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Finding Current and potential difference in a circuit

  1. Aug 13, 2008 #1
    1. The problem statement, all variables and given/known data
    The 5.00-V battery in the figure ( http://session.masteringphysics.com/problemAsset/1009413/11/yf_Figure_26_42.jpg ) is removed from the circuit and replaced by a 20.00-V battery, with its negative terminal next to point b. The rest of the circuit is as shown in the figure.
    a) Find the magnitude of current in the upper branch.
    b) Find the magnitude of current in the middle branch.
    c) Find the magnitude of current in the lower branch.
    d) Find the potential difference Vab of point a relative to point b.

    2. Relevant equations
    Kirkoff's loop law: V1 + V2 + V3 = 0
    Kirkoff's junction law: Current in = Current out

    3. The attempt at a solution
    For the top branch, I made current equal to I1, second branch I2, third branch (I1+I2).

    It's sort of hard to explain how I tried to do it, but here it goes:
    First I looked at the top 2 branches, going clockwise, starting with the first resistor at the left on the top branch.
    2(I1) - 10V + 3(I1) - 4(I2) + 20 - I2 = 0

    Since there are two variables, I did the same thing for the middle and bottom branch.
    Starting on the middle branch, the resistor on the right...
    4(I2) - 10(I1 + I2) + I2 - 20V = 0

    Isolating I2 from the second part I got I2 = -4 - 2I1.

    I substituted this into the first equation, but did not get the right answer.

    Not sure how to even start with finding the potential difference.

    Any explanation would be greatly appreciated!!
  2. jcsd
  3. Aug 14, 2008 #2


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    You know what the current is for the middle branch, so you know what the voltage drop or rise is across the middle resistor (4 ohm) to the right hand node. Call this Va.
    You also know what the current is in the top resistor (3 ohm), so you know the voltage drop or rise to the same right hand node. Call this Vb.

    Vab then is the difference between these. But be careful to keep your signs straight.

    What values did you get for the first 3 parts?
  4. Aug 14, 2008 #3
    well, that's my problem. I don't know how to find the current in any of the branches, and when I did try to find them, my answers were way off.
  5. Aug 14, 2008 #4


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    This is okay; but you have not said anything about the direction, which causes trouble later on. So if [itex]I_1[/itex] is to the left in the top branch, and [itex]I_2[/itex] is to the left in the middle branch, what is the direction of [itex](I_1+I_2)[/itex] in the bottom branch?

    If you have answered my question above, and since you're going clockwise around this bottom loop, do you see where the error in this equation is?
  6. Aug 14, 2008 #5
    hmmm I still seem to be getting the wrong answer.

    The direction of current would be flowing to the right in the bottom branch.

    In re-doing my calculations I got: 10V - 2(I1) + I2 - 20V + 4I2 - 3I1 = 0

    20V - I2 + 10(I1 + I2) - 4I2 = 0
    20V + 10I1 +52 = 0
    -2 - 0.5I2 = I1

    Plugging this part into the top equation still does not give me the right answer. Anyone know where i'm still going wrong?!
  7. Aug 14, 2008 #6


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    That sounds right.

    No, these are exactly the same as the equations in your original post; they have just been multiplied by a minus sign.

    Look back at the second equation in your original post:

    This equation is fine, except for the sign of one of the terms. Which sign is not right?
  8. Aug 14, 2008 #7


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    The way I find it useful to look at things to keep things straight is to take the active element, like in the case of the 20V battery and then say all the other voltages add up around the loop to that. That's not to say that care doesn't still need to be taken in keeping the signs straight in solving the equations or their effects in other equations.
  9. Aug 14, 2008 #8


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    Right, but to add up the voltages correctly you need to keep track of the signs of the voltages across the resistors, which is where Number1Ballar is going wrong.
  10. Aug 16, 2008 #9
    Applying Kirchoff's voltage law you will get these equations...
    10i1 + 30 i2 =20
    10i1 + 5 i2 = 5

    where i1 is the current flowing in the middle branch, i2 is the current flowing in the lower branch and i1+i2 is the current flowing in the upper branch...

    solving these eqns you will get i2=15/25
    so you can also easily calculate the pd b/w a n b.
  11. Aug 16, 2008 #10


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    Hi vishal_garg,

    No, I don't believe those answers are correct.
  12. Aug 16, 2008 #11
    alphysicist has you on the right track. The hard part of Kirchhoff circuits is keeping track of negative signs and voltage "drops" vs voltage "adds".

    Try using the PhET website from U. Colorado. There is an applet which allows you to create the circuit and watch current flow through the branches. You can even put in voltmeters and ammeters. It's kind of cheating since you can see the answers, but since you are here I will assume you will use it to check your work and not to cheat.
  13. Aug 16, 2008 #12


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    Thanks. Hadn't seen that before. Works pretty well.
  14. Aug 16, 2008 #13
    Hmm I don't know which sign is wrong.
    Here is the picture again, but I drew in positive and negative terminals on the resistors. I think my problem is understanding when to add and when to subtract. Is my image right? or am I completely off? It must be wrong, because in following this picture (with the +(ve) and -(ve) signs in this way) I keep getting the wrong answer.

  15. Aug 16, 2008 #14


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    Hi Number1Ballar! :smile:

    You're getting your signs wrong because you've left out the most important part out of the picture …

    you must always put all the unknowns in any picture (mechanics etc as well as circuits) …

    in this case, you should write I1 I2 and I3 on the picture, together with arrows to show the direction :smile:

    (it doesn't matter if you guess the direction wrong … the current will just come out as negative)
  16. Aug 16, 2008 #15
    Okay well I will try this again.

    Starting at point a, current going to the left, I got 10V - 2I1 + I2 - 20V + 4I2 - 3I1 = 0.
    Simplifying to: -10V - 5I1 + 5I2 = 0

    Then, starting from point b, current going to the left and assuming current in bottom branch is (I1 + I2) is: 20V - I2 + 10(I1 + I2) - 4I2 = 0.
    Simplifying to: 20V + 10I1 + 5I2 = 0.
    I2 = -4 - 2I1

    Substituting this into my first equation gives me a current of 2A, but the correct answer should be 0.4A.

    I'm actually a little confused with the answers that are given, it says 0.4A in top branch, 1.6A in middle branch, and 1.2A in lower branch. But if i'm making the current in the lower branch equal to I1 + I2....shouldn't the current be 2A? why is the first branch subtracted from the second branch?

    Sorry for dragging this thread out for so long...my professor didn't go over this concept in detail.
  17. Aug 16, 2008 #16


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    I think you may not be realizing some of the consequences of the choices you are making. With your first voltage equation, you have i1 going to the left in the top branch, and i2 going to the left in the middle branch. If you then say that the current in the bottom branch is (i1+i2), then that means that the current in the bottom branch is to the right. That means that the positive and negative symbols that you have on the bottom resistor (in the post #13) are backwards.

    (If the direction of the currents don't make sense, think about the point where all three branches come together on the left. Kirchoff's current rule says that the sum of the currents coming into the node equals the sum of the currents leaving the node.)

    Once you fix up the signs for the bottom resistor, which will change the sign in your second voltage equation, I think you'll get the right answer.
  18. Aug 16, 2008 #17
    Okay I think i'm finally starting to get this, thanks so much for your help!
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