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Homework Help: Kirchhoff's second law having exams soon,

  1. Oct 2, 2007 #1
    kirchhoff's second law....having exams soon, plz help

    1. The problem statement, all variables and given/known data

    i got this circuit, in the attachment below. i was asked to find:
    (1) I1 and I2
    (2) the voltmeter reading (the voltmeter has one terminal connected at the point between the 2 ohm resistor and the 12V battery, say point X, and the other terminal at the point between the 1 ohm resistor and the 3V battery, say point Y)
    (3) the power delivered by the 12 V cell
    (4) total power dissipated as heat by the circuit

    then, comment on the discrepancy between your answers (3) and (4)

    i solved parts (1) to (4)....but i'm not sure of the answers. and i don't understand what the last part asks for??? what discrepancy??

    2. Relevant equations

    Kirchhoff's second law
    P = IV

    3. The attempt at a solution

    (1) for the loop ABEFA

    (sum)E = (sum)IR

    i got the equation:
    10I1 + 8I2 = 12

    for loop CBEDC

    same method, and i got:

    8I1 + 9I2 = 3

    now i solved simultaneously
    I1 = 3.23 A and I2 = -2.54 A

    now, for I2, should i keep the minus sign?? and why i got the minus...i think it has something to do with the position of the two batteries....i took the 12V battery as reference, and hence got the 3V battery in the opposite direction.

    (2) i calculated the potential at X and at Y.

    both Vx and Vy should be zero since the negative terminals are always at a potential of zero.

    therefore pd = 0

    (3) i used the equation P = IV

    i used I1 as the current.

    giving me = approx. 39W

    (4) power dissipated = I^{2}R

    = [(I1)^{2}*2] + [(I1 + I2)^{2}*8] + [(I2)^{2}*1] = approx. 31W

    i can't find any discrepancy, maybe coz i'm too stupid for that??!!

    Attached Files:

    Last edited: Oct 2, 2007
  2. jcsd
  3. Oct 2, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    A minus sign for I2 means the current would be opposite the direction drawn, which means power goes 'into' the 3V battery, rather than out.

    3 * -2.54 ~ 7.6 W, which is about the discrepancy one is finding between 39 W from 12 V battery and 31 W dissipated by resistances.
  4. Oct 2, 2007 #3
    it means that the 12V is supplying about 39W and the circuit is losing 31W of it as heat??
    so only about 8W is useful??

    am i right?? and smething else.....is anyone having enough time to chck the other answers....coz i;m not so sure about them.....

    thank you and thnks astronuc
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