# Solving LR Circuit with Laplace Transform

I have a RLC circuit where the capacitor is connected in parallel with a resistance and inductance in series. The battery is connected "in parallel" with the capacitor and the RL branches. At t=0 the battery is disconnected from the circuit. I need to find the voltage across each element using the Laplace transform.

Here’s what I did:

$$E = RI + L\frac{dI}{dt}$$

Taking the transform I arrive at:

$$\frac{E}{s} = R ℒ(I) + L(s ℒ(I) - I_0)$$

With $I(t=0) = I_0$ $$=> ℒ(I) = \frac{E/L}{s(R/L + s)} + \frac{I_0/L}{R/L + s}$$

Which gives

$$I(t) = \frac{E}{R} + [I_0 - \frac{E}{R}]e^{-Rt/L}$$

The thing is this result looks kind of weird and I feel like I did something wrong. Could anybody let know what I should do about this?

• I don't see a term for capacitance anywhere, where did it go? – Alfred Centauri May 23 '17 at 23:14
• The voltage across the capacitance is the same as that across the R and L branch, – Y2H May 23 '17 at 23:52
• @Y2H - So you're saying (and your final equation above is saying) that the value of the capacitor in the circuit doesn't matter? Think about that. Does that make sense to you? – user93237 May 24 '17 at 0:22

$$E = RI + L\frac{dI}{dt}$$
would describe the time evolution of the inductor's current for $t<0$, not for $t>0$.
For $t<0$, one usually assumes that the circuit is at steady-state: this assumption is used to calculate the initial conditions, in this case, the voltage across the capacitor and the current through the inductor. The initial conditions are thus $v_C(0+)=v_C(0-)=E$, for the capacitor voltage $v_C(t)$, and $i_L(0+)=i_L(0-)=E/R$, for the inductor current $i_L(t)$, where $v_C(0-) = \lim_{t\rightarrow 0^-} v_C(t)$, $v_C(0+) =\lim_{t\rightarrow 0^+} v_C(t)$, etc. Recall that the voltage across a capacitor and the current through an inductor cannot change instantaneously.
For $t>0$ you should write the differential equation associated to the circuit realized at $t>0$, which is composed of three series-connected elements, and then transform it in the Laplace domain with the initial conditions I described in the previous paragraph.
• Hi, if you replace $E$ with $v_C$ in my answer, we are proceeding likewise. You will still get the same weird function for $I(t)$. – Y2H May 24 '17 at 18:20
• @Y2H No, $v_C$ is a function of time and you get a second-order differential equation. – Massimo Ortolano May 24 '17 at 18:46