An uncharged 4.7μF capacitor is charged to a pd of 12v through a 200 ohms resistor and then discharged through a 220 kiloohms resistor, calculate the time taken for the pd to fall from 12v to 3v.
I know theres the 'decay of charge' equation (with the exponential value, e) which I could use to find the time taken for volts to drop, but the textbook here hasn't even got to logarithms yet and it mentions nothing about logarithm in this whole chapter, so my hunch is that there must be another simpler way to do this question. However the page in the book does mention how to find 1% and 37% of the initial value of volts/charge, but obviously 3v from 12v is a different percentage difference.
EDIT: If there really isn't another method other than using logs, then I'm okay with that, since I get the same answer as in the back of my book. I just thought that there would be a simpler method that I've missed out, and I thought it was strange that they give absolutely no hint about logs/ln if they really are expecting me to use this method.

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 04062012 11:49
Last edited by internet tough guy; 04062012 at 13:16. 
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 04062012 12:26
Yes there is an alternative way.
If you know that the "half life" of the charge or pd on a capacitor is t_{½} = CR Ln 2
and that Ln 2 = 0.693 it becomes a simple calculation as you know C and R. In this case going from 12 to 3V is 2 half lives
12 to 6 to 3Last edited by Stonebridge; 04062012 at 16:26. 
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 04062012 12:40
(Original post by Stonebridge)
Yes there is an alternative way.
If you know that the "half life" of the charge or pd on a capacitor is t_{½} = CR Ln 2
and that Ln 2 = 0.639 it becomes a simple calculation as you know C and R. In this case going from 12 to 3V is 2 half lives
12 to 6 to 3
I really don't know what method this question is expecting me to use, because like I said, throughout this whole chapter theres no mention of logs/ln =, nor is there a hint that you should use it, in fact logs/ln are only brought up near the end of the book on the topic of nuclear energy. So I suspect that I'm not suppose to use logs/ln to do this question, and there must be alternative method.
btw the 'decay of charge' equation that I referred to in my OP, is this:
Q = Q0*e^(t/RC)
where, Q = charge its decreased to, Q0 = initial charge, e = exponential, t = time, R = circuit resistance, C = capacitance.
so yeah, when I first looked at the question, I did think about taking logs on both sides because I already know all the values except time, tLast edited by internet tough guy; 04062012 at 12:42. 
Foghorn Leghorn
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 04062012 14:25
Better watch because some marking schemes specifically want you to use a certain method to find the answer.

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 04062012 14:33
(Original post by Foghorn Leghorn)
Better watch because some marking schemes specifically want you to use a certain method to find the answer.
or do you also think that logs/ln is the only way to find the time, I'm starting to think this is the case, spent over an hour thinking and I still can't think of any other methods 
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 04062012 14:37
It's a standard formula in capacitor decay. The "half life" = 0.693 CR
There is normally no simple way of doing these questions, as they involve the exponential.
If this is an exam board question and the exam board specification includes the exponential and the use of logs then that's the way you solve it.
You asked if there was another way. There is. It's what I showed you.
The Log_{e} 2 comes from the log version of the formula by putting the charge as half the original value, and is normally accepted as a standard result.
How does your book explain how to find the time to drop to 1% of the charge without logs?
The "half life" is the time taken for the charge or pd to fall to half.
In this case it falls from 12 to 3.
That is two half lives.
One half life from 12V to 6V and a second one from 6V to 3V
I explained this in my first post. (Now highlighted.)
So you would do this question by writing
t =2x 0.693 CR
and subbing in for C and R
This is the only "simple" way of doing it.
Check if your exam board includes the formula in its spec. If not you have to use logs.
Edited to correct formula highlighted. (Oops!)Last edited by Stonebridge; 04062012 at 16:55. 
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 04062012 15:07
(Original post by internet tough guy)
so any idea how else I could do this question without using logs?
or do you also think that logs/ln is the only way to find the time, I'm starting to think this is the case, spent over an hour thinking and I still can't think of any other methods
Unfortunately I'm not as well versed with this as stonebridge, but he already provided the alternative. The point I was making is even if you used another method and got the answer correct you may not get the marks because they might specifically be looking for you to do it using logs.
Think about what the half life curve of the capacitor would look like on a voltage/time graph, it would seem it's decreasing exponentialy, hence why you would need to show your working using exponentials.

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 04062012 15:15
(Original post by Stonebridge)
It's a standard formula in capacitor decay. The "half life" = 0.639 CR
There is normally no simple way of doing these questions, as they involve the exponential.
If this is an exam board question and the exam board specification includes the exponential and the use of logs then that's the way you solve it.
You asked if there was another way. There is. It's what I showed you.
The Log_{e} 2 comes from the log version of the formula by putting the charge as half the original value, and is normally accepted as a standard result.
How does your book explain how to find the time to drop to 1% of the charge without logs?
The "half life" is the time taken for the charge or pd to fall to half.
In this case it falls from 12 to 3.
That is two half lives.
One half life from 12V to 6V and a second one from 6V to 3V
I explained this in my first post. (Now highlighted.)
So you would do this question by writing
2 t_{½} = 0.639 CR
and subbing in for C and R
This is the only "simple" way of doing it.
Check if your exam board includes the formula in its spec. If not you have to use logs.
You asked about how the book explains how to find the time to drop to 1% of the charge without logs, funny I was just going to ask about this because I'm not clear as to what they mean. Heres what they've written:
The time constant RC is the time taken in seconds for the capacitor to discharge to 37% of its initial charge. Given values of R and C, the time constant can be quickly calculated and used as an approximate measure of how quickly the capacitor discharges. However, as 5RC is the time taken to discharge by over 99%, 5RC is a better 'rule of thumb' estimate of the time taken for the capacitor to effectively discharge. So t = 5RC would give the time taken for the decay to fall to less than 1% of the initial value.
The bit in bold is what really baffles me, how did they come to the conclusion that 5RC gives an approx to 99% discharge?
Anyway, I did try your method, although I'm not familar with it so I may have done something wrong:
2 t_{½} = 0.639 CR
I subbed in C = 4.7μF and R = 220000
so t = (0.639*C*R)/2
I get t = 0.3303s
which is quite some way off the answer at the back of the book, 1.4s 
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 04062012 15:21
(Original post by Foghorn Leghorn)
Unfortunately I'm not as well versed with this as stonebridge, but he already provided the alternative. The point I was making is even if you used another method and got the answer correct you may not get the marks because they might specifically be looking for you to do it using logs.
Think about what the half life curve of the capacitor would look like on a voltage/time graph, it would seem it's decreasing exponentialy, hence why you would need to show your working using exponentials.

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 04062012 15:23
(Original post by internet tough guy)
I'm on AQAA exam board, and this question is just a summary question in the standard issue textbook.
You asked about how the book explains how to find the time to drop to 1% of the charge without logs, funny I was just going to ask about this because I'm not clear as to what they mean. Heres what they've written:
The time constant RC is the time taken in seconds for the capacitor to discharge to 37% of its initial charge. Given values of R and C, the time constant can be quickly calculated and used as an approximate measure of how quickly the capacitor discharges. However, as 5RC is the time taken to discharge by over 99%, 5RC is a better 'rule of thumb' estimate of the time taken for the capacitor to effectively discharge. So t = 5RC would give the time taken for the decay to fall to less than 1% of the initial value.
The bit in bold is what really baffles me, how did they come to the conclusion that 5RC gives an approx to 99% discharge?
Anyway, I did try your method, although I'm not familar with it so I may have done something wrong:
2 t_{½} = 0.639 CR
I subbed in C = 4.7μF and R = 220000
so t = (0.639*C*R)/2
I get t = 0.3303s
which is quite some way off the answer at the back of the book, 1.4s 
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 04062012 15:29
(Original post by Foghorn Leghorn)
Would it not be 2xLN(2)x220kx4.7micro? That gives you 1.4334s.
2 t½ = 0.639 CR
theres no exponentials here, so theres no need to take logs, I subbed in the values of C and R, and just solved for t½, did I make a mistake? 
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 04062012 15:29
(Original post by internet tough guy)
yeah, I don't have a problem with using logs, but like you said if they're looking for another method, then I won't get full marks for using the log/ln method. As I've said, taking ln on both sides and thereby getting rid of exponential, e, gets me the right answer, 1.4334 to be precise which when rounded down would give 1.4s as in the answer at the back. I just find it strange that they make no mention of logs/ln whatsoever in this chapter so I thought I'm suppose to use a different method.Last edited by Foghorn Leghorn; 04062012 at 15:31. 
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 04062012 15:34
(Original post by internet tough guy)
Thats what I did for my method, in the post you quoted me, I was using stonebridge's method in writing:
2 t½ = 0.639 CR
theres no exponentials here, so theres no need to take logs, I subbed in the values of C and R, and just solved for t½, did I make a mistake?
EDIT: also Ln(2)= 0.693Last edited by Foghorn Leghorn; 04062012 at 15:38. 
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 04062012 16:29
(Original post by Foghorn Leghorn)
According to wikibooks, Alevel physics allow the use of t(1/2)=RC Ln(2)
EDIT: also Ln(2)= 0.693
I've amended my earlier posts.
Yes it's 0.693 not 0.639
I also mislead you.
One half life is 0.693 CR so 2 half lives is 2 x 0.693 CRLast edited by Stonebridge; 04062012 at 16:44. 
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 04062012 16:36
(Original post by internet tough guy)
I'm on AQAA exam board, and this question is just a summary question in the standard issue textbook.
You asked about how the book explains how to find the time to drop to 1% of the charge without logs, funny I was just going to ask about this because I'm not clear as to what they mean. Heres what they've written:
The time constant RC is the time taken in seconds for the capacitor to discharge to 37% of its initial charge. Given values of R and C, the time constant can be quickly calculated and used as an approximate measure of how quickly the capacitor discharges. However, as 5RC is the time taken to discharge by over 99%, 5RC is a better 'rule of thumb' estimate of the time taken for the capacitor to effectively discharge. So t = 5RC would give the time taken for the decay to fall to less than 1% of the initial value.
The bit in bold is what really baffles me, how did they come to the conclusion that 5RC gives an approx to 99% discharge?
Anyway, I did try your method, although I'm not familar with it so I may have done something wrong:
2 t_{½} = 0.639 CR
I subbed in C = 4.7μF and R = 220000
so t = (0.639*C*R)/2
I get t = 0.3303s
which is quite some way off the answer at the back of the book, 1.4s
If you take 37% x 37% x 37% x 37% x 37% you get less than 1%
They have just taken an arbitrary value of 1% as the point where the capacitor is "effectively" discharged, and worked out that you need to decrease by 37% a total of 5 times to get just below that value.
The way I mentioned before using the half life will only work when the charge or pd has fallen an exact number of half lives. In this case from 12 to 3 is 2 half lives.
That is the clue to why I suspect you have to do the question this way, rather than calculate with logs. In this case there is an easier way. In most cases there isn't.
Edit
I had the last two digits of the number 0.693 the wrong way round.
You should get the right answer using the right number.
I mislead you with the formula. Apologies.
One half life is 0.693 CR so
two half lives is 2 x 0.693 CR
I've amended my original post (again!)
Apologies for rushing these last few posts and not checking what I wrote.Last edited by Stonebridge; 04062012 at 16:45. 
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 04062012 16:46
(Original post by Stonebridge)
What they've done is to point out that RC, the time constant, is the time taken for the charge to drop to 37% of the original. This comes from the exponential.
If you take 37% x 37% x 37% x 37% x 37% you get less than 1%
They have just taken an arbitrary value of 1% as the point where the capacitor is "effectively" discharged, and worked out that you need to decrease by 37% a total of 5 times to get just below that value.
The way I mentioned before using the half life will only work when the charge or pd has fallen an exact number of half lives. In this case from 12 to 3 is 2 half lives.
That is the clue to why I suspect you have to do the question this way, rather than calculate with logs. In this case there is an easier way. In most cases there isn't.
Edit
I had the last two digits of the number 0.693 the wrong way round.
You should get the right answer using the right number.
that formula is derived from the one you gave me: 2 t½ = 0.639 CR
if I multiply (0.693*C*R) by 2 instead of dividing it by two, then I get the right answer, the same as the one I got using logs. As it stands (dividing by 2), I get a similarly incorrect answer to before. 
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 04062012 16:51
To summarise, as I made a typo and another error in my posts through rushing and not checking.
The pd falls from 12 to 3V which is 2 "half lives".
This is the clue to the solution using the easier formula.
Use the fact that the half life time = 0.693 CR
(0.693 = Ln 2)
In this case we have 2 half lives so the time taken will be 2 x 0.693 CR 
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 04062012 16:52
(Original post by internet tough guy)
Ok, but could you check if this is right: t = (0.693*C*R)/2
that formula is derived from the one you gave me: 2 t½ = 0.639 CR
if I multiply (0.693*C*R) by 2 instead of dividing it by two, then I get the right answer, the same as the one I got using logs. As it stands (dividing by 2), I get a similarly incorrect answer to before.
Sorry I made a mistake through rushing the reply. Yes multiply by 2 for the 2 half lives.Last edited by Stonebridge; 04062012 at 16:57. 
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 04062012 16:57
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