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I know that phasors are used to simplify the calculations, I also get why in AC, current in a capacitor leads the voltage by 90°, and lags for an inductor (we can see that by differentiating/integrating), but how do we know it will solve differential equations too? Thinking of sine and cosine terms as vectors really helps in adding/subtracting them. (But I don't see how it helps in multiplying/dividing)

Suppose I have an RL circuit connected in series to an AC emf like this:

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Then, by phasors, we would say $\overline{Z} = \sqrt{R^2 + L^2 \omega^2}\angle\tan^{-1}(\omega L/R)$ or $\overline{i} = \displaystyle\frac{e_0}{\sqrt{R^2 + L^2 \omega^2}}\angle\tan^{-1}(-\omega L/R)$ as $\overline{i}=\displaystyle\frac{\overline{e}}{\overline{Z}}$ (Taking $\overline{e} = e_0\angle0$)

but without using phasors, we would have written $+e_0\sin(\omega t)-iR - L\frac{\text{d}i}{\text{d}t} = 0$ and try to solve that differential equation, right?

How do we know we're solving that differential equation by doing phasor algebra?

By the expression I got for $\overline{i}$, $i(t) = \displaystyle\frac{e_0}{\sqrt{R^2 + L^2 \omega^2}} \sin(\omega t - \tan^{-1}(\omega L/R))$

Then I thought, is it really the solution to the differential equation $+e_0\sin(\omega t)-iR - L\frac{\text{d}i}{\text{d}t} = 0$?

I looked up the solution to that equation and it's slightly different $$i(t) = \displaystyle\frac{e_0}{\sqrt{R^2 + L^2 \omega^2}} \sin(\omega t - \tan^{-1}(\omega L/R)) - c e^{-Rt/L}$$ According to this , the phasor answer is correct if c=0, which does not mean $i(0) = 0$, I checked the graph, but it isn't too different from the phasor answer for positive values of t. Why are we getting this extra term different from the phasor answer?

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$\sin(\omega t)$ can be represented as the imaginary part of $e^{i\omega t}$, which is a rotating vector in the complex plane (called phasors). Coincidentally, the solutions of the differential equations involved in the circuit come out to be of the form $e^{i\omega t}$, and so we can take the real part of the equation and prove it.

Now suppose $g$ is a solution of $+e_0\sin(\omega t)-iR - L\frac{\text{d}i}{\text{d}t} = 0$, and $f$ is a solution of $iR +L\frac{\text{d}i}{\text{d}t} = 0$, the you can easily verify that $g+\lambda f$ is also a solution $+e_0\sin(\omega t)-iR - L\frac{\text{d}i}{\text{d}t} = 0$ because it is a linear equation. We can solve the DE for $f$ and get $f=ce^{\frac{-Rt}{L}}$

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  • $\begingroup$ Ok, but the solution to the differential equation has an extra +e^-Rt/L term in it? $\endgroup$ – Rick Mar 28 '18 at 22:01
  • $\begingroup$ @Rick I've edited the answer $\endgroup$ – Prathyush Poduval Mar 29 '18 at 8:31

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