# Natural response RL circuit

1. Apr 11, 2017

### paulmdrdo

as you can see In the image I provided I have derived the equation for i(t) and v(t).

On circuit A I use a passive sign convention(current flowing from positive to negative) on the inductor, hence the equation and their corresponding graph.
i(t) is decaying exponentially, v(t) is decaying exponentially.

On circuit B I did use what I learned in physics, i.e the voltage induced across an inductor is always opposite to the increase/decrease in current
flowing throuhg it. Since the current i(t) is decreasing, the voltage should support that current hence the sign. when I perform kvl on circuit B I ended up having growth equation for both i(t) and v(t). Why is that?

Also in circuit A I expect to have same graph for both i(t) and v(t) (Although they are both decaying).
Please, kindly clear this up for me if you have time. Thanks!

Last edited by a moderator: Apr 18, 2017
2. Apr 11, 2017

### Svein

What is the final value of the current through the inductor (when the current does not vary any longer)?

3. Apr 18, 2017

### eq1

The signs are not consistent between the two circuits therefore you can't use v=Ldi/dt (using the symbols from the schematics) in both algebraic solutions. i.e. circuit a has positive current entering the + terminal of L. Circuit B has positive current leaving the + terminal of L and this is why it shows current growth.

Edit: I just realized you made the sign error on purpose. The issue is still a sign error. But you dropped it from the "physics" version of the equation. See the link below.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/induct.html#c1

Last edited: Apr 18, 2017