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Circuits and automotive battery Question

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  • Thread starter Matt1234
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  • #1
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Homework Statement


A constant current of 3 A for 4 hours is required to charge an automotive battery. If the terminal voltage is 10 + t/2 Volts, where t is in hours, how much energy is expended? (dw)

Homework Equations



i = dq/dt
v= dw/dq (w = work in joules)

p = iv


The Attempt at a Solution



[PLAIN]http://img203.imageshack.us/img203/2080/67548281.jpg [Broken]


please advise
 
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Answers and Replies

  • #2
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answer in the text is 475.2 Kilo Joules, i dont see how they get this, i keep getting 518.4.
 
  • #3
berkeman
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The term in your integral cannot be a constant. You need to account for the fact that the charging voltage has to increase over time in order to keep the charging current at 3A. The power is not a constant 36W...
 
  • #4
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The term in your integral cannot be a constant. You need to account for the fact that the charging voltage has to increase over time in order to keep the charging current at 3A. The power is not a constant 36W...
how do i setup such a question? i understand what your saying but not sure how to set this up
 
  • #5
Zryn
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Your assumption that P = V*I = 12*3 is incorrect, as the voltage changes over the 4 hours.

Trying finding the average voltage (any method will do, but I integrated the voltage over the time period and then divided it by the time period) and then using this number in your calculations.
 
  • #6
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im sorry im confused, i understand now that the voltage varies;
at t = 0 v = 10
at t =4 hours v= 12

i dont know how to setup this integral this is my most recent attempt:

[PLAIN]http://img88.imageshack.us/img88/4723/lastscaneb.jpg [Broken]
 
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  • #7
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Your assumption that P = V*I = 12*3 is incorrect, as the voltage changes over the 4 hours.

Trying finding the average voltage (any method will do, but I integrated the voltage over the time period and then divided it by the time period) and then using this number in your calculations.
following your steps i found the average voltage to be 11 volts.
 
  • #8
berkeman
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Since the voltage is a linear ramp, using the average voltage may work. What answer do you get using the average voltage to get the average power and integrate that?

I'd still do it the full way, with the current equation (versus time) multiplying the constant current inside the integral. The integral is over time after all, and your voltage is V(t).
 
  • #9
Zryn
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That integral you did looks good!

If you multiply the current by the integral of the voltage over 4 hours to get the power over 4 hours (what you just did), what bounds do you need to integrate the power between to get the work done?

The numbers are all there, its just in the understanding.
 
  • #10
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the bounds are the part im having difficulty with because in the formula for voltage time is in hours where as in i = dq/dt time is in seconds so im slightly confused.
 
  • #11
Zryn
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What purpose does doing the integral of voltage have?
 
  • #12
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i took the integral of the voltage to obtain the average voltage.
 
  • #13
Zryn
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No. The integral is the area under the curve i.e. the total summation of voltage over the entire period defined by the bounds of integration.

To get the average voltage, you have to ...
 
  • #14
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divide the total voltage over the time peroid by the upper limit.
 
  • #15
cepheid
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the bounds are the part im having difficulty with because in the formula for voltage time is in hours where as in i = dq/dt time is in seconds so im slightly confused.
Option 1:

Then don't multiply the integral through by the 3 A. Compute the integral (with the limits in units of hours), and get the answer in "volt-hours." Now you can convert the current into units of coulumbs/hr and multiply. Or you can convert the answer in volt-hours into an answer in "volt-seconds" and then multiply directly by the 3 A to get the answer.

Option 2:

Or you can convert *before* integrating. If v(t) = 10 V + t/2, both terms must have units of volts, which means that the constant factor of 1/2 multiplying the t must have units of hours/volt. You could convert this constant into units of seconds/volt, and integrate (also converting the upper limit of integration from 4 hrs into seconds). Then your answer will directly be in volt-seconds, which you can then multiply by 3 A.

Either way, you just have to be more careful about keeping track of your units by explicitly writing them out. And some sort of unit conversion will have to be done somewhere.

EDIT: Why the hell are you guys computing the average voltage anyway? To answer part (b), you need to compute the energy, which requires integrating p(t) with respect to time. Since p(t) is equal to i(t)*v(t), and i(t) is actually constant, you must integrate v(t) with respect to time (with the result of the integral being multiplied by the constant current value).

EDIT 2: Oh I see. The answer to my question is in what berkeman said at the beginning of post #8. Okay fine, but I agree with him that doing that is not any easier than computing the integral directly (since you still have to do that anyway!)
 
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  • #16
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i understand it now.

I appreciate your help Zryn, cepheid and berkeman. Thanks.
 
  • #17
Zryn
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EDIT: Why the hell are you guys computing the average voltage anyway? To answer part (b), you need to compute the energy, which requires integrating p(t) with respect to time. Since p(t) is equal to i(t)*v(t), and i(t) is actually constant, you must integrate v(t) with respect to time (with the result of the integral being multiplied by the constant current value).
What happens when you integrate v(t) with respect to time, multiply it with i(t) to get p(t), then integrate it over incorrect bounds to get work? You get the wrong answer, but also hopefully an explanation and understanding of where you went wrong.
 
  • #18
cepheid
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What happens when you integrate v(t) with respect to time, multiply it with i(t) to get p(t), then integrate it over incorrect bounds to get work? You get the wrong answer, but also hopefully an explanation and understanding of where you went wrong.
First of all, this thread is now over. Second of all, based on your quoted description, I don't think you understand what I am suggesting. I am not suggesting integrating the voltage, multiplying it by i, and then integrating *that* a second time, because that would just be completely wrong. HERE is what I was suggesting:

[tex] i(t) = I = \textrm{const} [/tex]

[tex] E = \int p(t)\,dt = \int i(t)v(t)\,dt = I\int v(t) dt [/tex]

You integrate v(t) once and you have your answer (at which point you just have to be careful about units when plugging in the numbers). It's simple and it doesn't require the EXTRA step of computing the average voltage, which accomplishes nothing.

Third of all, I wouldn't have used the incorrect bounds, and I very carefully instructed the OP on how to avoid getting the units wrong in my previous post.
 
  • #19
Zryn
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...based on your quoted description, I don't think you understand what I am suggesting. I am not suggesting integrating the voltage, multiplying it by i, and then integrating *that* a second time, because that would just be completely wrong...
integrating the voltage --> int 10 + t/2 from 0 to 4 = 4*[10t + t^2/4] = 44
multiplying it by i --> 44 * 3 = 132
integrating *that* a second time --> int 132 from 0 to 3600 = 3600*[132] = 475.2kJ

There was only a very slight problem with the OP's math and I was trying to work through it in a way that would (hopefully) show the problem, rather than say what the problem is and risk having the OP not actually understand, and then make the same mistake again, which seems to be quite common when people say they understand.

You integrate v(t) once and you have your answer (at which point you just have to be careful about units when plugging in the numbers). It's simple and it doesn't require the EXTRA step of computing the average voltage, which accomplishes nothing.
Well, the OP made a minor slip up with units and plugging in numbers, because it's not simple. Taking the extra steps to look at it in depth (hopefully) showed the problem, and hopefully it won't be made again. This accomplishes something, I believe.

I found when I was in university, that If I did the quick and easy math and made a simple mistake, I wouldn't get as many marks as compared to longer versions where the markers could see a simple mistake and the propagation of errors.

I also think that sometimes people who have been doing things for a long time can overlook the difficulty of doing things when you haven't mastered them, and it is generally better to do math with examples showing excruciating detail, rather than make the assumption that someone knows something and have them state at the end that they understand, because of terms such as 'its simple', 'its obvious' or 'its easy'.

So while this thread is over, please accept that there are many routes to the final goal, and doing things faster is not necessarily always the best thing, especially if you don't understand what is going on (I certainly don't most of the time :smile:).
 
  • #20
cepheid
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integrating the voltage --> int 10 + t/2 from 0 to 4 = 4*[10t + t^2/4] = 44
I agree that the integral gives you an answer of 44 Vh, but I don't think that this factor of 4 should be out front.

multiplying it by i --> 44 * 3 = 132
Okay, so you've multiplied 44 Vh * 3 A to give you a result of 132 Wh. This is already in units of energy. This is the answer. If you were to integrate it with respect to time a second time, that would be wrong, because your result would then have dimensions of energy*time.

integrating *that* a second time --> int 132 from 0 to 3600 = 3600*[132] = 475.2kJ
For some reason you are calling this integration, but it is NOT integration. It is merely unit conversion. What you have done is to take the 132 Wh and multiply it by 3600 s/h to get a result in Ws, also known as joules. This step is correct, which is why it leads to the correct answer, but it is certainly not integration.

I agree with everything you said in the rest of your post about how it is important to be a little more explicit with the steps of a solution when helping somebody who is just learning. However, the way in which you described those steps has not been correct in some cases.
 
  • #21
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guys thank you both so much, i wanted to prove to both of you because of your efforts I DO understand the problem. here is proof:

[PLAIN]http://img138.imageshack.us/img138/9182/lastscancs.jpg [Broken]

when i first did the problem i had a hunch that the units were not quite right, and didnt know how to handle them, now i do.

thanks again for both your efforts.
 
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  • #22
Zryn
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For some reason you are calling this integration, but it is NOT integration. It is merely unit conversion. What you have done is to take the 132 Wh and multiply it by 3600 s/h to get a result in Ws, also known as joules. This step is correct, which is why it leads to the correct answer, but it is certainly not integration
Too true, you're right.

I was mistakenly doing the second integral due to thinking I needed to integrate the function to get the work, but I had already done the work in finding the function :smile:.

...guys thank you both so much, i wanted to prove to both of you because of your efforts I DO understand the problem ...
Awesome!
 
  • #23
cepheid
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guys thank you both so much,
That's great, I'm really glad that you solved the problem and now fully understand the method. Nice work converting the current to units of coulombs/hr btw.

Zryn, sorry if I was a bit abrasive. You obviously helped the OP a lot, and everybody makes mistakes. Your suggesting an alternate method that delved into the mechanics of the problem a bit deeper probably did lead to greater understanding.
 

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