My, there sure has been a lot of head scratching, foot stomping, and traffic over this problem. Looks pretty straight forward to me. Assume va is the voltage at the positive terminal of the 12 volt source, vb is the voltage at the junction of R1, R2, and L1. And, ground is the junction R2, R3, and R5. Observe that once you calculate va, the time-constant and current present in R3 is easily found.Hello! Good evening to you all! I'm currently brushing up on RC & RL circuits in preparation for an upcoming exam, and I found this circuit to be rather complex in its design. I've had a limited understanding of natural and step responses for RC & RL circuits, and this circuit certainly hits it home. I'm supposed to find both the time constant and the current,, across thei3(t)R3resistor. My understanding of the the circuit is that for t < 0 and t = infinity, the inductor acts as a short circuit under DC conditions. However, between that, the inductor has voltage and current changing across time, so it does not act as a short circuit at t=0 or t>0.

View attachment 115480

So when I attempted to find the thevenin resistance,Req, for t > 0, I recognized that the current source is removed, whilst the voltage source is a wire. However, I don't know if this is true or not, but deactivating V1 allows a connection from the right end of the inductor to the junction between R1 and R4. Such that as the inductor is discharging, some current will flow through it, suggesting that R1 and R4 are not in series. Is this is correct, then would it be safe to presume that, from the perspective of the inductor, R3 is parallel to the series combination of R4 & R5? I'm not sure how the equivalent resistance can be found, based on how the inductor is positioned inside the RL circuit.

As for finding the currentit is mt understanding that a general solution is usually used with respect to step/natural responses for RC & RL circuits: \[ i(t)\;=\;i(t=\infty\;,\;or\;0+)\;+\;\lbrack i(t=0)\;-\;i(t=\infty\;,\;or\;0+)\rbrack\ast e^{-\frac t{L/R}} \]i3(t),

From this, I am under the assumption that this "could" be used to findHowever, what I'm confused about is on what happens to the circuit at t > 0 and t = 0 (or t < 0). I know that the current source is disregarded for both t < 0. However, when the switch is closed at t=0, I'm not sure how this changes the original circuit diagram. I'm also uncertain as to whether I should find the current across the inductor,i3(t).first, before solving foriL(t),i3(t).

I included an attachment of the attempts I've made so far, but I believe that I'm still approaching finding the equivalent resistance in the wrong way. Can someone help guide me through on how to tackle this kind of problem? Despite watching Youtube videos and referencing my textbook for problems related to this, I'e struggled with this for almost 5 hours nonstop and finally sought to ask for some much needed help. Thank you all.

It's smarter to use NODE analysis because only two equations are needed instead of four equations that the LOOP/MESH method requires.

Setting up the equations and using LaPlace transforms, we get:

Using the computer to do the grit work we get the solution for vA.

And, finding the inverse Laplace transform of vA, we get for va.

The exponent of "e" is 1.58824E+6. The reciprocal is the time-constant = 6.29628E-7.

The current present in R3 is

As was pointed out in a previous post, you can also easily get the time constant by calculating the resistance seen by the coil, which is (R2+R3||(R4+R5)||R2 = 1.58824 . The the time-constant is L/R = 6.29630E-7.

The formula you cited is f(t)=[f(0)-f(Inf)] e^-at + f(Inf) , which is good for a first order circuit with constant excitation. Ask if you have any questions.

Ratch

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