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How do I find voltage?

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi I'm trying to solve the following using mesh current method, instead of node voltage.

    For this I decided to use supermesh.

    I made loop 1 the leftmost loop, and loop 2 combination of the center and right loop.

    -150 + 20I1 + 80(I1-I2) = 0
    40I2 = 4I2 = 80 (I2-I1) = 0

    solving for this gives:

    I1 = 3.10 A
    I2 = 2 A


    I'm supposed to find v_1 and v_2. So since to find v_1, I do the following

    3.10 - 2 = 1.1
    V=IR = (1.1 A)(80 ohms) = 88 V <--------seems about right.

    So now I'm trying to find v_2.

    The only equations, I'm familiar is p = vi , p = i^2R v = ir. and since the current is given of 11.25 A for that portion.
    V_2 = (11.25)(R)

    I don't know how to proceed. I'm thinking about applying KVL, since voltage across a loop is 0.

    which would be on second loop

    -80(1.1 A)+40(2 A) + v_2 = 0
    v_2 = 8 V <-----------I'm not sure if this is right, or if anything I did is right. :(

    Circuit: http://imgur.com/LLno5.jpg
     

    Attached Files:

    Last edited: Oct 10, 2009
  2. jcsd
  3. Oct 10, 2009 #2
    V2 is the voltage across the current source. Ask yourself if there are other elements in the circuit that must also have the same drop in potential across them. In other words, are there elements connected in parallel with the current source? Hope that helps :)
     
  4. Oct 11, 2009 #3
    I'm not sure I understand. I see that the 5ohm resistor is in parallel to the 3 A current source.
     
  5. Oct 11, 2009 #4
    I'm guessing you meant the 4ohm resistor and not 5ohm since I don't see one in the circuit. The resistor in parallel with the current source has the same voltage across it as the current source. So if you know the current through that resistor, you can calculate the voltage drop (v=ir), which is the same voltage as V2.
     
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