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I have the following 1st order linear differential equation: $$L\frac{dI}{dt}+RI=E_0\sin(wt).$$ where $L$, $R$ and $E_0$ are constants. The goal here is to discuss the case when $t$ increases indefinitely.
What does this mean? How do you go about this?

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  • $\begingroup$ What do you think it means? What is your current understanding of the meaning of that differential equation? $\endgroup$ May 5 at 17:58
  • $\begingroup$ Are you supposed to solve it for $t \geq 0$ or the behaviour of the solution as $t$ increases? $\endgroup$
    – user789450
    May 5 at 18:14
  • $\begingroup$ I'd interpret that as "describe the behavior of the solution" (which solving the equation can help with). $\endgroup$ May 5 at 20:21
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The first step is to solve the differential equation. This is an inductor and resistor in series, excited by an AC voltage.

The easiest way to solve this is using the Laplace transform.

$$L\frac{dI}{dt} + RI = E_0 \sin{(\omega t)} \\ L(sI(s) - i(0)) + RI(s) = E_0 \frac{\omega}{\omega^2 + s^2} \\ (sL + R)I(s) = E_0 \frac{\omega}{\omega^2 + s^2} + Li(0) \\ I(s) = E_0 \frac{\omega}{(s^2+\omega^2)(sL+R)} + \frac{Li(0)}{sL+R} $$

Let's try partial fractions to break the terms of the denominator $$\frac{as + b}{s^2 + \omega^2} + \frac{c}{sL+R} = \frac{\omega}{(s^2+\omega^2)(sL+R)} \\ \implies s^2(aL + c) +s(aR+bL) + bR+c\omega^2 = \omega$$ Solving this gives us the following identities $a = \frac{-bL}{R}, c = \frac{bL^2}{R}, c = \frac{\omega - bR}{\omega^2}$, which in turn gives

$$a = \frac{-\omega L}{R^2 + \omega^2L^2} \\ b = \frac{\omega R}{R^2 + \omega^2L^2} \\ c = \frac{\omega L^2}{R^2 + \omega^2L^2}$$

So, we have $$I(s) = \frac{\omega}{R^2 + \omega^2L^2} \left(E_0\frac{-sL + R}{s^2+\omega^2} + \frac{L}{s+R/L}\right) + \frac{i(0)}{s + R/L}$$

Taking the inverse Laplace transform, (by looking at tables) $$I(t) = \frac{E_0}{R^2 + \omega^2L^2} \left[-\omega L\cos \omega t + R\sin{\omega t} + R e^{-\frac{R}{L}t}\right]+ i(0)e^{-\frac{R}{L}t} $$

As $t \rightarrow \infty$, the exponential terms will vanish, $$I(t) = \frac{E_0}{R^2 + \omega^2L^2} (R\sin{\omega t - \omega L \cos\omega t})$$ You can simplify this further by assuming $\frac{\omega L}{R} = \tan \theta.$ Then, the final expression is $$I(t) = \frac{E_0}{\sqrt{R^2 + \omega^2L^2}}(\sin \omega t \cos \theta - \sin \theta \cos \omega t) \\ = \frac{E_0}{\sqrt{R^2 + \omega^2L^2}} \sin (\omega t - \theta)$$

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  • $\begingroup$ The units don't match for your final solution. $\endgroup$ May 5 at 21:25
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    $\begingroup$ Thanks, I had made a mistake while taking the inverse Laplace transform. I fixed the error. $\endgroup$
    – orchi_d
    May 6 at 6:35
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You have a first order linear inhomogeneous equation. First solve the homogeneous version: $$L\frac{dI}{dt} + RI = 0$$ gives $$\frac{dI}{I} = -\frac{R}{L}dt$$ which integrates to $$\ln I = -\frac{R}{L}t + c_1$$ or $$I = Ce^{-Rt/L}$$

Now the inhomogeneous term is simple enough to solve with the method of undetermined coefficients: $$I_p = A \sin \omega t + B \cos \omega t$$ so $$I_p' = \omega A \cos \omega t - B \omega \sin \omega t$$ Plugging in we get $$L\omega A \cos \omega t -LB\omega \sin \omega t + AR \sin \omega t + BR \cos \omega t = E_0 \sin \omega t$$. and matching coefficients we find $$AR-LB\omega = E_0, L\omega A + BR = 0$$ Solve the second equation for $B = \frac{-L\omega A}{R}$ and substitute, and we end up with $$A = \frac{E_0R}{R^2 + L^2\omega^2}, B = \frac{-L\omega E_0}{R^2 + L^2\omega^2}$$ and finally $$I_{general} = Ce^{-Rt/L} + \frac{E_0R}{R^2 + L^2\omega^2} \sin \omega t + \frac{-L\omega E_0}{R^2 + L^2\omega^2} \cos \omega t$$ where $C$ depends on the initial condition. After a long time (as $t \to \infty$) the first term shrinks away leaving the last two terms. It is also possible to use a trigonometric identity to combine the two remaining terms into a single term with a phase angle, if that is preferred.

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