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Help using superposition to solve (simple?) circuit

  1. Mar 30, 2008 #1
    1. The problem statement, all variables and given/known data
    We're learning about superposition in class right now and I'm having some trouble visualizing how to make simplifications using series/parallel connections. I attached the picture of the circuit and we're being asked to solve for the voltages and currents through the resistors.

    2. Relevant equations
    I understand the superposition idea, we have to replace each voltage source with a short and each current source with a open circuit. So for this particular example, I would first replace the 7V source with a short, then solve for all voltages/currents, then set the 28V source to a short then do the same, then algebraically add the results for the final answer. I'm having trouble doing this though.

    For example, I first replace the 7V source with a short. I then have the 28V source, then the 3 resistors, would I be able to use the voltage divider rule to get the voltages through the 3 resistors, or should I use Kirchoff's laws? Please help and thanks in advance.

    Attached Files:

  2. jcsd
  3. Mar 31, 2008 #2


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    The simplest way to solve it is using the voltage divider rule, remembering of course that R2 and R3 are in parallel.
    But if you use Kirchoff's laws you will arrive to same result.
  4. Apr 8, 2008 #3
    Yeah, use the voltage divider rule...

    You'd get VR1 = [28v (4)] / [4 + (R2//R3)]

    then I through R1 = VR1/R1....


    VR2 = 28v - VR1

    then VR2 = VR3 since they are in parallel....

    and I of R2 is simply VR2/R2

    and I of R3 is VR3/R3...

    same goes when you use the 7v...

    then add I of R1 when you used 28v an I of R1 when you used the 7v....
    then you'd get the I through R1 when both of batteries are working...

    then do the same for others.
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