Circuit with two batteries, getting same numbers but different signs

Discussion in 'Homework Help' started by randomstudent, Sep 9, 2014.

  1. randomstudent

    Thread Starter New Member

    Sep 9, 2014
    Please help me out.

    I'm given a circuit with two batteries and I have to find out the currents and voltages.
    I've tried solving it using Kirchhoff's junction and loop laws and Ohm's law, and I've checked it in The numbers are the exact same, but the signs (+ -) are completly switched.

    What have I done wrong, I have no ideea... and redone it 3 times already ...

    Here's the circuit:


    Left battery: 10 v
    Right battery: 20 v
    R1=10 ohms
    R2=20 ohms
    R3=10 ohms
    R4=30 ohms
    R5=50 ohms

    Here's how I got the junctions (j1,j2,j3) and loops (A,B,C):

    j1: I0 - I1 - I2 = 0 => I0= I1 + I2
    j2: I1 - I3 - I4 = 0 => I1= I3 + I4
    j3: I4 + I3 - I5 = 0 => I5= I3 + I4
    (so I1 and I5 are the same)

    Loops (U - voltage, that's how europeans note it, I guess..):
    A: -U1 - U2 + 10 = 0 => U1 + U2 = 10 (Applying Ohm's U=I*R) => I0 + 2*I2 = 1 (I devided everything by 10)
    B: -U3 - 20 + U2 = 0 => -I1 + 2*I2 = 2
    C: -U4 - U5 + 20 = 0 => 3*I3 + 5*I3 = 2 => 8*I3=2 => I3 = 1/4 = 0.25 A

    I'm not going through the algebra because the numbers are correct, it's the signs that are switched (e.g. instead of -0.2 A, says its 0.2 A).

    Here are my results:

    I0 = -0.2 A
    I1 = - 0.8 A
    I2 = 0.6 A
    I3 = 0.25 A
    I4 = - 1.05 A
    I5 = - 0.8 A
    U1 = -2 v
    U2 = 12 v
    U3 = -8 v
    U4 = 7.5 v
    U5 = 12.5 v

    Enlighten me please!
  2. MrAl

    Well-Known Member

    Jun 17, 2014

    A quick check on I2...

    I2 can not be negative because there are only positive voltage sources available. It must be positive. Of course we are assuming this is in the realm of circuit analysis where we use conventional current flow and not actually physics where we might find the use of electron current flow, because using electron current flow the signs of the currents get reversed. The only way to be sure is to check the reference from which this problem came from.

    A quick check using a more direct Nodal Analysis with nodes v1, v2, v3, v4 and we dont need v4 because the circuit to the right of the 20v source does not affect the circuit to the left of the 20v source:
    and since v1=10 and v3=20 we get:
    and solving this single var equation for v2 we get:
    v2=12 volts

    This makes I2=12/20=0.6 amps not -0.6 amps so you got the right answer for that one at least.

    Now since 12v>10v that means I0 will be negative, and you got a negative result so that looks good.

    You should check your other results as well as they may actually be right.

    I did admire the fact that you labeled everything very concisely, and i would only add one tiny detail to your diagram. That is, i would label the voltages across each element with a long arrow where the tip of the arrow points to the most positive assumed voltage and the tail points to the most negative. This helps a little when writing the equations afterward. Thus instead of just a line or curved line, it would be a straight line (or slightly curved line if you prefer) with an arrowhead like:
    Sometimes plus and minus signs are used instead of a voltage arrow, but the plus and minus signs get too confusing while the long arrows are more clear. So in addition to current arrows that show current THROUGH each element you would then also have voltage arrows next to each element that show the voltage ACROSS each element.
    Last edited: Sep 9, 2014
    randomstudent likes this.
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
    j3: I4 + I3 - I5 = 0 => I5= I3 + I4

    I3 is the wrong sign since it must be towards the negative pole of the battery ie away from j3
  4. MrAl

    Well-Known Member

    Jun 17, 2014

    Maybe you could clarify what you are saying there. With the assumed current directions for I3 and I4 it looks like I5 comes out right.
    I think he just wanted to show that I5 was equal to I2, but he did not show all of his work unfortunately.
    I did not need I5 for my quick check so i ignored that part. In fact, i ignored I3 also because of the way the circuit was set up that result is too trivial and not needed for the solution to the left hand side of the 20v source.
  5. studiot

    AAC Fanatic!

    Nov 9, 2007
    Hi Mr Al

    I am not sure what the clockwise circles A, B and C represent to the OP.

    Are they meant to be three maxwell currents circulating in the three identified loops?

    In which case the current from J3 to the negative terminal is (Ib -Ic) since they are in the opposite direction

    However the OP has said he has used KCL at J3. Then the current from J3 to the negative terminal of the 20V battery must be towards the battery and equal to (I5-I3) at that junction.
  6. MrAl

    Well-Known Member

    Jun 17, 2014

    A, B and C, are simply "loops". They can be taken to be loop currents, but they are not used that way in his analysis, they are just taken to be loops.

    The function of the loop is to give us a clear direction to follow so are able to sum the voltages around the loop correctly as in:
    "The sum of the voltages around a closed path is equal to zero".
    Thus for loop A we have -10+U1+U2=0
    using the convention that the tip of a voltage arrow is positive and the tail negative so when we first encounter a tail while following the loop around the path the voltage is negative and when we first encounter a head the voltage is positive, with the assumed branch currents as drawn.
    We then attempt to solve for the BRANCH currents and voltages, not the loop currents. See the attachment for a clearer drawing.
    Since the circuit can be divided in half to the left of j3 and to the left of R2 and we are left with two cut branches only, the current I1 must be equal to I5, so if I1 is negative then I5 must be negative and if I1 positive then I5 must be positive.

    You could solve for the loop currents, but he doesnt seem to want to do that, although that would be a valid approach as well.
    Last edited: Sep 11, 2014
  7. randomstudent

    Thread Starter New Member

    Sep 9, 2014

    Sorry for the late reply, but fortunatly I've successfully passed my examen. Thanks for all your help!