Setting up a homogenous ODE using Kirchhoff's law of voltage

Thank you in advance for your help. I'm having trouble with the setup of this problem. The problem states:

The loop current I in a series RL circuit with constant voltage E0 satisfies LI' + RI = E0 (where I' is the first derivative of

I) by Kirchhoff's voltage law. Assume that R and L are constants. Asume that initially there is no current in the circuit.

a) Find the current as a function of time.

b) Find the steady state solution.

c) When will the current be (1-e^-1) times the steady state current?

Re: Setting up a homogenous ODE using Kirchhoff's law of voltage

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Originally Posted by

**nivek0078** Thank you in advance for your help. I'm having trouble with the setup of this problem. The problem states:

The loop current I in a series RL circuit with constant voltage E0 satisfies LI' + RI = E0 (where I' is the first derivative of

I) by Kirchhoff's voltage law. Assume that R and L are constants. Asume that initially there is no current in the circuit.

a) Find the current as a function of time.

b) Find the steady state solution.

c) When will the current be (1-e^-1) times the steady state current?

a) Solve , with I(0) = 0 (which is the meaning of "initially there is no current in the circuit"), and L, R, and E0 constant. If you don't know how to do that, then your trouble is that you don't understand the basics of differential equations. I'll assume that you do know how to do that, so that you've solved for I(t).

b) I believe that the "steady state solution" is asking what I(t) is "asymtopically equivalent to" for large time values. If you did (a) and understand what this means, it's not hard to determine.

Ex: If I(t) were a polynomial in t, then the steady state solution would be the highest power term of the polynomial.

Ex: If I(t) = 3t^2 -4t +7, then I(t) ~ 3t^2, because -4t +7 are eventually a miniscule fraction of I(t) once t is large.

Ex: If I(t) = 2t + 5sin(t), then I(t) ~ 2t, since 5sin(t) is a minscule fraction of I(t) when t is large (5sin(t) bounces between -5 and 5).

Ex: If I(t) = (3t +2)/(t+1), then I(t) ~ 3. That's because I(t) = (3t + 2)/(t+1) = (3(t+1) -3 +2)/(t+1) = (3(t+1)-1)/(t+1) = 3 - (1/(t+1)), and 1/(t+1) goes to 0 as t goes to infinity.

Ex: Any term going to 0 is discarded from the steady state solution. If I(t) = f(t) + g(t), and limit {t -> infinity) g(t) = 0, then the steady state solution for I(t) is the same as the steady state solution if I(t) has equalled *just* f(t). For the steady state, you can discard terms going to 0.

c) In part (a), you solved for a function I(t). In part (b), you determined a new function of t, the steady state function of I(t), let's call it SS(t).

(In other words, I(t) ~ SS(t)). Now part (c) asks "When will the current be (1-e^-1) times the steady state current?". Suppose that specific time(s) is labelled t'.

Then (c) is asking: "Find t' such that I(t') = (1-e^-1) SS(t')." So simply write down that equation and solve it for t'.

Note that, in general, there could be more than one solution. Note also that the domain here is t >= 0, so negative solutions must be discarded.

Re: Setting up a homogenous ODE using Kirchhoff's law of voltage

Thank you johnsome for the information.

This is open to anyone! I would like to check if my solution to part A) is correct. I got I(t)= E0 + C, if not what did i do wrong. Is the intergation factor e^t?

Re: Setting up a homogenous ODE using Kirchhoff's law of voltage

Quote:

Originally Posted by

**nivek0078** Thank you johnsome for the information.

This is open to anyone! I would like to check if my solution to part A) is correct. I got I(t)= E0 + C, if not what did i do wrong. Is the intergation factor e^t?

Could you not check it yourself? I(t)= E0+ C is a constant- the general solution to this equation is NOT constant! In particular, if I(t)= E0+ C then I'(t)= 0 so the equation becomes , NOT E0.