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Follz20
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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
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