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Consider the following circuit:

RL

When \$t > 0\$, the inductor \$L\$ begins to discharge. I already know from KVL that the sum of voltage drops in the loop containing the inductor is:

\$L\frac{di(t)}{dt} + Ri(t) = 0\$

From this, how could I derive the equation for the discharged current across the inductor i.e. at any time \$t\$:

\$i(t) = I_s e^{-\frac{t}{\tau}}\$

As I understand, it involves some integration, but I'm a bit stuck so if someone could help derive it for me, that would be much appreciated.

Thanks

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  • \$\begingroup\$ You don't so much prove it as derive it. This is more a mathematics question. Given a differential equation, come up with the time evolution. However, it's a simple enough one for engineers to do as well. Where do you want to start from, will tables of standard integrals do? \$\endgroup\$ – Neil_UK Dec 16 '20 at 9:56
  • \$\begingroup\$ Thanks or your answer. I only really know the basics of integration/differentiation (i.e. nothing further than differentiating \$2x^2 + 3x + 2\$ and trig functions). I just need someone to go through the mathematical process of deriving it, then I should be able to understand it better. \$\endgroup\$ – kendalmint24 Dec 16 '20 at 10:15
  • \$\begingroup\$ @kendalmint24 you'll need to know the derivative and integral of \$e^x\$, too, as well as chain rule of derivation, otherwise this will be impossible. \$\endgroup\$ – Marcus Müller Dec 16 '20 at 10:37
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Kirchoff's voltage law states that the algebraic sum of the voltages equals zero for any closed path (loop) in an electrical circuit. Writing the equation for the voltage loop with the inductor and resistor gives $$L \frac{di_L(t)}{dt}+Ri_L(t)=0 $$ Dividing with the resistance on both sides gives $$\frac{L}{R}\frac{di_L(t)}{dt}+i_L(t)=0 $$ The form of this differential equation indicates that that the solution for \$ i_L(t)\$ must be a function that has the same form as its first derivative. Such a function is an exponential function.

We make the guess that the solution is of the form \$i_L(t)=Ke^{st} \$ in which \$ K\$ and \$ s\$ are constants to be determined. Inserting the guess in the actual differential equation gives $$ \frac{L}{R}Kse^{st}+Ke^{st}=0$$ From this, we see that the equation is true if \$s=-\frac{R}{L} \$

Inserting this value for \$s \$ into \$i_L(t)=Ke^{st} \$ gives $$i_L(t)=Ke^{\frac{-tR}{L}} $$ Calling \$ \tau=\frac{L}{R} \$ gives $$i_L(t)=Ke^{\frac{-t}{\tau}} $$

All that's left is to determine \$K \$, and I will leave that to you. Hint: what is \$i_L(0) \$?

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  • \$\begingroup\$ Please consider changing "s" to a different variable, as it is, it is easily confused with the complex frequency as used in the Laplace transform. \$\endgroup\$ – S.s. Dec 16 '20 at 13:23
  • \$\begingroup\$ Thankyou! That was extremely helpful \$\endgroup\$ – kendalmint24 Dec 16 '20 at 16:16
  • \$\begingroup\$ The polarity you show for the emf across the inductor is wrong. Also, what is the implication of adding a ground connection? \$\endgroup\$ – Chu Dec 16 '20 at 16:42
  • \$\begingroup\$ @Chu I will remove the drawing, as it is not necessary. \$\endgroup\$ – Carl Dec 16 '20 at 21:24

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