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Circuit analysis

  1. Oct 23, 2008 #1
    i have the following circuit and i'm tryingto calculate I. (the values shown by the resistors are all ohms.) i can do this fine normally, but the addition of a resistor in the middle just complicates things. can anyone help me out? a picture of the crcuit has been attached.

    Attached Files:

  2. jcsd
  3. Oct 23, 2008 #2
    Well, what if there was no 22 ohm resistor? Calculate I, and find the voltages at the nodes where the 22 ohm resistor was. Then put the 22 ohm resistor back in the circuit. Can you see what affect it has now, if any?
  4. Oct 23, 2008 #3
    1 Amp.

    The 22 Ohm has no effect as there is no potential diff across it.

    It is like two 20 Ohm in parallel so 10 Ohms.
  5. Oct 23, 2008 #4
    ahh... is that because of the symmetry of the system? i mean, the resistor has equal current flowing into it in both directions, thus resulting in 0 current?

    how would i solve this if the system wasn't symmetrical? i.e. suppose all the resistors in the above examples were given values of 10 ohms, apart from, say, the resistor on the bottom right -- lets give that a value of 22 ohms.
  6. Oct 23, 2008 #5
    Symmetry yes but there is no current at all.. No V across an R then there is no I.

    You could write down all the equations based on Ohms law for each resisitor given that currents in and out of a node (junction point) must balance and voltages around a loop must add up to zero.

    OR you could transform 3 resistors which form a star (see better if re-drawn a bit) to a delta formation and add that to two remaining resistors.. It then becomes a much easier curcuit to analyse with two simple parallel pairs of resistors in series which you can reduce to a single resistor which is then in parallel with the remaining resistor.


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