R= kohm, C= uF,V= V

This java applet try to show the transient behavior that occurs when the capacitor is being charged and discharged.
1. You will find a circuit with 10 Volt Voltage source, 100kohms resistor and a 100uF capacitor in series.(k=103 , u=10-6 )
You can select different value of voltage/resistance/capacitance. (select value then Hit Update button!)
Or click within the cyan area(near battery), and drag the mouse up/down to change voltage.
2. Press Start to start the animation, Voltage of capacitor(Vc) and resistor(Vr) are shown near the elements.
3. There is a switch to control the flow path of the circuit.

4. If you click the mouse button twice,the timing t will be reset to zero.
5. The animation is suspended when you press the mouse button.
   If you click with left mosue button,animation continues when you release the mouse button.
   If you click with the right mouse button,you will need to press the mouse again to resume.
6. While the animation is suspended, move the mouse in the Vc(t) or T(t) plot area
Value of Vc, I and t. corresponse to mouse position will be displayed.
If you drag the mouse, It will show relative value.

If you really know what you are doing (You can change parameters in the applet)
Please enter reasonable value and Hit SETUP Button
R= kohm, C= uF, V= V

For example: Keep order of time constant (R*C) about several seconds.
Default value: R=100kohm, C=100uF give time constant = 10 seconds.

For the charging cycle: V_s=V_R+V_C=I R + \int \frac{I dt}{C} where V[sub]s[/sub] is the voltage from the power supply.
0=R \frac{dI}{dt}+ \frac{I}{C},  \frac{dI}{dt}=-\frac{I}{RC},   so the solution is I(t)=I_0 e^{-t/(RC)}
at t=0, V_c=0 so I(t=0)=I_0=V_s/R
The result is V_R(t)=I(t) R =V_s e^{-t/(RC)}, V_c(t)=V_s-V_R(t)= V_s (1- e^{-t/(RC)})