Question
Posted by Eileen on Sunday, April 13, 2008 at 10:05am.
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.
* Circuit Problem- Physics - drwls, Sunday, April 13, 2008 at 12:53pm
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
Okay I get it so far... but how does this change when an inductor is put there instead of a capacitor?
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.
* Circuit Problem- Physics - drwls, Sunday, April 13, 2008 at 12:53pm
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
Okay I get it so far... but how does this change when an inductor is put there instead of a capacitor?
Answers
drwls
Inductors behave in an opposite manner from capacitors. In steady state, they offer no resistance, but they do not allow an instantaneous change in current.
Eileen
So when switch is closed, the inductor acts as a block (or open circuit in that branch)?
And when it is opened after a long time, it acts as a bare wire?
And when it is opened after a long time, it acts as a bare wire?