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Homework Help: Need help on Ohm's Law

  1. Feb 12, 2005 #1
    Well, I'm quite idle in Lunar New year's holidays so I decide to do some exercises on Physics, I got stuck when solving the problem, then I tried with the suggestion on my text book, but still, I don't understand the problem, please help me out! btw, sorry, something has gotten on LaTex or I'm too bad at it, so I just place the raw syntax, sorry!
    Well, we have the figure:
    http://img.photobucket.com/albums/v649/maxpayne_lhp2/5.jpg
    And also, we have:
    [tex]\emf=48V, r(Internam resistance)=0\omega[/tex]
    [tex]R_1=2\onega[/tex]
    [tex]R_2=8\onega\][/tex]
    [tex]R_3=6\onega[/tex]
    [tex]R_4=16\onega[/tex]
    Well, they told me to find out the value of [tex]U_{MN}[/tex] (Well, in my native scientific system taugh in school,[tex] U_{MN} [/tex]stands for the potential diferece between M and N.
    Well, obviously, I found that
    [tex]U_{MN} = U_{MA}+U_{AN}[/tex]
    [tex]=U_{AN}-U_{AM}[/tex]
    Then, I tended to account for the Intensity of the current thru AB via the formula:
    [tex]I=\frac{\emf}{R_{AB}}[/tex]
    Then, I found out that I=8 (A)
    But soon, I saw something wrong and looked up in the suggestion corner, they told me:
    As r=0. we have
    [tex]U_{MN}=\emf.[/tex] Let the current flows thru AMB and ANB from A to B, then:
    [tex]U_{AN}=\frac{U_{AB}}{R_2+R_4} . R_2 = \frac{\emf}{R_2+R_4} . R_2= 16V[/tex]
    Do the same thing with [tex]U_{AM}[/tex]
    Though, I don't really understand what they mean, so please help me out! What did I do wrong?
    Thanks for your time and help!
     
    Last edited: Feb 12, 2005
  2. jcsd
  3. Feb 12, 2005 #2

    Andrew Mason

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    Homework Helper

    There is no resistance between M and N so there is no potential difference. M and N are at the same potential.

    To find the potential difference across R1 and R2, work out resistance for R1 and R2 in parallel: [itex]1/R_{1,2}=1/R_1 + 1/R_2[/itex] and for R3 and R4 in parallel: [itex]1/R_{3,4}=1/R_3 + 1/R_4[/itex]. Add [itex]R_{total}=R_{1,2}+R_{3,4}[/itex]. Then work out the total current using [itex]I = V/R_{total}[/itex]. That will allow you to work out the potential from A to M and from M to B.

    AM
     
  4. Feb 12, 2005 #3
    Ohh, sorry, I'm so clumsy, I didn't mean to draw a straight line connecting M and N, there was nothing connecting them at all. yes, they meant to cross the resistances.
    And that's helped. Thanks!
    BTW, I got difficulty when I need to go down for the next line, what's the syntax in LaTex for it? If you don't have time, please send me a link so that I can refer. Thanks alot!
     
    Last edited: Feb 12, 2005
  5. Feb 12, 2005 #4

    xanthym

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    Science Advisor

    Then the currents I_13 thru R_1 & R_3 and I_24 thru R_2 & R_4 are given by:

    [tex] :(1): \ \ \ \ I_{13} = \frac {E} {R_1 + R_3} [/tex]

    [tex] :(2): \ \ \ \ I_{24} = \frac {E} {R_2 + R_4} [/tex]

    from which you can determine all required voltage drops. The overall equivalent resistance R will then be:

    [tex] :(3): \ \ \ \ R = \frac {E} {I_{13} + I_{24}} = \frac {1} { \frac {1} {R_1 + R_3} + \frac {1} {R_2 + R_4} } [/tex]


    ~~
     
    Last edited: Feb 12, 2005
  6. Feb 12, 2005 #5
    Oh thanks, I got it, it's [tex]U_{MN}=4V[/tex]
    Thanks alot!
     
  7. Feb 12, 2005 #6
    Excuse me, still, I don't know what
    for? Please explain for me.
    Thanks!
    btw, if we figure the another way, as thru R3 and R4, is it the same?
    Thanks
     
  8. Feb 12, 2005 #7

    xanthym

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    The equivalent resistance R is the value of a SINGLE resistor that can REPLACE ALL the resistors R_1, R_2, R_3, and R_4 between your battery. In other words, this resistor R will draw the same current that ALL your other resistors draw TOGETHER when connected to a battery. It's useful to know R in terms of the OVERALL operation of ALL the resistors. However, you still need to use the individual resistor values to determine voltage drops between the individual resistors inside your circuit -- just like you were doing.
     
  9. Feb 12, 2005 #8
    Well, okay, Thanks!
    Thanks for all of your time and help!
     
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