# Time Constant / Capacitor Charge Time

by Ascetic Anchorite
Tags: capacitor, charge, constant, time
 P: 52 I know that T = CR (time constant = capacitor rating times resistance) but I do not know how to calculate (theoretically) the time taken for the final ~36.8% of the capacitor to charge. The first language of the lecturer is not English. I, and others, can barely understand a word he says! Thanks for any help.
 Emeritus Sci Advisor PF Gold P: 9,772 Do you know the equation which describes the charging of a capacitor?
 Mentor P: 41,563 The charge is an exponential function of time: $$Q = Q_{max}[1 - e^{(-t/RC)}]$$ Examine the behavior of this expression to get your answer. Hint: What's the charge after a time equal to one time constant RC? 2RC? ... and so on...
P: 52
Time Constant / Capacitor Charge Time

 Quote by Hootenanny Do you know the equation which describes the charging of a capacitor?
I don't. If this equation is in the exercise sheet I was given, then it is not clearly labeled. Most likely it is there but these notes / instructions are so ambiguous / poorly composed it is hard for me to make anything out at all (this is the main reason why I am late handing in this assignment. It was due in over four hours ago).
 Emeritus Sci Advisor PF Gold P: 9,772 The equation you need is the one posted above by Doc Al.
P: 52
 Quote by Doc Al The charge is an exponential function of time: $$Q = Q_{max}[1 - e^{(-t/RC)}]$$ Examine the behavior of this expression to get your answer. Hint: What's the charge after a time equal to one time constant RC? 2RC? ... and so on...
That equation is in the handout; it is labeled as 'the law for voltage'. And thanks, but I do not even understand your hint! Seriously, I would know much more about this subject if I had skipped lectures and stayed at home browsing library books.
 Emeritus Sci Advisor PF Gold P: 9,772 Okay, perhaps it is more apparent if we write the equation in the form of a ratio; $$\frac{Q}{Q_{max}} = 1 - \left[ e^{(-t/RC)}\right]$$ Now, if the capacitor is 63.2% charged what is the ratio Q/Qmax?
 P: 52 In the ‘voltage law’ equation in the exercise instructional notes, Qmax is V (supply voltage) and Q is Vc (voltage across capacitor). Therefore the ratio of Q/Qmax is supply voltage over capacitor voltage, which at 63.2% is 10V over 6.32V = 0.632V (I think). That seems to be as far as I can get. The ‘t’ in the formula represents a time. Just what time, I have no idea.
P: n/a
 Quote by Ascetic Anchorite In the ‘voltage law’ equation in the exercise instructional notes, Qmax is V (supply voltage) and Q is Vc (voltage across capacitor). Therefore the ratio of Q/Qmax is supply voltage over capacitor voltage, which at 63.2% is 10V over 6.32V = 0.632V (I think). That seems to be as far as I can get. The ‘t’ in the formula represents a time. Just what time, I have no idea.
t is the time since the capacitor started to charge. If you make t=0 in the formula, you see that at the start Q = 0 meaning that the capacitor is fully discharged. At t = infinity, Q = Qmax, meaning that the capacitor is fully charged.

Attributing to Q a certain value, you can use logarithms to calculate at which time this will happen.
 P: 52 This is one of those times when the 'don't directly answer it' rule of this forum serves as a serious impediment. There is no chance that I will be able to work this out on my own. I can be subjected to 20 posts of hints and I still doubt I would get it. I do not know why the time constant is significant. Why is 63.2% significant? Is the rate of charge linear up until that point and then exponential afterwards? What purpose does it serve? Why not just look at what the 100% charge time is, instead of messing around with 63.2% figures? What applications / uses is this time constant for? Why do all of the websites I have looked at not address these basic questions? Why do the many books I have looked at not address these questions? These 'learning' materials merely state that the time constant is 63.2% of the capacitor charge; they say nothing else! Stating that the time constant is 63.2% is hardly a comprehensive answer. Does the remaining 36.8% not count for anything? I am not being taught properly at the 'university', I cannot even make out the words the lecturer is saying. What chance is there? I have been on this course for about 5 weeks now. I should be competent with the basics by this stage but it is as if I have not even been to a single class. It is crap like this that makes me want quit. After all, what is the point of all this stress and pain when it leads nowhere at all? Show me a good teacher and I will show you a levitating pixie from another dimension. Where has the quality and standards gone? The teaching is rubbish, the books are poorly written. Ask a simple question, seek to learn basic information and end up spending days going round in circles in an agonizing maelstrom. I have sought to learn many things over the past year, and on every single occasion I have ended up spending far more time looking for [simple] answers than I should have needed to. Am I so aberrant?
Mentor
P: 41,563
 Quote by Ascetic Anchorite I do not know why the time constant is significant. Why is 63.2% significant?
The time constant tells you the shape (how sharply curved it is) of the exponential charging curve--it gives you a handy measure of how fast the capacitor will charge. Examine that charging equation again. Let's say that the time constant RC = 10 seconds. That means that 10 seconds after you apply the voltage, the fraction of charge on the capacitor will be:
$$1 - e^{-(10/10)} = 1 - e^{-1} = 1 - 0.368 = 0.638$$
That means it will have 63.8% of its final (total) charge.

What about after 20 seconds (2 time constants)? The fraction of charge will be:
$$1 - e^{-(20/10)} = 1 - e^{-2} = 1 - 0.368*0.368 = 0.865$$
thus 86.5% charged.

As time goes on, each additional second adds less and less total charge.

 Is the rate of charge linear up until that point and then exponential afterwards?
No. It's always exponential. (Note that the charging equation takes the form of 1 minus a decreasing exponential.)

 What purpose does it serve? Why not just look at what the 100% charge time is, instead of messing around with 63.2% figures?
Once you understand how the exponential function works, you'll see that it takes forever to get to 100% charge. So that's not very useful. Much more useful is a measure of how fast it takes to get to a certain percentage of the final charge. 63.2% is just used because it appears when the exponential term has the mathematically simple form of 1/e. (Mathematicians like simple exponents like -1. ) You could use a different number such as the time it takes to get to 90% charged (which would be some multiple of the standard time constant, about 2.3*RC).

Read more about charging capacitors and time constants here: Charging a Capacitor
 P: n/a The 100% time is not significant because it is infinity. The 63.2% figure is important because if you replace t by the time constant in the formula you get: $$Q = Q_{max}[1 - e^{-\frac{RC}{RC}}] = Q_{max}[1 - e^{-1}] = Q_{max}[1 - 0.368] = 0.632Q_{max}$$ edit to add: DocAl beat me by 2 minutes.
 P: 52 Thank you, Doc. I will spend some time looking at and thinking about what you have posted, and also at the site you linked to. If I am finally able to understand this (I feel closer now) then I will post back.
 P: 52 Thank you as well, SGT.
 P: 52 According to the lab instructions sheet the voltage across the capacitor will not reach the maximum voltage applied to the circuit until e^-t/CR becomes zero. It says that for e^-t/CR o become zero t must be infinite. Yet the online calculator, linked to by Doc, says that the capacitor has reached 0.1 microfarads (100%) after 0.0050 seconds. I programmed into that online calculator a resistance of 10000 ohms, 10 volts across the circuit, 0.01 microfarads and a time of 0.0050. I also used my calculator (pocket) to see what the answer would be when e^-t/CR is e^-50/CR = 0 (where CR = 0.0001). The answer given is zero. The answer was also zero when I stated the time as 100. 50 and 100 are some way from infinity.
P: n/a
 Quote by Ascetic Anchorite According to the lab instructions sheet the voltage across the capacitor will not reach the maximum voltage applied to the circuit until e^-t/CR becomes zero. It says that for e^-t/CR o become zero t must be infinite. Yet the online calculator, linked to by Doc, says that the capacitor has reached 0.1 microfarads (100%) after 0.0050 seconds. I programmed into that online calculator a resistance of 10000 ohms, 10 volts across the circuit, 0.01 microfarads and a time of 0.0050. I also used my calculator (pocket) to see what the answer would be when e^-t/CR is e^-50/CR = 0 (where CR = 0.0001). The answer given is zero. The answer was also zero when I stated the time as 100. 50 and 100 are some way from infinity.
After 5 time constants, the value of the exponential is less than 0.01. For each additional time constant the value approximately halves. For a time constant of 0.0001 s, the 1% value is reached after only 0.5 ms. In 5 ms, or 50 time constants the final value is practically indistinguishable from zero.
Infinity is a theoretical value. In practice we use 5 time constants as the time to reach final value.
 P: 52 I see. So the calculator was lying to me, or rather, being practical. Thanks.
 P: 3 Hi 38.6% is the factor from the the time given to capacitor for charging. T (time constant) always shows capacitor voltages decay (discharging) time with respect to the given charging time . you will always find (in practicals) that the difference between charging time and decay time is 72%. TRY IT

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