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Homework Help: 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.




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


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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

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


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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

    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)]

    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


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    Homework Helper

    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.

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


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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


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    Homework Helper

    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.

    Last edited: Jul 19, 2014
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