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Combining current sources

  1. Jul 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Combining the current sources of the given circuit. I've attached the circuit diagram below.
    circuit 1.PNG
    3. The attempt at a solution

    Since the sources are in parallel I can just add up the values to get an equivalent source and same for the resistors. I'm curious in the proof of this fact. With two current sources and one resistor, I can see how the sources can be combined. But for cases like this, how does the 36A affect the 20 ohms resistor? How do the currents circulate in the circuit?
  2. jcsd
  3. Jul 3, 2016 #2
    Current will want to flow across the path of less resistance. So a resistor with high resistance will have less current going through it, while there is more current flow across a resistor with less resistance. The only time current will not flow through a resistor is if there is a shorted wire in parallel with the circuit.
    Imagine driving on a 3-lane road. The first lane on the left side is congested, the middle lane has 1-2 cars moving quickly, and the rightmost side is closed due to construction..The movement of the cars can be analogous to current flow in a circuit, and of course, we'd want to take the lane that is moving and isn't congested..
    The 20Ω and 5Ω will have the same voltage drop across them because they're connected in parallel. Each current sources will have current flowing across both resistors until their voltages are same.
    Formula to keep in mind: V=I*R
    Hope this somewhat helps :-p
    Last edited: Jul 3, 2016
  4. Jul 4, 2016 #3


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    Staff: Mentor

    The upper horizontal line is one single node, so to make the arrangement clearer you could redraw the schematic showing that node as a single point, with 5 wires going to it. Likewise for the lower horizontal line.

    Current doesn't accumulate at a node; KCL says that at any node, current in = current out.

    The 6A source pumps 6A into the upper node, the 36A source pumps a further 36A in, while the 12A source pumps 12A out of that upper node. Applying KCL to the upper node tells you how much current must, therefore, be leaving that node via other paths, i.e., through the resistors.

    you could just as correctly call it a 12A current sink
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