A 9v circuit charges a capacitor to steady state. Then the switch is opened.

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

Very kind drwls, thanks a lot.

Charlie