What were the questions? What is the circuit?
A capacitor acts as an open circuit to DC, but has no resistance.
I am trying to do these 2 problems. I'm starting out with the RC Circuit (2nd one):
h t t p ://img201.imageshack.us/my.php?image=circuit5ij7.png
*delete spaces or link won't load!
a) The capacitor acts as a bare wire with no resistance, so:
I=V/R=V/(4/5R)
b) Capacitor acts as an open circuit, so:
I=V/R=V/(4R)
Are those right so far?
I need a little help on c, d, and e.
3 answers
The questions is in the link I gave above. Here it is again (you have to delete the spaces in 'h t t p :'):
h t t p ://img201.imageshack.us/my.php?image=circuit5ij7.png
h t t p ://img201.imageshack.us/my.php?image=circuit5ij7.png
Sorry, I forgot to check out the link with the circuit and the questions. That was clever of you to add the spaces in h t t p so the URL could be displayed.
(a) Initially, C looks like zero resistance, so the effective resistance seen but the battery is 4R/5
(b) steady state I = V/R, since the capacitor cannot allow DC current
(c) current = 0 through the battery, but V/(5R) through 4R, and the capacitor discharges.
(d) Solve exp[-t/(5RC)] = 0.5 , for t.
5RC is the series resistance that the capacitor sees while discharging
(e) (1/2) C V^2 is the energy dissipated in resistors while discharging
(a) Initially, C looks like zero resistance, so the effective resistance seen but the battery is 4R/5
(b) steady state I = V/R, since the capacitor cannot allow DC current
(c) current = 0 through the battery, but V/(5R) through 4R, and the capacitor discharges.
(d) Solve exp[-t/(5RC)] = 0.5 , for t.
5RC is the series resistance that the capacitor sees while discharging
(e) (1/2) C V^2 is the energy dissipated in resistors while discharging