I am given:

Kirchhoff’s voltage law states that the sum of the voltage drops across an inductor, L dI/dt, and across a resistor, IR, must be the same as the voltage source, E(t), applied to the circuit. The resulting ODE model is LdI/dt+IR=E(t). If a 12-volt battery is connected to an RL circuit with a ½ henry inductor, L, and a 10 ohm resistor, R , find the current I(t) given that I(0)=0.

Ok, so I understand the formula is $E(t)=L(dI/dt)+IR$, but what is confusing me is with the mention of I(t). How do I begin to solve this?


  • $\begingroup$ All I know is what you see there, unfortunately. I feel that this problem wasn't explained very clearly... $\endgroup$ – westhe32nd Jan 25 '14 at 23:07
  • $\begingroup$ If $E(t)$ is the voltage source, and the voltage source is a $12$-volt battery, then it seems like $E(t)=12$. Right? $\endgroup$ – Potato Jan 25 '14 at 23:07
  • $\begingroup$ I can't believe I missed that! $\endgroup$ – westhe32nd Jan 25 '14 at 23:08
  • $\begingroup$ Can you take it from there? $\endgroup$ – Potato Jan 25 '14 at 23:10
  • 1
    $\begingroup$ Context is good, but this isn't ECE-SE, so the lead is somewhat buried. The essentials are: you have a differential equation $Lf^\prime +Rf-E=0$ with boundary condition $f(0)=0$. I propose editing the question to reflect that (even though a complete answer already exists). $\endgroup$ – Jonathan Y. Jan 25 '14 at 23:16

Hint: this is a separable equation:

$$L(dI/dt) = E - IR$$

So, we can separate and integrate:

$$\int \dfrac{L}{E-IR}~dI = \int dt$$


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