You are Here: Home

# Is there another way to do this capacitor question without using logs?

Announcements Posted on
Take our survey to be in with the chance of winning a £50 Amazon voucher or one of 5 x £10 Amazon vouchers 28-05-2016
1. An uncharged 4.7μF capacitor is charged to a pd of 12v through a 200 ohms resistor and then discharged through a 220 kilo-ohms 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.
2. Yes there is an alternative way.
If you know that the "half life" of the charge or pd on a capacitor is t½ = 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 3
3. (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 t½ = 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
But I don't know the ''half-life'' though, and how did you get ln 2?

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, t
4. Better watch because some marking schemes specifically want you to use a certain method to find the answer.
5. (Original post by Foghorn Leghorn)
Better watch because some marking schemes specifically want you to use a certain method to find the answer.
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
6. 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 Loge 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!)
7. (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.

8. (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 Loge 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 t½ = 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.
I'm on AQA-A 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 t½ = 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
9. (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.

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.
10. (Original post by internet tough guy)
I'm on AQA-A 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 t½ = 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
Would it not be 2xLN(2)x220kx4.7micro? That gives you 1.4334s.
11. (Original post by Foghorn Leghorn)
Would it not be 2xLN(2)x220kx4.7micro? That gives you 1.4334s.
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?
12. (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.
Sorry I can't answer this for because I don't know, I just know exam boards can be a pain in the arse when it comes to methology. It could be that they expect you to already know how to interchange between exponentials and logs/ln therefore that's why they didn't mention it. But I wouldn't take my word on that one.
13. (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?
According to wikibooks, A-level physics allow the use of t(1/2)=RC Ln(2)

EDIT: also Ln(2)= 0.693
14. (Original post by Foghorn Leghorn)
According to wikibooks, A-level physics allow the use of t(1/2)=RC Ln(2)

EDIT: also Ln(2)= 0.693
Yes. Apologies for getting the last 2 digits round the wrong way.
I've amended my earlier posts.
Yes it's 0.693 not 0.639

One half life is 0.693 CR so 2 half lives is 2 x 0.693 CR
15. (Original post by internet tough guy)
I'm on AQA-A 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 t½ = 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
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.
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.
16. (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.
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.
17. 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
18. (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.
See my other post.
Sorry I made a mistake through rushing the reply. Yes multiply by 2 for the 2 half lives.
19. (Original post by Stonebridge)
See my other post.
oh ok, thanks anyway for offering this alternative method

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: June 4, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### Don't be a half-term hermit

How to revise this week and still have a life

### Subjects you're happy to never do again

Poll
Study resources
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.