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Find total power in a circuit with two power supplies

  1. Nov 22, 2016 #1
    1. The problem statement, all variables and given/known data

    (a) Apply Nodal Analysis, writing down the equations which determine VA and VB, the voltages at nodes A and B [8]
    (b) Find values for VA and VB. [4]
    (c) By calculating the currents supplied by the 44 V and 2 V sources, or by any other means, calculate the total power dissipated by this circuit.
    circuit.png


    2. Relevant equations


    3. The attempt at a solution
    (a)

    (Va-44)/4 + (Va- 0)/6 + (Va-VB)/1 = 0 ..... Eq 1
    (Vb -Va)/1 + (Vb -0)/3 + (VB-2)/2 = 0 ..... Eq 2

    (b)

    Eq 1 ...
    x 12
    3Va -132 + 2Va +12Va - 12 Vb = 0
    17 Va - 12Vb = 132

    Eq 2 ... x6
    6Vb - 6Va + 2Vb + 3 Vb -6 = 0
    11Vb - 6Va = 6

    Eq 1 ... x 11
    187Va - 132 Vb = 1452
    Eq 1 ... x 12
    -72 Va + 132 Vb = 66

    115Va = 1518
    Va = 13.2

    =>
    Vb = 7.7 aprox

    How do I go about doing (c)?
     
  2. jcsd
  3. Nov 22, 2016 #2

    gneill

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    Staff: Mentor

    Note the polarity of the 2 V source.
     
  4. Nov 22, 2016 #3
    Ok. So, should my (Vb - 2)/2 be (Vb +2)/2 ?
     
  5. Nov 22, 2016 #4

    gneill

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    Yes.
     
  6. Nov 22, 2016 #5
    Ok, thank you for pointing that out. So Va = 12.052 and Vb = 6.028 (approximately)

    How do I go about doing C?
     
  7. Nov 22, 2016 #6

    gneill

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    Or, 12 V and 6 V exactly :smile:
    Where is the power coming from? Where is it going?
     
  8. Nov 22, 2016 #7
    Haha, that's my problem. We haven't done to much on power in my lectures yet.

    I'm assuming the power flows from the two voltage sources and is used up by the resistors as per joules law.

    I'm just not sure how to get the current produced by each voltage source....

    Do you calculate the voltage drop from 44V -> A and then divide by the resistance of 4 ?
    And do something similar for the 2V source?
     
  9. Nov 22, 2016 #8

    gneill

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    Correct.
    Yup. In fact, that's very much like what you did to write the node equations, right? They're just KCL, summing the branch currents.
     
  10. Nov 22, 2016 #9
    Ok. Cool. So the current in produced by 44V source is 8A and the 2V is 2A?

    To get the total power dissipated by the circuit, do I get the total resistance in the circuit and multiply it by the sum of the two currents?
     
  11. Nov 22, 2016 #10

    gneill

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    One of those currents is correct :wink:
    How would you find this "total resistance" when you have two sources? I suppose you could use the superposition theorem and do all the work twice...too much work for me! Or, you'd need the individual currents in every resistor. Again more work than is necessary.

    Once you have the correct currents of the voltage sources, figure out how much power they each provide. That's where all the power is coming from, after all.
     
  12. Nov 22, 2016 #11
    Ok, I think the 44V current is definitely correct. Is the 2V source -2A?

    Ah, so I just do P=VI for both of them and sum them together?
     
  13. Nov 22, 2016 #12

    gneill

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    Staff: Mentor

     
  14. Nov 22, 2016 #13
    Ah sugar. I forgot about that. So is it (6+2)/2 = 4A?
    (We didn't talk about polarity too much in lectures)
     
  15. Nov 22, 2016 #14

    gneill

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    Yup. Always take a second to check polarities before you start writing equations. Make it a habit. They like to slip a reversed polarity in every once in a while to see if you're paying attention :wink: :smile:
     
  16. Nov 22, 2016 #15
    Thanks. I always make silly little mistakes like that. So, the total power dissipated by the circuit is 44(8) +2(4) = 360 Watts.

    Thank you so much for your hep. My university doesn't post the solutions to past papers (which is where that question is from). It can be quiet annoying when you're just not quiet sure of an answer. So this has been a great help. I'll be sure to tell my class mates about this website :smile:
     
  17. Nov 22, 2016 #16

    gneill

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    You're welcome. Good luck in your studies!
     
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