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dashin00 dashin00
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11 years ago
Hi - I've never done any form of maths or physics before and I've been given this question (8 marks!!)

A capacitor of capacitance C is discharging through a resistance R. In terms of the time constant t=RC  (the 't' looks a bit different though - it's like a 't' but without the top bit)  when will the stored energy in the capacitance be half of its initial value?
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wrote...
11 years ago
applying kirchoffs loop law at some time t
wrote...
11 years ago
Pete, if you are new to this, it may be slightly difficult.
It may not be 't' but Tau, the symbol usually used to denote time-constant.

Voltage across a capacitor that is being discharged thru a resistor follows an exponential decay curve.
V(t)=Vo * exp(-t/RC),  where Vo is voltage at time t=0
Energy is= 0.5 CV^2
So energy is half when voltage is 1/sqrt(2) of initial voltage. ie V(t)=0.707 Vo
0.707=exp(-t/RC)
solving, t=RC ln(sqrt(2))
Maybe you should read up why the capacitor voltage follows that equation. I'll give you a hint. The discharge rate or current is proportional to capacitor voltage. As discharge occurs, the voltage drops and so does the discharge rate. This differential equation has such an exponential solution
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