Asked by charlie
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.
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.
Answers
Answered by
drwls
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
= 4.8*10^-8 F = 0.048 uF
Capacitances in Farads tend to be small
Answered by
bobpursley
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.
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.
Answered by
charlie
Very kind drwls, thanks a lot.
Charlie
Charlie
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