How to Solve This Complex Current Electricity Circuit Problem?

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SUMMARY

The discussion centers on solving a complex current electricity circuit problem involving multiple batteries and resistances. Participants clarify the correct application of the current formula, emphasizing the need for proper parentheses in the equation. The final equation derived for current (i) is i = {2n - N} / N, indicating the relationship between the number of batteries and total resistance. The conversation concludes with an acknowledgment of the peculiar nature of the circuit where potential differences are zero, yet current can still flow.

PREREQUISITES
  • Understanding of Ohm's Law and its application in circuit analysis
  • Familiarity with the concepts of electromotive force (emf) and internal resistance
  • Knowledge of circuit components such as batteries and resistors
  • Ability to manipulate algebraic equations involving electrical parameters
NEXT STEPS
  • Study the implications of internal resistance in battery circuits
  • Learn about Kirchhoff's laws for circuit analysis
  • Explore advanced circuit analysis techniques using Thevenin's and Norton's theorems
  • Investigate the behavior of circuits with short-circuited components
USEFUL FOR

Electrical engineering students, circuit designers, and anyone interested in understanding complex electrical circuits and their behavior in practical applications.

coldblood
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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.
 
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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 said:
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?
 
coldblood said:
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:
 
NascentOxygen said:
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?
 
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:
coldblood said:
in the equation, i = [E1 + E2 +------+En] [strike]- [E1 + E2 +------+E N-n][/color][/strike][/color] / R1 + R2 + ------- + RN
total emf / total resistance

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

Special Thanks To
NascentOxygen and ehild
 
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
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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