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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    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

  2. radar
    • 5 years ago
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    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????

  3. anonymous
    • 5 years ago
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    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?

  4. anonymous
    • 5 years ago
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    to radar: C is used as a constant in this case, the constant for integration, not capacitance.

  5. radar
    • 5 years ago
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    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!

  6. anonymous
    • 5 years ago
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    As an alternative, can this equation be solved using Laplace Transforms?

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