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Help with this confusing circuit Using superposition find voltages!

  1. Mar 7, 2013 #1
    Help with this confusing circuit!! Using superposition find voltages!

    The problem is

    "In the following circuit, find the voltages at A,B, and C. Use or confirm with superposition."

    I was able to find the current through the top one ohm resistors as .333A (sorry didnt label on my picture). but what do i do next???


    PICTURE:
    http://imageshack.us/a/img689/7025/circuit3a.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 7, 2013 #2

    mfb

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    Staff: Mentor

    You could try to analyze the lower part, and ignore the upper part for a while (as you know its contribution already). What is the current flowing through the resistor below B?
     
  4. Mar 7, 2013 #3
    There is a little trick here. Actually the circuit is quite simple (in fact, as simple as a circuit can be).

    Hint: If you have two paths in a circuit and one of them has o resistance, all charges will flow trough that path.
     
  5. Mar 7, 2013 #4

    VVS

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    Vvs

    It is very simple. Try using KCL and KVL.

    For example:

    I is flowing through the top resistor and 1A is flowing in the bottom left branch. Applying KCL at node A. I+1A will flow through the 1ohm resistor from Node A to Node B. Multiply that current by the 1 Ohm Resistor will give you the voltage drop across the 1ohm resistor so B=A-(I+1A)*1.

    Continue analysis in a similar fashion for the rest of the circuit.

    hope this helps
     
  6. Mar 7, 2013 #5
    What you say is usually ok, but not for this paticular problem. All the current will flow only trough the resistor at the top of the circuit. As for the bottom part, all the current will flow trough the resistance-free cable.
     
  7. Mar 7, 2013 #6
    sorry I do not understand. Which two paths and which has 0 resistance?
     
  8. Mar 8, 2013 #7

    mfb

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    Staff: Mentor

    Did you check the direction of the 1A source to the left?
    There will be a current flow in the interior of the circuit.
     
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