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How does one calculate the time constant of a discharging capacitor?

Hello All,

I'm currently working through a core practical (CPAC11 to be precise for the Edexcel students) and it requires charging a capacitor and discharging it and timing how long it takes to reach a certain voltage e.g. from 6V to 5V, 5V to 4V etc.

I've got the data into Excel and I've made a graph with time on the y-axis and voltage on the x and you're then required to determine a value for the time constant of the discharge but I don't really know how I would go about doing that.

Could I perhaps receive an explanation on how I'd go about working it out please? Thanks in advance!!
Afaik you get given the formula for charge remaining on a capacitor during discharge in the formula sheet

it's Q=Q0 e(-t/RC)

and since the PD across a capacitor at any instant is given by C=Q/V you can write
V=V0 e(-t/RC)
since you've got measurements of voltage and time it'll be convenient to rearrange this formula a bit

(will help if you use the log laws, do you know how to do that?)
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ErtugrulBey
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Original post by Joinedup
Afaik you get given the formula for charge remaining on a capacitor during discharge in the formula sheet

it's Q=Q0 e(-t/RC)

and since the PD across a capacitor at any instant is given by C=Q/V you can write
V=V0 e(-t/RC)
since you've got measurements of voltage and time it'll be convenient to rearrange this formula a bit

(will help if you use the log laws, do you know how to do that?)

So would I need to sub in my values for voltage (6V at t=0 and then 6x0.63, 6x0.63^x) and then rearrange for RC?

I think I managed to get a time constant using the graph by just doing 63% of 6V and 63% of the output and so on and I got 15.25s but I'm not sure if it's correct. It then asks me to find a value for capacitance using my value for the time constant, how would I go about using the time constant in this case? I don't really understand how the time constant relates to the capacitance since I don't have a value for charge
typed in what I thought was a lovely explanation and the forum dumped it...

RC = time constant (in seconds) ... in a discharge it's the time taken for the PD on the capacitor to fall from it's original value to 0.37 of it's original value.
if the resistance is known and the time constant can be estimated from the graph you can find the value of the capacitor with algebra

0.37 V0 (2sf) is the fraction of the original PD that's *remaining* across the capacitor after 1RC
0.63 V0 is the fraction of the original PD across the capacitor that's *gone* after 1RC

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