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Calculate the Current in each branch of the circuit?

  1. Mar 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the current in each branch of the circuit bellow?
    The Circuit: http://picasaweb.google.com.au/hilly1989/ElectronicsDiagrams#5317672833283799074 [Broken]

    Calculated currents using Multisim:
    http://picasaweb.google.com.au/hilly1989/ElectronicsDiagrams#5317672834037978066 [Broken]
    http://picasaweb.google.com.au/hilly1989/ElectronicsDiagrams#5317672827677160434 [Broken]
    http://picasaweb.google.com.au/hilly1989/ElectronicsDiagrams#5317672823464210754 [Broken]

    2. Relevant equations
    Ohm's Law V=IR
    KVL: Sum of all voltages must = 0
    KCL: Sum of current going in = sum of current going out


    3. The attempt at a solution
    Basically we have to show the math behind the simulated circuits.
    I know i have 8 unknowns in the circuit. And i think i have been able to get 8 independent equations.

    I1=I2+I3
    -12+V5+V2+V3+V4+4=0
    4+V4+V3+V1=0
    V1=8*I3
    V2=3*I1
    V3=5*I2
    V4=I2
    V5=I1

    (Note: I've ignored the Batteries being labeled as V1 and V2)

    We have been taught to obtain equations using Ohm's Law, KVL and KCL.
    I have tried to solve these equation but am unable to.
    Are these equations correct and am i on the right track?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 27, 2009 #2

    tiny-tim

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    Hi hilly1989! :smile:

    Nice diagrams btw! :wink:
    -12 in the second equation is wrong … you missed one of the batteries :redface:
     
  4. Mar 27, 2009 #3
    Thanks for the reply.

    (2) -12+V5+V2+V3+V4+4=0

    I have the second battery at the end of this equation as "+4".
    Should this be a negative - if the my 12v Battery is positive?
     
  5. Mar 28, 2009 #4
    you're gravitating toward mesh analysis, a method that uses kVL to solve for currents in circuits that form loops with mostly batteries (current sources can be handles with a super-mesh).

    To solve the system of equations, try to use your KVL to form a system of 2 equations that are in terms of two currents. Then, using good old plug in chug or an RREF matrix with your calculator, solve for the two currents. Note, you have two unknown currents by this point and two equations. Of course, with two currents, the third can always be obtained.

    If I understand your question, yes. When you write the voltage of a source, you follow the direction of the current. The sign you hit first is the sign of the voltage if the voltages are positive in the diagram. Reversing the direction of the current would reverse the sign of every voltage source that relies on that current during the construction of KVL equations. This is because of the passive sign convention. Current comes out the positive of a source and into the positive of a resistor. If your battery is 5V but your current is entering the the positive terminal, you need to reverse the terminals to satisfy the convention, which makes the voltage negative of whatever it was. So the voltage would be -5 * the current. If the voltage was in that same initial polarity but labeled -5, to satisfy the convention you would flip the terminals and make it 5v.

    a simple rule to follow: make all voltage sources positive numbers. Don't forget to change the terminals appropriately. (changing -5 to 5 would reverse the + and - sing on the diagram). Then, whatever sign you hit first as you follow the current will be the sign.


    Also, your second equation (-12+V5+V2+V3+V4+4=0) is correct.
     
    Last edited: Mar 28, 2009
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