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Find power dissipated in circuit with a ground

  1. Jul 3, 2015 #1
    1. The problem statement, all variables and given/known data
    1. Find the power dissipated by 12 ohm resistor
    2. What is the potential at points a,b,c,d?

    2. Relevant equations
    P=I^2R
    I=V/R

    3. The attempt at a solution
    For a) 3v-IR-6v-IR=0
    9v=I(18)
    I=0.5
    P=0.5^2 * 12 = 3 watts
    But not sure if the ground changes anything

    b) unsure if ground changes anything
     

    Attached Files:

  2. jcsd
  3. Jul 3, 2015 #2
    Hint : What is the voltage btw point b,a and b,c and a,d ? Groud represents 0V
     
  4. Jul 3, 2015 #3
    Between bc is 6v and between ad is 3v? if my current calculation is correct, then is v between ab is 3? But are those the potential differences or the actual potentials? For example does the 6 ohm resistor use up all its potential such that it is 0 at point b and point c gains 6v?
     
  5. Jul 3, 2015 #4
    Between bc is 6v and between ad is 3v? if my current calculation is correct, then is v between ab is 3? But are those the potential differences or the actual potentials? For example does the 6 ohm resistor use up all its potential such that it is 0 at point b and point c gains 6v?
     
  6. Jul 3, 2015 #5
    Edit : for mistakes, Vad = 3V, Va = 0 since it's connected to the ground so Vd = 3, Vbc = 6V just as you said, according to your calculation P = RI^2 = V^2/R = 3, you're almost there !,
     
    Last edited: Jul 3, 2015
  7. Jul 3, 2015 #6

    SammyS

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    screen-shot-2015-07-03-at-5-40-04-pm-png.85508.png
    The ground changes nothing.

    Your solution is correct.
     
  8. Jul 3, 2015 #7

    ehild

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    The "ground" means that the potential is zero at point a. But that does not change the potential differences between two points of the circuit.
     
  9. Jul 4, 2015 #8

    CWatters

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    You got the right answer but this equation is inconsistent.

    +3v implies you are summing the voltages anticlockwise so to be consistent it should be +6v not -6V...

    +3V - (I*12) +6V - (I*6) = 0
     
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