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Simple voltage problem

Discussion in 'Homework Help' started by Shmurk, Jun 29, 2009.

  1. Shmurk

    Thread Starter Member

    Aug 13, 2007
    I need to find the potential between the node C (junction of R2, R3, and R4) and the ground.
    This is a "basic electricity" problem (without using the node method or something else) but I have two problems.
    Here's what I found:
    let i1 (respectively i2, i3, i4) be the current through the resistor R1 (respectively R2, R3, R4)
    Code ( (Unknown Language)):
    1. i1 = i2
    2. KCL: i2 + i4 = i3
    4. so: i1 + i4 = i3
    but no current is passing through the battery V2, so by applying KVL to the left loop:
    Code ( (Unknown Language)):
    1. V1 = (R1 + R2 + R3) * i1
    (that's my first problem: I hope it's "i1"...)

    My second problem is: I have some intuition that "i4 = 0".
    Is it because no current is passing through V2? so we would have:
    Code ( (Unknown Language)):
    1. i3 = i(through V1) = i1 (???)
    And then, the final part of the solution:
    Code ( (Unknown Language)):
    1. i1 = i2 = i3
    3. V(R3) = R3 * i3 = R3 * (V1 / (R1 + R2 + R3))
    4. and V(C) = V(R3) + V2
    Is it correct according to you?
  2. steveb

    Senior Member

    Jul 3, 2008
    You have good intuition and you basically have it right, but you need to be careful with the precise order of your logic and the correct reasons for your shortcuts.

    You correctly noted that the i4 is zero and there is no current in V2. This should be your first step and you should justify it by noting that there is no return path to ground (hence V2 has no current) and V3 has no return path that goes through R4 or ground (hence i4 is zero). Note that if any part of the right hand circuit had a gound in it, you wouldn't be able to make these simplifications.

    One you make those statements the answer becomes trivial. Your statement that i1=i2=i3=V1/(R1+R2+R3) becomes clear. Then establishing the value of Vc is straightforward as you noted.

    This is a great example where a little thinking ahead of time greatly simplifies the problem. Try it by brute force with no forethought, and you get a big messy equation that takes a lot more time to solve.
    Last edited: Jun 29, 2009
  3. Shmurk

    Thread Starter Member

    Aug 13, 2007
    Thank you for the answer, it's clear, and you're awesome!
    That's what I "thought" without having a real explanation for it.
    That, I understand. It would add one big loop including all the voltage sources at once.

    Well, I'll keep on reading, thank you again!