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

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.

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

2 Answers 2


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.

  • \$\begingroup\$ Arrow indicates the positive side of the voltage. So there's no sign error. \$\endgroup\$ Jun 1, 2020 at 10:28
  • \$\begingroup\$ 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? \$\endgroup\$ Jun 1, 2020 at 15:12
  • 2
    \$\begingroup\$ Yes, it's called "passive sign convention". \$\endgroup\$ 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.

  • 2
    \$\begingroup\$ Arrow indicates the positive side of the voltage. So there's no sign error. \$\endgroup\$ Jun 1, 2020 at 10:28
  • \$\begingroup\$ 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. \$\endgroup\$
    – Hufi
    Jun 1, 2020 at 10:40
  • \$\begingroup\$ Please google "passive sign convention" and you'll get what I'm trying to say. \$\endgroup\$ Jun 1, 2020 at 10:59
  • \$\begingroup\$ 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. \$\endgroup\$
    – Hufi
    Jun 1, 2020 at 16:47
  • \$\begingroup\$ 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. \$\endgroup\$
    – jDAQ
    Jun 2, 2020 at 2:42

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