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Find the voltage across AB

  1. Oct 8, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-10-8_20-23-30.png



    2. Relevant equations
    V = IR

    3. The attempt at a solution
    I used nodal analysis.
    Let voltage at central point be Vx
    So
    1) (10 - Vx )/ 2 is current entering from left branch.
    2) Current entering junction from middle branch is 4 Vs
    Vs is current in eqn 1 times 2 ohm . This is (10 - Vx ) /2 * 2
    Multiply this with 4
    and we get current entering from middle branch is 4(10 - Vx)
    3) Current leaving junction from rightmost branch is Vx / 8
    Equation is 1 + 2 = 3
    Which is (10 - Vx)/2 + 4(10 - Vx) = Vx/8
    This gives 9(10 -Vx)/2 = Vx/8
    This gives Vx = 360/37.
    So Vab is half of Vx which is 180/37V.

    I did the same using another method.
    let current through left most 2 ohm resistance be Ia
    Eqn 1 applying KVL in left loop, 10 = 2 Ia - 4 Vs * 2. Negative sign cause we're taking negative direction of current as mention in figure.
    Now Vs = 2 Ia
    So eqn 1 becomes. 10 = 2 Ia - [4 (2 Ia)] * 2
    Solving this we get 10 = -14 Ia.
    So Ia = - 10/14.
    Vs = 2 Ia = -20/14/
    4Vs = - 80/14.
    Current in right loop is sum of currents entering junction = Ia + 4Vs = -10/14 - 80/14 = -90/14.
    Voltage drop at AB is -90/14 * 4 = -360/14.
    This doesn't match answer I got above.
     
    Last edited: Oct 8, 2016
  2. jcsd
  3. Oct 8, 2016 #2

    gneill

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    Staff: Mentor

    Your first solution is correct.

    Your second solution is not correct because you treated the current source as a voltage source. You can't know what the potential difference will be across a current source so KVL doesn't work for that loop. You might have tried a supermesh approach, but it would not be any simpler than the nodal analysis method.
     
  4. Oct 9, 2016 #3
    Thanks for the help. So we cannot measure voltage across a current source?
     
  5. Oct 9, 2016 #4

    gneill

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    Staff: Mentor

    Well, we can measure it once it's in operation. Or deduce it once we have analyzed the circuit so that we can use KVL to find out what it has to be to balance the equations. But we can't tell beforehand what it will be. Remember, a current source will produce whatever potential difference is required to maintain its current into its load. That could be anything, even zero or a negative value.
     
  6. Oct 9, 2016 #5
    So we cannot use KVL in a loop that has a current source as the voltage across it is varying and unknown.
    Till now we thought that if we're using KVL in a loop we just jump over the current source.
     
  7. Oct 9, 2016 #6

    gneill

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    Staff: Mentor

    Well it may not be varying but it will be unknown until you solve for it.
    You probably mean that you employed a supermesh, forming a loop around the current source by amalgamating the loops that it borders? Or, you can simply assign the potential across the current source a variable and work it into your equations that way. But you'll need to add another equation since you need as many equations as you have unknowns.
     
    Last edited: Oct 10, 2016
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