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Current Electricity Problem

  1. Jul 14, 2014 #1
    Hi friend I am Stuck in a problem. Please help me in solving this. Thank you all in advance.
    The problem is as follows.

    https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-xpf1/t1.0-9/10270329_1576533655907072_4352617381833809587_n.jpg
    https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-xfa1/t1.0-9/q71/s720x720/10500551_1576533745907063_1098072757182948910_n.jpg

    Solution

    https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-xfp1/t1.0-9/q71/s720x720/10524686_1576533929240378_2650154413940968208_n.jpg

    Friends Please help me in solving this. I'll appreciate the help. Thank you all.
     
  2. jcsd
  3. Jul 15, 2014 #2

    NascentOxygen

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    Staff: Mentor

    This is equivalent to a vertical stack of cells with the two ends shorted together. So the current that flows is not given by that equation you wrote, viz.,

    I = ((E1 + E2 + ...) - ( E1 + E2 ... ))/ R
     
  4. Jul 16, 2014 #3
    NascentOxygen:

    Please tell me how we'll find the current in this circuit?
     
  5. Jul 16, 2014 #4

    NascentOxygen

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    Staff: Mentor

    You have written it correctly ... the first line under your text "Battery,"

    Use that equation. But do it correctly this time. :wink:
     
  6. Jul 18, 2014 #5
    in the equation, i = [E1 + E2 +------+En] - [E1 + E2 +------+E N-n] / R1 + R2 + ------- + RN

    here E = αR
    Hence,

    i = [αR1 + αR2 +-------+ αRn] - [αR1 + αR2 +-------+ αR N-n] / [αR1 + αR2 +-------+ αRN]

    typical to solve this, so If I assume all the resistances identical, then

    i = α{[R + R +-------+ R(n times)] - [R + R +-------+ R (N-n times)] / α[R + R +-------+ R(N times)]

    hence,
    i = α{[nR]-[(N-n)R]} / αNR

    i = {n - N + n} R / NR

    i = {2n - N} / N

    Is this wrong. It is giving some value?
     
  7. Jul 18, 2014 #6

    ehild

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    Homework Helper
    Gold Member

    You have a single loop. You wrote correctly somewhere that i= total emf/ total resistance.

    What is the total emf? What is the total resistance? What is the current in terms of α?

    And do not avoid parentheses. What you wrote i = [E1 + E2 +------+En] - [E1 + E2 +------+E N-n] / R1 + R2 + ------- + RN means [E1 + E2 +------+En] - ([E1 + E2 +------+E N-n]/R1) +R2 + ------- + RN , that you subtract Amps from Volts and add Ohms.


    ehild
     
    Last edited: Jul 18, 2014
  8. Jul 18, 2014 #7

    NascentOxygen

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    Staff: Mentor

    total emf / total resistance

    As ehild noted, extra brackets are needed around your denominator
     
  9. Jul 18, 2014 #8
    Thank you friends,
    I have got the answer. I was doing a silly mistake.

    Special Thanks To
    NascentOxygen and ehild
     
  10. Jul 19, 2014 #9
    The current i in the loop = α .

    But this means that potential diference between the terminals of any battery given by V= E -iR = 0 .This also means that potential difference between any two points in the circuit is 0 .

    How would current even flow in the circuit ?
     
  11. Jul 19, 2014 #10

    ehild

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    Gold Member

    Yes, it is a strange circuit.
    All batteries drive current through their internal resistance and all these currents are equal: All batteries are as if short-circuited. But current can flow between two points at the same potential.

    ehild
     
    Last edited: Jul 19, 2014
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