it in what you wrote! :approve:
     
  5. Jul 20, 2016 #4
    We discussed that total voltage is zero and current out is equal to current in.
     
  6. Jul 20, 2016 #5
    I read through the link you posted. So I would have to calculate the current for each battery, and then add them together? Would it be like separating the problem into two different problems, each with one of the two batteries? Maybe if you could get me started it would help?
     
  7. Jul 20, 2016 #6

    Delta²

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    I don't see right now how this problem can be reduced to a series or parallel configuration or a mix.

    However it is not so hard to solve you apply Kirchoff's Current Law in the top (or bottom) junction and you ll have
    ##I_7=I_8+I_4## (1)
    (noticing that the current ##I_7## that flows through the 7ohm resistor is the same as that through the 12V battery, and similarly the ##I_8## current is that same as that flows through the 9V battery.
    So we have equation (1) that has 3 unknowns, we still need 2 more equations to extract from the circuit, so that we ll have total 3 equations with 3 unknowns.

    Applying Kirchoff's Voltage Law (KVL) in the left loop with the 12V battery, the 7ohm and 4ohm resistors we get
    ##12-4I_4-7I_7=0## (2)

    You can similarly apply KVL in the loop with the 9V battery, the 8ohm and 4ohm resistor and get the (3) equation.

    And then you have to solve a system of 3 equations with 3 unknowns.
     
  8. Jul 20, 2016 #7
    Ok, apparently I don't know KVL as well as I thought. How would you divide up this problem? Is there ever a point where you would include both batteries in a loop "equation"?
     
  9. Jul 20, 2016 #8

    CWatters

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    First step is to mark up the circuit with arrows indicating the three currents. It does not matter which direction they point. Later one or more may turn out to be negative which means the current flows the opposite way to your initial choice.

    Then add arrows for the voltages across each resistor. These must be consistent with your current arrows.

    Then write KVL and KCL equations.
     
  10. Jul 20, 2016 #9
    So would the currents be
    12V + 9V - I8 - I7
    12V - 4I - 7I
    9V - 4I - 8I
    ?
     
  11. Jul 20, 2016 #10

    cnh1995

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    Current through 7 ohm and 8 ohm is not same. Write KVL in the form of two equations with two unknown currents.
     
  12. Jul 20, 2016 #11
    I'm not sure how to write the equations. Normally I would write the equation according to the path the electrons would flow, but since there's two batteries I'm not sure where things start and end and what flows against what. Sorry, I'm just having a little trouble getting this.
     
  13. Jul 20, 2016 #12

    cnh1995

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    There are two loops in the circuit. Start from any battery at any terminal. For example, say current through 8 ohm is I1 towards left and current through 4 ohm is I2 downward. So, current through 7 ohm is I1-I2, towards left. You can choose any directions for I1 and I2. Using these two unknowns, write KVL equations for two loops.
     
  14. Jul 21, 2016 #13

    epenguin

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    Yes.
     
  15. Jul 21, 2016 #14

    CWatters

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    Re: KVL/KCL... You can't do that if you don't know which way electrons flow until the end. You appear to have skipped the first step which is to mark up the circuit to define what you mean by +ve current flow in each loop. The direction is arbitrary. You should then use that to write the equations. Later if a current turns out to be negative you know that your initial assumption about the direction of current was "wrong". It is possible to skip straight to equations BUT it is very easy to make errors particularly with the sign of a voltage or current.

    Re: The superposition theorem is probably better at solving this problem but you should understand both methods.
     
  16. Jul 21, 2016 #15

    CWatters

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    For an example KVL/KCL see..
     
  17. Jul 21, 2016 #16

    CWatters

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    and an example solved using superposition..
     
  18. Jul 21, 2016 #17
    Thankyou everyone for all the help! I was able to get the right answer and check it in class! Thanks a ton everyone!
     
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