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Capacitors - m having problem calculating the potential difference

  1. Apr 16, 2013 #1
    Q) In the adjacent circuit, find the potential difference across AB
    qxuomb.jpg

    Fine , now i did the charge distribution as follows
    xpyjcm.png

    Now in the first loop (sorry i forgot to mention the points in my image)
    applying kirchoff's loop law
    10 = q1/8 + q/8

    or, q + q1 = 80

    Next in the second loop

    q1/8 + (q - q1)/8 + (q - q1)/8 = 0

    or, 2q - q1 = 0
    or, q1 = 2q

    gives, q = 80/3 uc

    Again in the second loop
    VA - (q - q1)/8 - q1/8 = VB

    => VA - VB = q/8 = 10/3 V
    but answer says its 2V.

    Definitely i m wrong but where???

    Any help is appreciated..!!
     
  2. jcsd
  3. Apr 16, 2013 #2
    I'm not sure I follow the q-q1 arguments... and kirchhoff's loop rule seems to be producing more variables then equations for me, so I would suggest something else.

    1) Ignoring AB for the moment, can you reduce the circuit into a simpler circuit with equivalent capacitance? Clearly draw out the transformation, because you'll need it to find Vab later!

    The following rules might be useful:

    [itex]1/C_{1} + 1/C_{2} = 1/C_{eq}[/itex]
    [itex]C_{1} + C_{2} = C_{eq}[/itex]

    But I'm leaving it to you to know when to use which!
     
  4. Apr 16, 2013 #3
    i don't knw why i always get those suggestions i probably knw.. :P

    dear what i want you to do is check what's wrong in my soln...
    if u cud please point out my mistake, that would help me clear 100 doubts of mine...
    i obviously have an alternate soln.. the one you posted using it...
    but they don't actually work in complex capacitor problems.
    Thank you!!
     
  5. Apr 16, 2013 #4

    gneill

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    Staff: Mentor

    Check your equation for the second loop. It looks like you're going counterclockwise around the loop. Be sure to use the "leading" charge expression (expression for the charge on the first plate encountered on your KVL walk) for each capacitor. Could be you used the wrong one for the first term in your equation...
     
  6. Apr 16, 2013 #5
    i think i m pretty correct in my soln..
    Rather the answer given is wrong...!!
     
  7. Apr 17, 2013 #6

    gneill

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    Staff: Mentor

    "Pretty correct" is not quite the same thing as "correct" :smile:

    The answer should be 2V as advertised.
     
  8. Apr 17, 2013 #7
    yup u r not "pretty right" u r "right" this time and me to.. lol.. :P

    got it finally...

    q + q1 = 80 and 2q = 3q1 and simply, Va - Vb = (q - q1)/8 = 16/8 = 2V.

    Thanks!!
     
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