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Capacitors

If you're charging/discharging a capacitor under constant current.. how do you treat this in calculations? What changes? :redface:

Thanks,
TheRandomer

Reply 1

amywalters

TheRandomer

If you're charging/discharging a capacitor under constant current.. how do you treat this in calculations? What changes? :redface:

Thanks,
TheRandomer

Well charging a capacitance is storing energy inside it, so you'll have your equation;
Q=CV, and W=1/2QV

Discharging a capacitor has an exponential relationship, so at any time during the discharge, potential difference, current and charge can be found using the equations;
Q= Qo e^(-t/RC)
V=Vo e^(-t/RC)
or I=Io e^(-t/RC)

..where Qo, Vo and Io are the charge, potential difference and current when t=O, and t=RC is the time constant, and e is the exponential constant.

Reply 2

TheRandomer

Yeah but if it's CONSTANT current, what happens?

Reply 3

zedliv

that means that you can use Q=It without finding the area under the graph

Reply 4

bobbles_lass

yeah - done practially with a variable resistor, though it's a bit fiddly...
then your Q/t graph is just a rectangle rather than the exponential decay one

Reply 5

Mehh

bobbles_lass

yeah - done practially with a variable resistor, though it's a bit fiddly...
then your Q/t graph is just a rectangle rather than the exponential decay one

Triangle. Rectangular in the limiting case where the current = 0, kinda defeats the point really.

What might be interesting however is the energy vs time graph. Which I do believe is a quadratic equation.