
#1
Jan1611, 07:34 PM

P: 14

1. The problem statement, all variables and given/known data
A toaster, which uses 6 alkaline D batteries, must produce at least 900 W of thermal energy. D cell batteries have a capacity of about 5 Ah. How long can the batteries run the toaster? 2. Relevant equations 3. The attempt at a solution Well I know that Power is (J/s) and that ah can be converted to Coloumbs. However when I solve out 900(J/s)/10,800C, I get .00833V/s. So how do I finish this problem and solve for time? I feel like I'm close to the answer, but I can't piece together the last step. I would greatly appreciate any help. Thanks! 



#2
Jan1611, 09:03 PM

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You've been provided with a total amount of battery charge. To figure out how long that charge will last, you need to know the rate at which that charge is "used up." So answer me the following questions:
1. What is the quantity that is a measure of the rate of flow of charge called? 2. How can you compute the quantity in part 1 from the power consumption, which is given (hint: you'll need to know something about 6 D cell batteries in series). 



#3
Jan1711, 01:27 PM

P: 14

1. The rate of flow is the current, is it not? This is the change in charge over the change in time.
2. The D cells in a series all have the same current, but add together to give the total voltage and resistance. After looking at the units attached to the different quantities, do I need to use resistance at all? Or current? Sorry, I feel like all the pieces are here, I just don't know how to fit them together. 



#4
Jan1711, 01:51 PM

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Electricity of a Toaster
For #1, yes, current is what I meant.
For #2, let me put it this way. You know the power. You need to know the current. What is the relationship between the power consumed by the load and the current flowing across the load? Hint: how much energy is lost per unit of charge in going across the load and what is this called? 



#5
Jan1711, 01:56 PM

P: 14

I think this makes sense now. Are you referring to the fact that Resistance is equal to potential difference/current?




#6
Jan1711, 02:15 PM

P: 14

Would it be 30 hours? As in 900W/30AH=30H? Or am I missing it here?




#7
Jan1711, 09:26 PM

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This is the relationship between power and current. With what I said in the next quote, I was trying to steer you towards this relationship by explaining why this relationship is true: Now, a Dcell battery has a voltage of 1.5 V. So, six of them in series would have a total voltage across them of 9 V (this is what I meant in my first post when I said that you would need to know something about Dcell batteries). Now you can calculate the current from I = P/V. Once you know the current, you can combine it in a certain way with the total amount of charge to figure out how long that charge will last. (Think about how the units have to work out). 



#8
Jan1711, 10:16 PM

P: 14

Many thanks for this, it has made the problem much more clear. Using the formula, I obtained I=100A, which is the necessary current that the batteries must supply to the toaster. Since each battery has a capacitance of 5ah, I believe I need to divide the current by the capacitance. This would yield 100A/5ah, which would mean that the batteries could supply 20 hours of power to the toaster.
One lingering question though (assuming I did this correctly that is), why wouldn't I multiply the batteries capacitance together and then divide by that number? Thanks so much, you've been a great help! 



#9
Jan1711, 10:19 PM

P: 14

Wait after further reflection I think I've done this backwards. I should have divided the 5AH by the 100A, giving me .05H and then multiplied this by 6 to get .3 or 18 minutes. Is this the right way?




#10
Jan1711, 10:41 PM

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I'm afraid that there are numerous problems with this:
[tex] \frac{100~\textrm{A}}{5~\textrm{Ah}} = 20~\textrm{h}^{1} [/tex] Remember how I said that you would know which was the right way to combine the charge and the current because of how the units would have to work out. Well, you got an answer in units of hours^{1}, so obviously dividing the current by the charge was not the right way to do it! Even without paying attention to the units, it should be obvious that this answer can't be right. Intuitively, does it make sense that if you draw 100 A (which is a CRAZY amount of current), that 6 small batteries can last almost a full day? It doesn't make any sense, especially since 5 amphours of charge means that if you draw 5 A from the battery, it will last 1 hour (and if you draw 1 A from the battery, it will last 5 hours). So HOW could it possibly last 20 hours at 100 A? 



#11
Jan1811, 12:00 AM

P: 14

Thanks I understand that I messed up. Did you not see my second post though? I realized I needed to decide the 5AH by the 100A which yielded 18 min.




#12
Jan1811, 12:12 AM

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#13
Jan1811, 12:45 AM

P: 14

Right I figured that after your most recent explaination. I think I'm good now. Thanks so much. You were incredibly helpful.



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