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Terminal Voltage Question

  1. Jul 8, 2013 #1
    1. The problem statement, all variables and given/known data

    I've attached the diagrams for both questions

    Question 24
    Determine the terminal voltage of each battery

    Question 25
    a) What is the potential difference between points a and d
    b) What is the terminal voltage of each battery?

    2. Relevant equations

    V = E - Ir

    3. The attempt at a solution

    I was able to solve Q24 and Q25 but only after seeing the answer and changing the signs of my equations so I'm just wondering why this is the case:

    Q24:
    I drew the current is in a clockwise direction
    -I(1.0) + 18V - I(6.6) - 12 - I(2.0) = 0
    I = 0.625A

    For the 18V battery: V = 18 - (0.625 * 1.0) = 17.4V
    For the 12 V battery: V = 12 + (0.625 * 2.0) = 13.3V

    --> The explanation for why the 12V battery is E + Ir is because there is a voltage gain going across the internal resistance from left to right....which I understand

    But for Q25:
    From an earlier example, I1 = -0.858A; I2 = 2.58A; I3 = 1.73A

    a) potential difference Vad = Vd - Va = -I1(30) = -25.7V --> how come you don't use the negative value of current 1, since that was what I calculated from the previous example?

    b) 80V battery: the terminal voltage is the potential difference from point g to e
    45 V battery: the terminal voltage is the potential difference from point d to b

    V(80) = E1 - (I2)r = 80 - (2.58 * 1.0) = 77.4V

    V(45) = E2 - (I3)r = 45 - (1.73 * 1.0) = 43.3V

    So in this case, how come V(45) is not also E2 + (I3)r? Is it because it is going in the same direction as the current (I3) and in the previous question, it is going against the current?
     

    Attached Files:

  2. jcsd
  3. Jul 8, 2013 #2

    mfb

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    Staff: Mentor

    I agree.
    d has certainly a higher potential than a.

    Right, the current direction is different.
     
  4. Jul 8, 2013 #3
    But can you tell me why that is?
     
  5. Jul 8, 2013 #4

    mfb

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    Staff: Mentor

    The current direction (as drawn in the sketch) is an arbitrary definition, you can choose it as you like. If you choose the opposite direction, the current value and all signs for it change. The voltages stay the same.
     
  6. Jul 8, 2013 #5
    So for Q25 part a, I just take the value of I1 as 0.858 and not the negative sign?

    And about the directions:
    for Q24, the 12 V battery is V = E + Ir because it is in the opposite direction of the current BUT
    for Q25, The 45 V battery is V = E - Ir because it is in the same direction as the current?
     
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