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I'm just starting a differential equations class and so far I am completely lost with these application problems. Specifically, in this one problem asks to find a differential equation for the current of a circuit. The problem:

Determine a differential equation for the current $i(t)$ if the resistance is $R$, the inductance is $L$, and the impressed voltage is $E(t)$.

The examples in the book mention using Kirchoff's second law, which the book defines as "the impressed voltage $E(t)$ on a closed loop must equal the sum of the voltage drops in the loop." But the book does not give an example of how to set up a DE using this law so I'm pretty lost. How do I even get started?

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The voltage across a resistance $R$ with current $i(t)$ flowing through it is $i(t)R$ from Ohm's law. The voltage across an inductance $L$ with current $i(t)$ flowing through it is $L\frac{di(t)}{dt}$. If you had a capactitance $C$ with current $i(t)$, the voltage across it would be $\frac{1}{C}\int_{-\infty}^{t}i(t)dt$. I assume you have a circuit with a resistance $R$ in series with an inductor $L$ and an applied voltage $E(t)$. in that case we can write a single loop equation starting from the positive side of $E(t)$:

$$i(t)R + L\frac{di(t)}{dt} - E(t) = 0$$.

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  • $\begingroup$ That's exactly what the answer is. I just don't really understand. How do you know the $L\frac{di(t)}{dt}$ part? Is that just something you are supposed to memorize? Voltage is equal to inductance times the change in current? $\endgroup$ – Sabien Jun 10 '15 at 3:16
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    $\begingroup$ As far as network theory is concerned, you can take the definition of an inductor as a device whose voltage is proportional to the rate of change of current. In electromagnetics, this can be derived from Faraday's law of inductance. An inductor is typically a coil of wire, and a change in the current through it induces a voltage which is proportional to it's inductance. A capacitor is a device who's current is proportional to the rate of change of the voltage. A resistor is a device who's voltage is proportional to the current. $\endgroup$ – BruceZ Jun 10 '15 at 4:24
  • $\begingroup$ Thanks, that explanation really clears it up. I'm a little disappointed that I was expected to know this - no electronics knowledge is a prerequisite for the DE class I'm in, but I may have simply missed this explanation somewhere in the book. Appreciate the clarification. $\endgroup$ – Sabien Jun 10 '15 at 5:31

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