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EE Solving a circuit using KCL and KVL

  1. Dec 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Find V0 and i0.
    Untitled.png

    2. Relevant equations
    Total current at each node is = 0.
    Total voltage around each loop is = 0.

    3. The attempt at a solution
    I can't seem to be able to analyze this circuit. That i0/4 is really messing me up. I am unsure how to apply both KCL and KVL on this practice problem. I can solve other circuit problems that have at least two resistors through each loop, but this one doesn't contain two resistors through each loop. A walk-through through this problem would be nice.

    20151225_200305.jpg
     
  2. jcsd
  3. Dec 25, 2015 #2

    NascentOxygen

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    Staff: Mentor

    You have 9A entering the top node, so equate this to the 3 currents leaving it.

    The voltage across the 2Ω resistor convenently gives you an expression for vo .

    http://www.imageshack.com/a/img109/4666/holly1756.gif [Broken]
     
    Last edited by a moderator: May 7, 2017
  4. Dec 25, 2015 #3

    I get: 9A = (5/4)i
    Solving i = 7.2A

    And how do you determine the current passing through the middle loop with two currents by each other?
     
    Last edited by a moderator: May 7, 2017
  5. Dec 26, 2015 #4

    cnh1995

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    Homework Helper

    There are 3 currents leaving the top node which add to 9A. They are io, io/4 and current through the 8Ω resistor, say i.
    Can you write expression for i in terms of io?
    Hint: voltage across the 2Ω resistor is vo.
     
    Last edited: Dec 26, 2015
  6. Dec 26, 2015 #5

    NascentOxygen

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    Staff: Mentor

    It is worth pointing out that the labelling in this problem runs somewhat contrary to convention. It is customary that where a voltage across an element is labelled ##v_o## then the current through that same element will be known as ##i_o## ; but such convention has not been followed in this figure, and---while not wrong---this could cause some confusion.

    http://www.imageshack.com/a/img109/4666/holly1756.gif [Broken]
     
    Last edited by a moderator: May 7, 2017
  7. Dec 26, 2015 #6
    The direction of 8Ω resistor is defined by its voltage sign. You can write all currents with the correct direction.
     
  8. Dec 26, 2015 #7
    Problem Solving
    Ah I think I see it. Correct me if i'm wrong.

    Since the 8Ω element is in parallel with the 2Ω element and the i0/4 element, their voltages must equal each other.

    So we can set the voltage of the 8Ω element equal to the 2Ω so : 2i0 = 8iv0
    solving for iv0, we get iv0 = i0/4.

    Adding up the currents leaving the node: i0/4 + i0/4 + i0 = 9A
    We get i0 = 6A.

    Since we know i0 now, we can solve for v0. So, v0 = i0/4 * 8Ω : giving us v0 = 12V.

    Questions
    1. How did we know the direction of the current passing through the 8Ω element?

    2. This doesn't seem logical to me, but if we can set the voltages of elements equal to each other, can we set the voltage of the i0/4 varying current element equal to the voltage 8Ω element? Or are we only allowed to set voltage producing elements equal to each other when in parallel? Thus, the current element doesn't produce a voltage so it wouldn't be allowed.
     
  9. Dec 26, 2015 #8

    NascentOxygen

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    Staff: Mentor

    Current through a passive element flows from its higher potential end to its lower potential end, so once the + and - signs for voltage are marked on an element this determines the direction you will consider current in that element to be flowing. (It will happen from time to time that this voltage or current turns out to be a negative quantity, this simply means that any initial assumption you made had the direction reversed. This is to be expected, as before you apply any maths it often is not obvious what the polarity will be.)

    All elements in parallel share an identical voltage. There are no exceptions.
    http://www.imageshack.com/a/img109/4666/holly1756.gif [Broken]
     
    Last edited by a moderator: May 7, 2017
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