parallel rc discharge
Hi,
Yes, there is a difference. In the calculation shown by Bunalmis the current i(s) is assumed to be same through R and C, which is not true for a parallel circuit. Intially, The capacitor hogs all the current, and Ir through R will be zero, but, as the capacitor gets charged as per Vc(t) = (1/C)∫ ic(t)dt, the the resistor starts diverting a current Ir(t) through it such that Ir(t) = Vc(t)÷R. and this process continues till the capacitor gets charged fully to the applied voltage Vi. Thus the capacitor charges at a slower and slower rate as times goes and this is how the time constant comes into picture.
Here we have to note that by assuming a finite i(t), we have implicitely included an internal resistance for our voltage source, as otherwise an infinite current would have charged the capacitor instantly to the applied voltage Vi. So to calculate the time constant, you have to consider this source resistance also. Once you include the source resistance Rs, you will notice that the current i(t) itself is time dependent and is given by i(t) = (Vi-Vc)÷Rs, which keeps on dimnishing as the capacitor voltage Vc increases. Thus the overall time constant in this case is increased. At t = 0, the output voltage Vo = 0, and at t = ∞ the output voltage is given by Vo = Vi*R÷(Rs+R).
Regards,
Laktronics
Added after 2 hours 40 minutes:
Hi,
There is a correction in what is written above, that is for a voltage source when an Rs is assumed, the overall time constant IS NOT INCREASED, IT IS ACTUALLY REDUCED since Rs appears in parallel with R and the time constant becomes parallel combination of R with Rs multiplied by C. The expression of output voltage Vo in this case is given by Vo = [Vi*R÷(Rs+R)](1-e^-t/Tc) where Tc = (RsR/Rs+R)C .
This is because, a voltage source with a series Rs is equivalent to a current source with a parallel Rs and a current value = V/Rs. So, Rs appears in parallel with R.
Now if you assume that the parallel R,C is connected to a current source of strength I, then the time contant will be just R*C. In this case, the current through the capacitor Ic = I*e^-t/RC and voltage Vc = IR*(1-e^-t/RC).
Regards,
Laktronics