A 9v circuit charges a capacitor to steady state. Then the switch is opened.
All resistance units in kilo-ohms:
2 rsistors, 20 and 40kO in parallel with a single 20 kO resistor. Total R is 3.33x10^4 Ohms.
Am I close?
If so, and the time constant is 1.6x10^-3 seconds, then is the capacitance C= 1.6x10^-3/3.33x10^4 Farads? It seems very small.
The combined resistance of the circuit it correct. C = time constant/R
= 4.8*10^-8 F = 0.048 uF
Capacitances in Farads tend to be small
The total resistance is
20*40/60 + 20=80/6 + 20= 33.3Kohm
RC= 1.6ms
C= 1.6ms/33.3kohm= you are right.
Small capacitors make small RC time constants, which make for circuits good for high pass filters.
Very kind drwls, thanks a lot.
Charlie
Yes, you have made some correct calculations, but let me explain in a step-by-step manner to ensure clarity.
First, let's determine the total resistance (R) of the circuit. You have correctly calculated it as 3.33x10^4 ohms. To find this value, you correctly considered the parallel combination of the two resistors (20kΩ and 40kΩ) and added the third resistor (20kΩ) in series.
Now, to calculate the capacitance (C), we can use the time constant formula: τ = R * C, where τ is the time constant. In your case, the time constant is given as 1.6x10^-3 seconds.
Rearranging the formula, we have C = τ / R. Substituting the values, we get:
C = 1.6x10^-3 / 3.33x10^4
= 4.81x10^-8
So, the capacitance of the circuit is approximately 4.81x10^-8 farads. It might seem small when expressed in farads, but it's a common range for capacitors used in electronic circuits.
Therefore, your calculations are correct, and the capacitance you obtained is accurate based on the given information.