L (dI/dt) + RI = E
It’s instructive to first solve this form where E is presumed constant. Begin by rearranging the DE to get:
L
dI = (E – RI) dt so that: ò LdI/ (E – RI) = ò dt
Let u = (E – RI) , Then: du =  RdI and dI = du/ R
So:

L/R ò du/u = ò dt
Now,
u = E – RI and this has a (+ve) value (e.g. uo = E) at t = 0 and we assume it
remains positive so çu
ç = u and:
(L/R)
ln u = t + C
Or:
(L/R) ln (E – RI) = t + C
The
constant of integration C can be found knowing that I = 0 at t = 0 hence:
C
=  (L/R) ln E
After
suitable substitutions and some algebra:
L/R
[ln E – ln (E – RI)] = t
ln
[E/ E – RI] = Rt/ L
Taking
natural logs:
E/
(E – RI) = exp (Rt/L)
Which
can also be written: (E
– RI)/ E = exp ( Rt/L)
Solving
for I: I
= E/R [1  exp (Rt/L)]
Then
the current (for t > 0) always increases toward the steady state value of
E/R. Now,
what about the solution if we let E = E_{o} sin wt? (Oscillatory voltage).
It
should be fairly clear from past articles –blog posts that the solution must
have the form: A sin wt
+ B cos wt
Using
the integrating factor, the solution we obtain (note this is not a
higher order DE!) is:
I
exp (at) = ò exp (at) E dt +
C
Then
one will obtain:
I
= E_{o} { RL sin wt  wL ^{2} cos wt}/ (R^{2}
+ w^{2} L ^{2 })
+ k exp (Rt/L)
This
can also be done using the Laplace transform technique, after rearranging the
equation to get:
dI/dt
+ (R/L) I = E_{o} sin wt
Then
write the transformed eqn. as, using a = R/L:
s
I(s) + a I(s) = E_{o} w/ (s^{2} + w^{2} )
Whence:
I(s)
[ s + a] = E_{o} w/ (s^{2} + w^{2} )
I(s)
= E_{o} w /
(s + iw) (s  iw) (s + a)
For
which a partial fractions decomposition can be used, followed by the inverse
Laplace transform.
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