Question

Assume a circuit like that in the picture is used to charge a 470-uF capacitor to 50 volts, which is then discharged through a 1.8-k ohhm resistor. How do I find the time constant. I know it has something to do with T=RC...

Answer 1

http://user-content.enotes.com/b664caf15141c6753b3bc6dae03e108baac123b0.jpg

Answer 2

When you discharge the capacitor you simply deal with RC circuit( you open the switch)
Kirchhoff loop rule:
\[\frac{ dq }{dt } R+\frac{ q }{ C }=0 \Rightarrow q=q_0e^{-\frac{ t }{ RC }}\]
where \[\tau=RC\]
is the time constant.
In our case it's simply 1.8*10^3 *470*10^-6=0.847 seconds

Answer 3

I guess you don't have to change K-ohms to just ohms?

Answer 4

I changed it 1.8 K-ohms = 1.8 * 10^3 ohms