Open circuit voltage & EMF help

In summary, the conversation discusses various aspects of batteries, including open circuit voltage, current flow, and internal resistance. It also presents specific problems for which solutions are sought. The solution to the first problem involves using V=IR to find the open circuit voltage and current flow. In the second problem, the cells are connected in parallel, resulting in a total voltage of 1.2 volts and a current flow of 0.12A in each cell. The third problem involves finding the internal resistance of a carbon cell using the formula R(internal)=(E-V)/I.
  • #1
Follz20
4
0
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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  • #2
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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Related to Open circuit voltage & EMF help

1. What is open circuit voltage?

Open circuit voltage is the voltage difference between the two terminals of an electrical circuit when there is no current flowing through it. It is also known as the potential difference or electromotive force (EMF) of the circuit.

2. How is open circuit voltage different from closed circuit voltage?

Open circuit voltage is measured when there is no current flowing through the circuit, while closed circuit voltage is measured when the circuit is complete and current is flowing through it. Open circuit voltage is typically higher than closed circuit voltage due to the absence of current.

3. What factors affect open circuit voltage?

The main factors that affect open circuit voltage are the type of material used in the circuit, the temperature of the circuit, and the presence of any external electromagnetic fields. These factors can cause fluctuations in the voltage and affect the overall performance of the circuit.

4. How is open circuit voltage measured?

Open circuit voltage can be measured using a voltmeter, which is connected to the two terminals of the circuit. The voltmeter will display the voltage difference between the two terminals, indicating the open circuit voltage.

5. Why is open circuit voltage important?

Open circuit voltage is an important concept in understanding the behavior of electrical circuits. It is used to determine the maximum voltage that a circuit can generate and helps in designing and optimizing circuit components for better performance. It is also a crucial parameter in determining the efficiency of a circuit.

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