Question

Step by step solution for the diff eq. for a LR circuit.

L*dI/dT + I*R = V
where V = Imax*R

Answer 1

LdI/dT + I*R =V\[dI/dT + R*I/L =V/L\]
replace d/dt with D
\[(D+R/L)I=V/L\]
THEREFORE m+R/L=0 GIVES m=R/L
\[Ce ^{-RT/L} \] is the general solution , and the particular solution is \[(V/L)*(1/(D+R/L))\]
now 1/(D+R/L) is L/R
so your particular solution becomes V/R;
\[I=Ce ^{-RT/L}+V/R\]
using the initial condition that at T=0;I=0; this gives C=-V/R

Answer 2

Collectedsoul, did this answer your question? I was wondering where C came into this probllelm. I was under the impression you were working with inductive electromotive force as a result of varying current. I don't have an answer for you but was wondering if this was a satisfactory solution. I am assuming the C is for capacitance????

Answer 3

to dipankarstudy:

A couple of qs more,
1) Why is the particular solution (V/L)∗(1/(D+R/L))?

2) Can you explain in a little more detail why 1/(D+R/L) = L/R?

Answer 4

to radar:

C is used as a constant in this case, the constant for integration, not capacitance.

Answer 5

O.K. I must admit I was over my head. My career was in electronics, but primarily maintenance and modifications of FAA Long Range Radars. Never got too much into the theory. But, am interested!

Answer 6

As an alternative, can this equation be solved using Laplace Transforms?