# a capacitor problem

Discussion in 'General Electronics Chat' started by shiva bharadwaj, Jan 24, 2010.

Sep 29, 2008
19
0
Hi guys,,i have a question regarding capacitor charging and discharging,,this may seem pretty ordinary but i am very confused with charging and discharging action
consider an initially uncharged capacitor,connect it across a (constant current source of 1 ampere which is in series with 0.5Ω resistor),this is done in the time interval 0<t<T,,so according to the formula Vc=1/c∫i dt, voltage across the capacitor would be "T" volts,right,,?(at t=T)
now in the interval T<t<2T the capacitor is disconnected from circuit and is connected across a (1 Ampere current source which is in series with 1Ω resistor) but the direction of current source is "opposite" to that of the previous one(previous interval),,,,,
now what is the new voltage across the capacitor,,???????(AT t=2T)

can you please explain me the capacitor behaviour,particularly in the interval T<t<2T,,since it is already charged to T volts.

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,564
1,281
When you charge or discharge capacitor through current source then voltage on capacitor will bye change linearly (changes at a constant rate).
Ic=C*dV/dt
So for example 1mA and 100uF
ΔU=I*t/C=10V per second.

Sep 29, 2008
19
0
thank you....so in interval T<t<2T,,the capacitor should start discharging from "T" volts ,
and at t=2T,,voltage should be zero,right??

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,564
1,281
Yes, should be zero. But if you still apply the current source then capacitor is now start to charging in opposite direction.
And series resistor does not affect the voltage or time on the capacitor