# Find the differential equation relating $v(t)$ and $i(t)$

Find the differential equation relating $$\v(t)\$$ and $$\i(t)\$$ of the AC circuit in the following figure.

My approah:
Considering the inductor:
$$v(t) = L\frac{di_{L}}{dt}$$ $$i_L = \frac{1}{L}\int v_{(t)} \,\mathrm{dt}$$

Considering the capacitor: $$i_{C} = C\frac{dv_{(t)}}{dt}$$

$$\therefore i_{(t)} = i_{L} + i_{C}$$ $$i_{(t)} = \frac{1}{L}\int v_{(t)} \,\mathrm{dt} + C\frac{dv_{(t)}}{dt}$$ $$\frac{d^2v_{(t)}}{dt^2} - \frac{1}{C}\frac{di_{(t)}}{dt} + \frac{1}{LC}v_{(t)} = 0$$

Is this correct? Please correct me.

• May want to use $\text{d}t^2$ in the denominator of the first term.
– jonk
Jun 1, 2020 at 7:02
• @jonk thanks for pointing that. Jun 2, 2020 at 1:00

I think there is a sign error in the current equation and therefore sign errors in the final result. If the direction of v(t) was flipped, the result would be correct i think.

• Arrow indicates the positive side of the voltage. So there's no sign error. Jun 1, 2020 at 10:28
• Interesting, i didn't know there was any controversy in the convention about the voltage arrow direction. I learned that the arrow is pointing from positive to negative potential. So i would argue that the positive potential would be in the right side of the figure. Is there any international norm or highly reliable textbook covering this? Jun 1, 2020 at 15:12
• Yes, it's called "passive sign convention". Jun 1, 2020 at 15:21

There is a sign error in your solution. Note, that i(t) and v(t) are not on the same direction. Thus i(t) = -(i_L(t) + i_C(t)).

You could verify your solution by applying Laplace transform. The complex impedance of L and C in parallel is s*L/(s^2*LC+1) and therefore we have V=-sL/(s^2*L*C+1)*I (the negative sign is due to the different directions of V and I). A little bit of algebra gives

s^2*V + 1/(L*C)V = - sI/C

Remembering that the multiplication with s is equal to taking the derivative, you end up with the differential equation.

• Arrow indicates the positive side of the voltage. So there's no sign error. Jun 1, 2020 at 10:28
• Yes, there is a sign error. If both voltage and current arrows point in the same direction, we have U=ZI. If they point in different directions the equation is U=-ZI.
– Hufi
Jun 1, 2020 at 10:40
• Please google "passive sign convention" and you'll get what I'm trying to say. Jun 1, 2020 at 10:59
• Obviously, not the same convention is used in all the countries! In Europe (or at least in the German speaking countries), if the voltage arrow points from A to B, the voltage V_AB is the difference of the potentials at A minus the potential at B. This means, that if V_AB>0 then the potential at point A is higher than that at point B. It seems that in the USA (or may be the English speaking countries), the contrary convention is used.
– Hufi
Jun 1, 2020 at 16:47
• I have never seen this German convention usage of the arrow for specifying $v(t)$, so I agree with @RohatKılıç regarding it not having a sign error. But in the end it is up to the OP to figure what notating is being used in the book he's using.
– jDAQ
Jun 2, 2020 at 2:42