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ganeshie8

  • 2 years ago

Imagine a battery shorted with a very long conducting wire. Q : if the wire is uniform through out the loop, can i use V = E.dl directly (to know potential drop between two points) ? If yes, how can we extend the same concept to a loop containing few resistors in series ? I am stuck in Lec.9... pls help..

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  1. quarkine
    • 2 years ago
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    use the fact that I=constant for different R 's so E will vary with R directly for all R's(resistors)..

  2. ganeshie8
    • 2 years ago
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    not clear.. you mean loop length is constant ? and, equation V = E.dl doesnt say E varies with R... can you pls explain some more... ?

  3. quarkine
    • 2 years ago
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    i quote "Q : if the wire is uniform through out the loop, can i use V = E.dl directly (to know potential drop between two points) ?" ans : yes because the resistance is uniformly distributed. for discrete resistances in series the voltage is distributed between the resistances as the conducting wire has negligible resistance.. |dw:1339526221153:dw| since same current through both the resistors,therefore the V around each is in inverse ratio of the resistance values of the resistors.

  4. nick67
    • 2 years ago
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    If you have a uniform wire with circular section you can use the formula: \[R = \rho * L / S\] where R is the resistance in Ohm, p is the resistivity in Ohm*m, L is the wire length in meters, S is the wire section in meters^2. This way the total resistance in a uniform wire is directly proportional to its length.

  5. ganeshie8
    • 2 years ago
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    quarkine, for discrete resistances i have difficulty in understanding how voltage is distributed between the resistances only. you jumped directly to the resistance/voltage relation. still i dont see connection between V = E.dl & ohms law :( nick67 : i used lot of ohm's law & KCL/KVL in DC circuit analysis.. but i never developed good intuition of it. i understand collisions with atoms and energy losing stuff. but thinking of how potential difference develops across resistor is scary..

  6. locnguyen
    • 2 years ago
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    i think 2 part of your question are not related because resistance have nothing to do in the equation \[V = E \Delta l\] You can only derive Ohm's law with the help of this equation, and beside, the resulting Ohm's law is true for the wire, but cannot apply for resistors, since they normally have fixed values produced. the second part of the question is just simly Ohm's law.

  7. quarkine
    • 2 years ago
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    |dw:1339610745648:dw| here let R1>R2.. to move a charge from B to A you have to do more work than from C to B.. you cant use the formula V = E.dl as it is Wrong and is actually dV=E.dl and cant use dV=E.dl either because E=dV/dl is not a continuous (actually it is not differentiable) function around the loop as the value of E is different in loop's wires and in the resistor (which can be seen from the fact that a loop wire of length dl has different value of dV (almost 0) around its ends than the same length inside a resistor wire (much larger value))

  8. ganeshie8
    • 2 years ago
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    quarkine, this is the the best explanation :) now i understand fully why we cannot use dV = E.dl in series network... really appreciate the help thanks a lot!!

  9. quarkine
    • 2 years ago
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    glad to help :).it was a good question..

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