# Current Electricity Problem

1. Jul 14, 2014

### 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

2. Jul 15, 2014

### 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

3. Jul 16, 2014

### coldblood

NascentOxygen:

Please tell me how we'll find the current in this circuit?

4. Jul 16, 2014

### Staff: Mentor

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

Use that equation. But do it correctly this time.

5. Jul 18, 2014

### coldblood

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. Jul 18, 2014

### ehild

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
7. Jul 18, 2014

### Staff: Mentor

total emf / total resistance

As ehild noted, extra brackets are needed around your denominator

8. Jul 18, 2014

### coldblood

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

Special Thanks To
NascentOxygen and ehild

9. Jul 19, 2014

### 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 ?

10. Jul 19, 2014

### ehild

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