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?

Answer 1

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.

Answer 2

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?