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Electric Circuit help

  1. Mar 19, 2005 #1
    Electric Circuit help!!

    Ok here's my question:

    For the cirucit shown in the drawing(attached), find the current (I) through the 2.00 ohm resistor and the voltage (V) of the battery to the left of this resistor.

    Ok, should I be using Kirchoff's drop=rise, and junction rules?

    It doesn't seem to work because my voltage is also unknown. Could I possibly add them resistors in series/parallel?

    Any directions appreciated. Thanks :confused:
     

    Attached Files:

  2. jcsd
  3. Mar 19, 2005 #2

    xanthym

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    Your attachment is not downloadable. You need to resubmit.
    Note the following msg above your attachment:
    "Attachments Pending Approval"


    ~~
     
  4. Mar 19, 2005 #3
    Hmm, I can download it without problems. Well here it is again.
     

    Attached Files:

  5. Mar 20, 2005 #4
    Ok for those who can't get it:


    4ohm 8ohm
    |---/\/\/\-----/\/\/\---|
    | ---->3amps |
    |__/\/\/\_______|l____|
    | 6ohm 24V |
    | <--- I=? |
    |___|l________/\/\/\__|
    V=? 2ohm

    That's the best I can do with keyboard lol.
     
  6. Mar 20, 2005 #5
    Darn, picture got messed up. sorry. The I=? is suppose to be far right, so with the 24 V.
     
  7. Mar 20, 2005 #6

    xanthym

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    Which branch (top or middle) does the "3 amps" refer to??
    (Hint: Enclose such diagrams within CODE & /CODE tags to preserve spacing.)

    ~~
     
  8. Mar 20, 2005 #7
    It refers to the top one, and the one you have to find refers to the bottom one, sorry. I'm not too familiar with tags but I'll keep that in mind thanks.
     
  9. Mar 20, 2005 #8

    xanthym

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    Code (Text):


         4ohm       8ohm
    |---/\/\/\-----/\/\/\---|
    |     3amps ---->       |
    |                       |
    |__/\/\/\_______|l______|
    |   6ohm        24V     |
    |                       |
    |            <---- [COLOR=Red]I=?[/COLOR]  |
    |___|l________/\/\/\____|
        [COLOR=Red]V=?[/COLOR]        2ohm
     
    The solution is obtained using Kirchoff's Laws. For this purpose, the following are established:
    {(+) Current Direction in Top Branch} = {Left-to-Right} = IT
    {(+) Current Direction in Middle Branch} = {Right-to-Left} = IM
    {(+) Current Direction in Bottom Branch} = {Right-to-Left} = IB

    We first note the Top Branch has equiv resistance RT=(4 + 8)=(12 Ω) thru which flows (3 amps), thereby indicating a voltage drop (using Ohm's Law) of ΔV=(-)(3 amps)*(12 ohms)=(-36 V). Thus, applying Kirchoff's Laws:
    (1) ---> Voltage Loop Clockwise Top & Middle Branches
    Begin top-left corner (clockwise):
    (-36 V) + (24 V) - (6 ohms)*IM = 0
    ::: ⇒ IM = (-2 amps)

    (2) ---> Current Node Left Side Middle Branch
    Into node is (+):
    IM + IB - IT = 0
    ::: ⇒ (-2 amps) + IB - (3 amps) = 0
    ::: ⇒ IB = (5 amps)

    (3) ---> Voltage Loop Clockwise Top & Bottom Branches
    Begin top-left corner (clockwise):
    (-36 V) - (2 ohms)*IB + VB = 0
    ::: ⇒ (-36 V) - (2 ohms)*(5 amps) + VB = 0
    ::: ⇒ VB = (46 V)



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    Last edited: Mar 20, 2005
  10. Mar 21, 2005 #9
    IM = (-2 amps)

    what does negative currents mean? flow the opposite direction?
     
  11. Mar 21, 2005 #10

    xanthym

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    A "negative current" means the computed current flows in the direction opposite to the direction defined to be the (+) direction when setting up the solution. For the middle branch (to which Im refers), the (+) direction was initially defined to be {Right-to-Left}. Thus, the actual computed current of (-2 amps) will flow {Left-to-Right}. The value of Im=(-2 amps) automatically indicates this and must be used whenever the middle branch current value is required in Kirchoff's Laws.
    (Note: The (+) current direction can initially be defined in any direction that is convenient to solving the problem. However, once defined, this initial definition must always be maintained throughout the solution.)


    ~~
     
    Last edited: Mar 21, 2005
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