anonymous
  • anonymous
Step by step solution for the diff eq. for a LR circuit. L*dI/dT + I*R = V where V = Imax*R
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
radar
  • radar
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????
anonymous
  • anonymous
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?

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anonymous
  • anonymous
to radar: C is used as a constant in this case, the constant for integration, not capacitance.
radar
  • radar
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!
anonymous
  • anonymous
As an alternative, can this equation be solved using Laplace Transforms?

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