Solving R-L circuit differential equation

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mANmAN01

Joined Jun 26, 2020
1
Given that Ri+Ldi/dt=Vsin(wt). The answer given here is i= (V/Z)[sin(wt- Φ) + ke^(-t/τ)]. I could not solve this equation further. The figure is the tried approach to solve the question.
 

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ZCochran98

Joined Jul 24, 2018
88
One mistake you've made - an easy one to do - is you've stated V = Vsin(omega*t) and just substituted it in. If you had a constant voltage, you could do that, but because V depends on t, you can't just do that - your integration will become incorrect.

From there, there are a couple of ways to solve it: you could either solve for the homogeneous solution (Li'(t) + Ri(t) = 0), which is easy, and then toil over the particular solution (Li'(t) + Ri(t) = Vsin(omega*t)), which I don't recommend because of how tedious it is; you could solve using Laplace transforms (if you know them); or you could use the explicit solution to a first-order ODE. If you know Laplace transforms, I recommend that one. If you don't know them, then use the explicit solution one (link if you don't remember how to do so - start reading from page 9 of the linked PDF).
 

MrAl

Joined Jun 17, 2014
7,766
Given that Ri+Ldi/dt=Vsin(wt). The answer given here is i= (V/Z)[sin(wt- Φ) + ke^(-t/τ)]. I could not solve this equation further. The figure is the tried approach to solve the question.
Hi,

Something does not look right although you might explain.

First, where did you get that differential equation from?
Second, do you know what a rectifier circuit does?
Then, if you make the inductor value very very small you get a full wave rectifier with almost pure resistive load which means the time constant is very very small. A very very small time constant or just a resistor load with no inductor leads to an output current that is NOT purely sinusoidal it is a full wave rectified sine.
 
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