# Physics

When discharging a capacitor it will
a) loose half of its charge in equal time intervals
b) loose a third of its charge in equal time intervals
c) loose a quarter of its charge in equal time intervals
d) gain half of its charge in equal time intervals

I know that d) isn't correct. I believe it's a).

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1. a, b and c are all correct. The time intervals are different in each case.

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posted by drwls
2. c = q/v
v = q/c
dv/dt = (1/c) dq/dt = -(1/c) i (negative because the discharge current is decreasing the q on the capacitor)
Now let's discharge this capacitor through a resistance R
v = i R so i = v/R
put those together
dv/dt = -(1/c) v/R
dv/dt = -(1/Rc)V
dv/v = -(1/Rc_) dt
ln v = k -1/(RC) t where k is a constant of integration
v = e^[k -(1/Rc)t ] = e^k e^-(t/Rc)
v = initial v * e^-(t/Rc)
call initial v = Vo
v = Vo e^-(t/Rc)
i = (Vo/R) e^(-t/Rc)
q = Vo c e^(-t/Rc)
So the chare is exponentially decreasing with time
If it loses half its charge in T seconds, it will lose half again in T seconds, and half again in another T
so charge versus time table would be
t = 0, q = Vo c
t = T, q = (1/2) Vo c
t = 2T, q = (1/2)(1/2) = 1/4 Vo c
t = 3 T, q = (1/2)(1/4) = (1/8) Vo c
etc, so it loses 1/2 of its MOST RECENT charge in each "half life"
NOW
exactly the same logic applies to the 1/3 rule
t = 0, q = Vo c
t = T1, q = (2/3)Vo c
t = 2 T1, q = (2/3)(2/3) = (4/9) Vo c
t = 2 T1, q = (2/3)(4/9) = (8/27) Vo c
etc
in other words since the decay of charge is exponential, it loses the same fraction of the most recent value (not the original value) in every time interval no matter if 1/2 or 1/3 or 15/33 or whatever.

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posted by Damon

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