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Our circuit, where R1 = 3 Ohms, R2 = 1 Ohm, R3 = 2 Ohms, L = 5 milliHenry, and I = 0.01 Ampere.

Trying to figure out the differential equation for this circuit, as well as the current for the inductor when the "switch" (Imagine one going to R3/R1) has been turned from R1 position to R3, for both the initial switching iL(0) and iL(t) (t > 0) when it has been switched for a longer time period, and the voltage for uL(t) when t > 0.

I know that $${d i (t) \over dt} + {R \over L} i(t) = {V \over L}$$ which applies for a basic RL-circuit with only one resistor. However, as we can see, we have multiple resistors in this circuit. I thought of simply adding R2 and R3 together via parallel calculation, but I do not seem to get a reasonable equation out of it. Do i have to take R1 in consideration even though it does not affect the circuit?

Figuring out iL(0) was simple enough, it didn't really need calculating. Just sharing the known current for R1 and R2, so I ended up with the value of 0.0025 for iL(0).

iL(t), uL(t) and the differential equation are a bit trickier since it's a new subject for me. Any tips are appreciated.

  • \$\begingroup\$ StackExchange is not meant for answering homework questions, and this question clearly resembles what appears to be a homework question. There should be plenty of resources online to help you with your question, or you can ask a friend/professor! \$\endgroup\$ – ConcernedHobbit Apr 16 at 16:24
  • \$\begingroup\$ What transformation could you make to I1 and R2 to possibly make it a little easier for you to see how to write the equation? \$\endgroup\$ – John D Apr 16 at 16:32
  • \$\begingroup\$ @JohnD I suppose I could try to turn it into a basic RL-circuit by trying to make R3 the only resistor left in the circuit, but I am not sure how. Perhaps by making the R in the differential equation equal to R3-R2? \$\endgroup\$ – TootsieRoll Apr 16 at 17:24
  • \$\begingroup\$ @TootsieRoll Are you familiar with Thevenin equivalents? \$\endgroup\$ – John D Apr 16 at 17:32
  • \$\begingroup\$ @JohnD No, I am not. \$\endgroup\$ – TootsieRoll Apr 16 at 17:36

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