# B Time constant and capacitors

1. Jun 6, 2016

### Hannah7h

So the rate at which a capacitor charges and discharges is dependent on resistance in a circuit and the magnitude of capacitance of the capacitor? So the time constant is equal to RC. So using this equation where Q=Qoe-t/RC ,time constant is the time taken (when the capacitor is discharging) for charge on a capacitor (Q) to decrease to 37% of Qo ( i.e. charge on the capacitor when it is is fully charged). But I've been reading around and what I don't get is how the time constant is also equal to the time taken for the charge (Q) on a charging capacitor to increase by 63% of Qo . If this makes any sense, would be good if someone could maybe explain it mathematically aswell i.e. how 'e' is involved?

2. Jun 6, 2016

### cnh1995

Have you studied calculus?

3. Jun 6, 2016

### Hannah7h

Um haven't studied it in much detail, but I could give it a go

4. Jun 6, 2016

### cnh1995

Well, KVL equation for an RC circuit is a differential equation. Solving that differential equation, you get exponential expressions for charge, current and voltage.

5. Jun 6, 2016

### Hannah7h

Ok so... how does this relate to the 63% of Qo?

6. Jun 6, 2016

### cnh1995

Or in other words, time taken to lose 63% of Q0.
You can see that the time taken to gain a charge of 0.63Qo (while charging) is equal to the time taken to lose the same charge of 0.63Q0(while discharging). It's obvious, isn't it?

7. Jun 6, 2016

### cnh1995

If you are asking where 63% comes from, it comes from the constant 'e-1' in the solution to the differential equation of the RC circuit.

8. Jun 6, 2016

### lychette

the time constant is the time taken to get to within 37% of the final value.
In decay the final value is 0%!
In growth the final value is 100%

9. Jun 6, 2016

### Hannah7h

Ok that was pretty obvious now I look at it, thank you for explaining it otherwise I probably wouldn't have got there lol

10. Jun 6, 2016

### Hannah7h

Yep this makes sense, thanks!