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Finding source current of a circuit

  1. Sep 4, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm doing a practice problem and I'm asked to find source voltage and source current of the circuit below.

    2. Relevant equations
    i=v/R
    i1+i2+i3..=0 (KCL)
    v1+v2+v3..=0(KVL)

    3. The attempt at a solution
    So I started by calculating the current for the 2kΩ resistor, which I said 5V/2kΩ=2.5 mA.

    Now I used it on the far right 8kΩ resistor to calculate the voltage, I calculated the voltage to be 8kΩ*2.5mA=20V.

    I want to use both voltages, 5V and 20V, to calculate the KVL for the far right loop.
    So now I have the 20kΩ resistor + 5V +20V should all equal zero, but I'm unsure of which values are positive or negative to finish my KVL loop. How can I know if the voltage is positive or negative with the information I was given?



    So n
     

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  3. Sep 5, 2014 #2

    CWatters

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    No need to formally apply KVL. The voltage drop across the 2+8 combination also appears across the 20k and with same polarity...because they are in parallel.
     
  4. Sep 5, 2014 #3

    CWatters

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    PS: You could add that to the drawing and then check it must be correct using KVL.
     
  5. Sep 5, 2014 #4

    NascentOxygen

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    You are told the polarity of the voltage across the 2k, so that tells you the direction of current in the 2k---which is likewise the direction of current in the 8k. When you add the series voltages it gives the voltage across the 20k, including polarity, so this allows you to determine the current, in magnitude and direction, through the 20k. You can then sum those currents and you have the current, in magnitude and direction, through the 5k. Knowing the current, you can calculate .....
     
  6. Sep 5, 2014 #5
    Okay so I added up the voltages(5V and 20V) and now have 25V in parallel. I used that to calculate the current, 25V/20=1.25mA. I used KCL to say 2.5mA-1.25mA=1.25mA in the top. From here, I'm not sure what the diagonal resistor in the middle represents. Is that in parallel or series? I'm not sure how to use KCL with a loop like that.
     
  7. Sep 5, 2014 #6
    Update:
    I added the currents 1.25mA+2.5mA to get 3.75mA going into the 5kΩ resistor.

    5V*3.75mA=43.75V, which is parallel to 15kΩ.

    43.75V/15kΩ=2.91mA
    43.75V/10kΩ=4.375mA

    Adding the currents...
    3.75mA+2.91mA+4.375mA=11.04mA, is
    11.04mA*10kΩ=110.4V+43.75V=154.2 V vs

    Still confused about reading circuits properly with diagonal resistors, and the polarity of "current" in general.
     
  8. Sep 5, 2014 #7

    NascentOxygen

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    Does your calculator agree with this equation? In an equation, one side must equal the other side.

    It is.

    Equations are supposed to demonstrate equality---equality of one side with the other. You can't just add a bit extra to one side when the mood takes you, yet still show it as an equation!

    Your method looks right, though you may have missed that second 10k resistor. Your cavalier treatment of the humble equals sign needs addressing!
     
  9. Sep 6, 2014 #8

    CWatters

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    Have a think about the difference (if there is any) between a joint/dot and a line on a circuit diagram.

    Then look carefully at that diagonal resistor. Look at the electrical nodes it shares with other resistors. Does it make any difference (electrically) if it's drawn diagonally or vertically :-)
     
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