# Current Electricity Problem

coldblood
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

Staff Emeritus
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

coldblood
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?

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

Use that equation. But do it correctly this time. 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?

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.

ehild

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Staff Emeritus
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

coldblood
Thank you friends,
I have got the answer. I was doing a silly mistake.

Special Thanks To
NascentOxygen and ehild

Vibhor
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 ?

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.

ehild

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