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?
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.
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?
When you replace the capacitor with an inductor in the RC circuit, the behavior of the circuit changes. Here's how it affects each part of the circuit:
(a) Initially, the inductor acts as a short circuit, so the effective resistance seen by the battery is still 4R/5, just like in the case of a capacitor.
(b) In the steady state, the inductor allows DC current to flow and acts as a short circuit, so the current is simply I = V/R, just like in the case of a capacitor.
(c) When the circuit is first connected, there is a transient behavior as the current starts to build up in the inductor. This is because the inductor resists changes in current. The equation for the transient behavior, known as the inductor's time constant, is given by L di/dt = V, where L is the inductance of the inductor. This equation describes how the current in the inductor changes over time.
(d) To solve for the time when the current reaches a certain value, you can use the equation L di/dt = V. Rearrange the equation and then integrate both sides to find the time.
(e) The energy dissipated in the resistors when the inductor is present can be calculated using the formula (1/2) LI^2, where L is the inductance and I is the current flowing through the resistors.
So, the main difference when replacing the capacitor with an inductor is that the inductor resists changes in current, leading to transient behavior and affecting the calculation of energy dissipation.