# quick KVL question

Discussion in 'Homework Help' started by ihaveaquestion, May 9, 2009.

1. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
http://img154.imageshack.us/img154/4533/111222111122.jpg
1) If I try to do KVL in the loop w/ the voltage source, R1 and R2 I get:

-10 + iR1 + iR2 = 0

i (R1+R2) = 10

i = 10 / R1 + R2 = 10/3

But this value of current appears nowhere in the problems, i.e.
it's not the current of i2 or i3 at t = 0+

Why is that?

2) Also, after a long amount of time as t goes to infinity, the inductor acts like a short... the voltage of R1 and L are the same, but ALL the current goes through L and none through R1... So my question is, the inductor only ACTS like a short and isn't ACTUALLY one right? Because if it were actually one then it would get rid of R1 and i2 would be the same as i1 L (at t infinity)

2. ### steinar96 Active Member

Apr 18, 2009
239
4
1)

You can't use KVL like that in this example. If you were gonna use KVL you would have to do it for all the meshes. And solve a simultanious equation containing differential quantities. Which i suspect is out of your scope at the moment.

2) Have you seen a typical inductor. It's a coil of wire wound around some material core. The wire conducts just like an ordinary wire so an inductor is actually a short. There is a catch though which originates in the magnetic field the charges flowing create when moving in a circular fashion around the material core. The field at any time tries to maintain its current strength. So if the current trough the inductor increases it spans a voltage across itself with polarity to resist the change in current (change of magnetic flux creates an electric field).

So technicly the inductor actually is a short in the sense that its just a wire wound in circles. But the consequence is that it resists change in current trough itself.

So when there is no change in current the inductor is neither aquiring nor spending energy in attempt to keep it constant. That is why the current flows trough a inductor just like a normal wire when the current is not changing. When it begins to change the inductor absorbs or expends energy in attempt to keep it constant.

So when the current is constant the inductor acts as a short and can be depicted as a normal wire in a circuit. In the circuit you linked to above when the current is constant and has been for a long time no current flows trough R1 because the inductor opposes no current trough itself (acts just like a normal wire).
At that time when the current is constant and has been for a long time the capacitor is not conducting trough itself so it acts as a circuit breaker. Meaning current in that circuit will flow only trough the inductor and the R2 resistor.