Natural response LR circuit

[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!

 

[CataM]

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.

Yes, the voltage supports it, but you have to still support it in the paper.
With the voltage polarity you have drawn in circuit B, V= -L*di/dt , if di/dt <0 (which it is), then the voltage is + => supporting the current.
You have said: I will support the current with the + sign in a nice drawn circuit, but my mathematics will not support it anymore.

If you apply the Faraday-Lenz law, you should be very careful with the signs because the voltage polarity of the inductor is such that it is opposed to the source. If there is no source in the circuit, the voltage polarity is such that maintains the flow of current in the same way it was charged, but the mathematical equation must still be the Farady-Lenz equation.

If you simply remember that the current flows in the same direction it was charged to assign properly the initial condition, then you can forget the Faraday-Lenz equation and apply the Circuit Theory equation as in circuit A.