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, 100
kohms resistor and a 100
uF capacitor in series.(k=10
3 ,
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.
Click near the (Red) switch to toggle between charging stage and discharging stage.
horizontal position -> charging
vertical position -> discharging.
Red curve shows voltage as a function of time. V(t)
Blue curve shows current as a function of time.I(t)
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.
For the charging cycle: $V_s=V_R+V_C=I R + int frac{I dt}{C} $ where V
s 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)})$
