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Homework Help: Calculate resistance between terminals 1 and 2, 1 and 4...

  1. Nov 16, 2017 #1
    Screen Shot 2017-11-16 at 21.21.28.png
    Calculate resistance between terminals 1 and 2, 1 and 4, 1 and 3, 2 and 4, 3 and 4, and 2 and 3

    My attempt of solution:
    For R between 1 and 2 i done: (R1 + R2) || (R1 +R2 + R3)

    The problem is:
    Im not sure how to find which is in series and whats in parallel so I'm confused and I think my solution/working is wrong
  2. jcsd
  3. Nov 16, 2017 #2


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    Series resistors share the same current. So start at on end and keep moving along the circuit. If you go through two resistors without going through a node (solid circle in this case), you have two resistors in series. So the 10Ω resistor and the 50 Ω resistor are not in series with anything because when you move past either one, you encounter a node.

    Parallel resistors share the same voltage across them. You need to remember that straight lines in a circuit are equipotentials. Therefore the electric potential is the same along any straight line until you have to cross a circuit element (resistor, capacitor, etc) in which case there is a voltage drop or rise.

    On edit: When you calculate equivalent resistances in the case, you may imagine a battery connected across the points mentioned in the problem. In that case, it may be possible that resistors 10Ω and 50Ω will be in series. I suggest that you redraw the circuit with the battery drawn in so you can see what's going on. (My thanks to @Tom.G for pointing this out.)
    Last edited: Nov 17, 2017
  4. Nov 16, 2017 #3

    The Electrician

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    There is another unlabeled essential node. The junction of R1, R2 and R3 is a node. You could label that node "5" and calculate the various resistances between that node and all the others. Include those in the work you turn in and you could really impress your instructor. :)):wink:
  5. Nov 16, 2017 #4
    Whenever you crisscross wires, like the top lead of R3 and the right lead of R4, you make it confusing to see the actual circuit. I find it easier to work with a redrawn circuit where none of the wires cross each other. like:

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