The time constant of an RC circuit is the time it takes the capacitor to charge or discharge 63.2% (= 1 - 1/e) of the way when the circuit is open or closed.
The time constant equals RC.
In your case, that is 1050 seconds or 17.5 minutes. You don't need to know the voltage.
A resistor 3 Mega ohms and a capacitor of 350 mirco F are connected in series to a 50V power supply. Calculate the time constant.
Do I need to solve for an actual number? I'm confused
3 answers
Thanks so much..okay for part b it saysto find the current at a time when the charge on the capacitor has aquired 90% of its maximum value.
Do I use the same equation or time constant for this?
Do I use the same equation or time constant for this?
The charge on the capacitor after closing the circuit is
Q = C V [1 - e^(-t/RC)]
Note the appearance of the RC "time constant" in this equation. I showwed earlier that RC = 1050 seconds
The maximum value of the charge (which occurs when t approaches infinity)is
Qmax = CV
When Q = 0.90*CV,
[1 - e^(-t/RC)] = 0.9
e^(-t/RC) = 0.1
t/RC = 2.3026
t = 2418 seconds = 40.3 minutes
Q = C V [1 - e^(-t/RC)]
Note the appearance of the RC "time constant" in this equation. I showwed earlier that RC = 1050 seconds
The maximum value of the charge (which occurs when t approaches infinity)is
Qmax = CV
When Q = 0.90*CV,
[1 - e^(-t/RC)] = 0.9
e^(-t/RC) = 0.1
t/RC = 2.3026
t = 2418 seconds = 40.3 minutes