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
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.
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?
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
To calculate the time constant, you need to know the values of the resistor (R) and the capacitor (C) in the given circuit.
In this case, the resistor value is given as 3 Mega ohms (3 MΩ) and the capacitor value is given as 350 micro farads (350 µF).
The formula for the time constant (τ) in an RC circuit is:
τ = R * C
where:
τ = time constant
R = resistance in ohms
C = capacitance in farads
By substituting the given values into the formula, you can calculate the time constant as follows:
τ = (3 MΩ) * (350 µF)
Note that for consistent units, we need to convert 3 MΩ to ohms and 350 µF to farads before performing the multiplication.
Converting:
3 MΩ = 3 * 10^6 Ω
350 µF = 350 * 10^(-6) F
Now we can calculate:
τ = (3 * 10^6 Ω) * (350 * 10^(-6) F)
= (3 * 350) * (10^6 * 10^(-6)) Ω * F
= 1050 * 10^(6-6) Ω * F
= 1050 Ω * F
Therefore, the time constant (τ) in this series RC circuit is 1050 Ω * F.
Remember, the time constant is a quantity, not an actual time value. It represents the time it takes for the voltage across the capacitor to charge (or discharge) to approximately 63.2% of its total value, which is equal to 1 - 1/e (where e is the mathematical constant approximately equal to 2.71828).