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A level physics time constant

Please could somebody tell me if this is right, I’m a bit confused!

When it says changing the initial voltage doesn’t affect the time constant of capacitor discharge, does it mean the voltage of the power supply or the capacitor? Wait I think those are actually equal?

If it means Vc then because C is constant (C=Q/V) if Vc is increased/decreased, the charge stored needs to also increase/decrease proportionally - so the time constant doesn’t change as the it takes the same amount of time for the same amount of charge to decay to 37% of its original value (providing the resistance is also constant).

Reply 1

Pangol

What you say it right, but there is no need to consider the initial p.d. across the capacitor or the charge on it at all. As you say, the time constant is the amount of time that it takes for the charge on the capacitor (or the p.d. across the capacitor) to fall to 1/e of its original value (I think 1/e is easier to remember than the percentage), whatever that original value was.

Reply 2

Mavs04

Original post by Pangol

Thanks! Ik it’s unlikely to come up I just remember stuff better when I understand why.

Reply 3

Pangol

Original post by Mavs04

Thanks! Ik it’s unlikely to come up I just remember stuff better when I understand why.

If you are happy with the capacitor discharge equation, X = X₀e^-t/RC, where X is either charge Q or p.d. V, with X₀ being the initial charge/p.d., then it follows from that. Put t = RC (so that we see what happens after a time of one time constant has passed), and you immediately get that X = X₀/e - that is, the charge/p.d. has fallen to its initial value divided by e, without it mattering what the original value of X was.