Current Electricity Problem

  • Thread starter coldblood
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  • #1
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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.
 

Answers and Replies

  • #2
NascentOxygen
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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
 
  • #3
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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

NascentOxygen:

Please tell me how we'll find the current in this circuit?
 
  • #4
NascentOxygen
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Please tell me how we'll find the current in this circuit?
You have written it correctly ... the first line under your text "Battery,"

Use that equation. But do it correctly this time. :wink:
 
  • #5
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You have written it correctly ... the first line under your text "Battery,"

Use that equation. But do it correctly this time. :wink:

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?
 
  • #6
ehild
Homework Helper
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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:
  • #7
NascentOxygen
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in the equation, i = [E1 + E2 +------+En] [strike]- [E1 + E2 +------+E N-n][/strike] / R1 + R2 + ------- + RN
total emf / total resistance

As ehild noted, extra brackets are needed around your denominator
 
  • #8
133
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Thank you friends,
I have got the answer. I was doing a silly mistake.

Special Thanks To
NascentOxygen and ehild
 
  • #9
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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 ?
 
  • #10
ehild
Homework Helper
15,543
1,914
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:

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