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Homework Help: Series Circuit with two batteries

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data
    A. Make a prediction about the direction of current flow through each resistor. (6 points)
    B. What current flows through each resistor? (15 points)
    C. What is the voltage drop across R2? (4 points)

    2. Relevant equations

    3. The attempt at a solution
    I'm sort of lost as to how to go about the equation. Normally I could tell the direction of the current when given a positive or negative sign near the battery.

    Attached Files:

  2. jcsd
  3. Mar 15, 2013 #2

    rude man

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    The wider end of the battery is +.
  4. Mar 15, 2013 #3
    So the top half of the circuit goes counter clockwise while the bottom half goes clockwise?
  5. Mar 16, 2013 #4


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    Certainly, the bottom loop goes clockwise.

    While you may be correct regarding the top loop, I wouldn't hazard a guess without further analysis. (analysis of the circuit, not my personal analysis with my shrink!)
  6. Mar 16, 2013 #5


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    One way to analyse this type of circuit is to mark on the circuit arrows showing which direction you are assuming the current and voltage drops to be in the resistors. These have to be self consistent but it doesn't matter if you are wrong!

    Then you apply KCL and KVL to write a series of simultaneous equations. It's important that these are written correctly taking into account the direction of the arrows.

    When you solve these simultaneous equations some values may come out -ve. That just means you got the direction of the arrow wrong when you made your assumption. No matter. Don't change the arrows because that will make your working out appear wrong.
  7. Mar 18, 2013 #6
    So I should solve the circuit by guessing the type of direction the current is flowing? Then the answer will tell me whether I'm wrong or not?
  8. Mar 18, 2013 #7

    rude man

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    If your answer is negative you picked the wrong direction, but your answer is still correct since two minuses make a plus! In other words, a current going in one direction is the same as the negative of the same current going in the other direction.
  9. Mar 18, 2013 #8
    Ok. Well I solved it out and after making some changes to the direction, I think I got it right. The top half will flow clockwise while the bottom will flow counter-clockwise. That is what I got. And regarding the current, I got the Current in R1 to be 3 amperes, Current in R2 to be 1.5 amperes, and Current in R3 to be 2 amperes. Does that sound right?
  10. Mar 18, 2013 #9


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    Sorry it can't be right.

    There is no way to make 3, 1.5 and 2 add up to zero as required by KCL at (for example) the node at the top left.
  11. Mar 18, 2013 #10
    Oh ok. I found that I had three equations for only two unknowns. Let me double check my equations first.
    Last edited: Mar 19, 2013
  12. Mar 19, 2013 #11
    After trying some equations for an hour, I'm still lost. The only problem with my equations is the unknowns of the current. I don't know where to begin to find it. I realized that I actually need three equations for the three unknown currents.

    i) I1R1-E1-I2R2

    I started from the top left and the middle left.
  13. Mar 19, 2013 #12


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    Can you mark up the diagram with your initial assumptions about the voltage drops and currents.

    I would start by marking the top left node Va and the bottom right node Vb=0V (arbitrary choice). Then apply KCL to one of these nodes.

    I prefer to write that in the form..

    I1 +I3 -I2 = 0

    Hint: You can write equations for each current in terms of Va.
  14. Mar 19, 2013 #13
    I labeled top right resistor R1, the middle battery E1, middle resistor R2, bottom left battery E2, and bottom resistor R3. I didn't include Va or Vb anywhere because I never understood that stuff
  15. Mar 19, 2013 #14


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    I've redrawn the circuit and labeled up the currents. I made the arbitrary decision that they all flow into node Vb which clearly can't be true...

    Applying KCL to node Vb you get

    IR3 + IR2 + IR1 = 0 ............... (eqn 1)

    What we would like are equations for each of those currents. One way to do that is as follows...

    Looking at the bottom row of the drawing you have the path...

    Node Va --- 6V bat --- R3 ---- Node Vb

    Using Ohms law we can write an equation for IR3 as follows..

    IR3 = (Va - 6V - Vb)/R3

    Since we have choosen Vb = 0 that reduces to

    IR3 = (Va - 6V)/R3 ..................................... (eqn 2)

    Using the same approach you can write..

    IR2 = (Va - -12V)/R2 = (Va+12)/R2 ................. (eqn 3)
    IR1 = Va/R1 ................................................. (eqn 4)

    Now put eqn 2,3,4 into 1...

    (Va - 6V)/R3 + (Va+12)/R2 + Va/R1 = 0

    Substituting values for the resistors that reduces to

    Va = -24/11 Volts

    Then each current can be calculated individually...

    IR1 = Va/R1 = -1.091 A

    IR2 = (Va - -12)/R2 = 2.455 A

    IR3 = (Va - 6)/R3 = -1.364

    As a check add up the currents...

    -1.091 + 2.455 -1.364 = 0

    Note that...

    IR1 is negative so the current in the top is flowing in the oposite direction to my arrow (eg anti clockwise!)

    IR3 is also negative so the current in the bottom loop is flowing in the oposite direction to my arrow (eg clockwise!).

    Perhaps revise the prediction, draw the arrows the other way where necessary and repeat the calculations :-)

    Attached Files:

  16. Mar 19, 2013 #15
    Thank you so much! That actually made sense in the end. Now about the third question, its about the voltage drop across R2. I used the V=IR equation substituting R2 with 4 ohms and I2 with 2.455 A and found that it was 9.82 V. Does that sound right?
  17. Mar 20, 2013 #16


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    It might be :-)

    There is an easy way to check the drop on R2 because I calculated Va and Vb. The voltage drops down that branch must add to Va-Vb.
  18. Mar 20, 2013 #17
    Oh ok that makes sense. Thanks for all of your help. :)
    Last edited: Mar 20, 2013
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