Open circuit voltage & EMF help

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SUMMARY

The discussion focuses on calculating open circuit voltage and current for nickel cadmium cells in both series and parallel configurations. For six 1.2-volt cells in series, the open circuit voltage is confirmed to be 7.2 volts, with a current of 0.72A flowing through a 10-ohm load. When the cells are reconfigured in parallel, the open circuit voltage remains at 1.2 volts, with 0.12A flowing through each cell. Additionally, the internal resistance of a carbon cell is determined using the equation R(internal) = (E - V)/I, where E is the open circuit voltage and V is the voltage under load.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Knowledge of series and parallel circuit configurations
  • Familiarity with internal resistance concepts
  • Basic algebra for manipulating equations
NEXT STEPS
  • Research the implications of internal resistance on battery performance
  • Learn about the characteristics of nickel cadmium cells
  • Study the differences between series and parallel circuits in detail
  • Explore advanced battery management systems and their applications
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Students studying electrical engineering, hobbyists working with batteries, and professionals involved in battery technology and circuit design.

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Any help appreciated with these!

Homework Statement

1. A battery is made up from a series combination of six 1.2 volt nickel cadmium cells supplying a 10ohm load-

a) What is the open circuit voltage?
b) Ignoring internal resistance, what is the current flowing in each cell?

2. The nickel cadmium cells in Q1 are reconnected in parallel and are connected to the same load-

a) What is open circuit voltage?
b) Ignoring internal resistance, what is the current flowing in each cell?

3. Determine the internal resistance of a carbon cell if the open circuit voltage is 1.70 volts and when a 10 ohm resistor is connected across the battery terminals the voltage falls to 1.26 volts.Relevant equations

I=V/R
V=IR

The attempt at a solution

1a- Do i simply use V=IR which would be 6x1.2 = 7.2 V? Is that the correct answer?

1b- Using I=V/R, is it simply 7.2/10 = 0.72A (current doesn't change through a resistor in a series circuit right?)?

2a- Do the voltages of each cell coalesce within the parallel circuit or is the total voltage simply V=IR = 1.2V?

2b- Since I(current) isn't the same through each resistor in a parallel circuit, would it be 0.12A flowing through each cell?

3- I'm really not sure how to solve for this. Transposing the E(emf)= V - IR(internal) equation, I get R(internal) = (E - V)/I. But I'm not sure which value (ie 1.26 or 1.7V) goes where or how to solve this problem.

Any help is very much appreciated! Cheers
 
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1a) 7.2 is correct, but you don't need to use V=IR...
1b) yes and yes (you can *kind* of think about it as the current having only one route to take. There are better explanations though).

2a) yes. (ask yourself what the voltage would be if one battery where to stop working?)
2b) yes.

3) try http://en.wikipedia.org/wiki/Internal_resistance"
 
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