enter image description here

I tried to find the voltage on the capacitor as a function of time, Vc(t).
In the initial condition (t<0) the switch was closed and the current source was open circuit, so I found the initial conditions for the capacitor and the inductor ( VC(0)=65v, and IL(0)=1A).
At t=0, the switch will open and the current source connect to circuit like the picture here.
I'd tried to write the differential equation of the circuit and got something weird:

enter image description here

I know that the equation of RLC circuit 2d must be positive, otherwise, I will get one of the roots positive which is impossible.
Please, I need your help, what is important to me is to proof the equation so I can show the lecturer that he was wrong, but I'm not sure if what I did was right.
Here is the solution of the lecturer:
(A1 and A2 easy to find)

enter image description here

the decition of current in the inductor is true: enter image description here

  • \$\begingroup\$ Well, first of all, your initial conditions are wrong. \$\endgroup\$ – Jan Feb 9 at 20:01
  • 1
    \$\begingroup\$ @Jan why ??? the initial condition are true, the switch is closed and there is no current source, so in s.s the capacitor is open-circuit and the inductor short circuit, the voltage across the capacitor is a voltage divider, and the current through the inductor is voltage source divided by 2 resistors (1A) \$\endgroup\$ – Knowledge Feb 9 at 20:06
  • \$\begingroup\$ The schematic is not clear the current source is open circuit for t<0. Guess that is confusing \$\endgroup\$ – Huisman Feb 9 at 20:30
  • \$\begingroup\$ @Huisman but i fixed it... why no one can help me \$\endgroup\$ – Knowledge Feb 9 at 21:50
  • \$\begingroup\$ @Knowledge I get your lecturer's result. \$\endgroup\$ – jonk Feb 9 at 23:21

Since everyone seems to have been variously confused at different times about your question, let me re-draw and re-phrase it:


simulate this circuit – Schematic created using CircuitLab

(The above reflects the case after the switch opens and after the current source has been added, both at \$t=0\$.)

On the left is your new circuit (without initial conditions showing, as I think it is unnecessary in order to answer your question.) On the right is the same circuit through a simple, obvious Thevenin transformation about which I'm sure you agree. I prefer the right side because it makes it a little easier to answer your question about the lecturer's answer.

Let's use KCL for the node marked as \$V_t\$:

$$C\,\:\frac{\text{d}\,V_{\left(t\right)}}{\text{d}t}+\frac1{L}\int V_{\left(t\right)} \:\text{d}t=\frac1{L}\int V_{s\left(t\right)} \:\text{d}t$$

(I've placed out-flowing currents on the left and in-flowing currents on the right.)

Some development results in:

$$\begin{align*} \frac{\text{d}\,V_{\left(t\right)}}{\text{d}t}+\frac1{L\,C}\int V_{\left(t\right)} \:\text{d}t&=\frac1{L\,C}\int V_{s\left(t\right)} \:\text{d}t\\\\ \frac{\text{d}^2\,V_{\left(t\right)}}{\text{d}t^2}+\frac{V_{\left(t\right)}}{L\,C}&=\frac{V_{s\left(t\right)}}{L\,C}\\\\ \end{align*}$$

But we also know that the current, \$I_{\left(t\right)}\$, is everywhere the same throughout the right-hand circuit. So it follows that:

$$V_{s\left(t\right)}=39\:\text{V}-R\,I_{\left(t\right)} = 39\:\text{V}-R\,C\,\:\frac{\text{d}\,V_{\left(t\right)}}{\text{d}t}$$

Substituting in:

$$\begin{align*} \frac{\text{d}^2\,V_{\left(t\right)}}{\text{d}t^2}+\frac{V_{\left(t\right)}}{L\,C}&=\frac{1}{L\,C}\left(39\:\text{V}-R\,C\,\:\frac{\text{d}\,V_{\left(t\right)}}{\text{d}t}\right)\\\\ \frac{\text{d}^2\,V_{\left(t\right)}}{\text{d}t^2}+\frac{R}{L}\:\frac{\text{d}\,V_{\left(t\right)}}{\text{d}t}+\frac{V_{\left(t\right)}}{L\,C}&=\frac{39\:\text{V}}{L\,C} \end{align*}$$

Which is just a 2nd order diff-eq I'm sure you can solve.

But just looking at the characteristic equation, (\$s^2+650\,s+1\times 10^5=0\$), you would easily find \$s=-400\$ and \$s=-250\$. Which is a huge clue for setting up the general solution.

The lecturer's general solution is correct.

|improve this answer|||||
  • \$\begingroup\$ My problem is me how to decide the direction of current in the inductor and capacitor? when the switch is closed, I agree with this that direction of current in the capacitor as what you wrote, but the current direction of the inductor, before the switch is opened, into the left side ... because the voltage source is alone in circuit and easy to see it, am i right? \$\endgroup\$ – Knowledge Feb 10 at 10:37
  • \$\begingroup\$ I add another picture in the post, maybe what confuse me is to decide the direction of the currents before and after ... it always confuses me! \$\endgroup\$ – Knowledge Feb 10 at 10:40
  • \$\begingroup\$ @Knowledge It's a matter of taking a convention and sticking with it during analysis. The voltage at \$V_{\left(t\right)}\$ rises when the current direction is down as shown in the right-side diagram. So just stay with that convention. This also sets the direction through \$R\$ for analysis purposes. It's pretty basic. Yes, I completely ignored your initial condition direction. But that's just a sign change in your case, when analyzing the constants of integration. It doesn't change the analysis or the characteristic equations that develop. \$\endgroup\$ – jonk Feb 10 at 10:42
  • \$\begingroup\$ So this is that we were taught that the direction of the current we set before opening the circuit switch, we must continue with the same direction even after opening the circuit switch because I listened to the lecturer's instruction, the circuit analysis made that one of the signs in the equation I came up with was negative, The main problem here is the sign of one of coefficients , is that what causes problems, your opinion? \$\endgroup\$ – Knowledge Feb 10 at 10:51
  • \$\begingroup\$ @Knowledge If you solve out the general equation's specific constants of integration correctly, then you will find that the current will indeed be in that direction at \$t=0\$ and that this particular oppositely-directed current through \$R\$ will indeed lead to the right initial voltage for \$V_{s\left(t\right)}\$, too. There is no conflict here. None at all. In fact, I recommend that you also solve out \$V_{s\left(t\right)}\$ so that you can see this initial direction more easily. \$\endgroup\$ – jonk Feb 10 at 10:55

One way to get the initial condition is through the Thevenin equivalent circuit of the current source and the resistor in parallel of it. Then, the initial DC voltage of the cap and DC current of the inductor can be solved easily, considering L1 as a short, and C1 as an open circuit.

enter image description here

enter image description here

|improve this answer|||||
  • \$\begingroup\$ Don't forget that the switch opens at \$t=0\$. \$\endgroup\$ – jonk Feb 10 at 0:07
  • \$\begingroup\$ True, but to calculate the initial condition, which is the voltage of the cap before opening the switch, the circuit is right. \$\endgroup\$ – X J Feb 10 at 0:11
  • \$\begingroup\$ I think the OP already knows the initial conditions, since they are shown on the drawing already. The OP wants to confirm (or refute) the general solution provided by the lecturer after \$t=0\$. \$\endgroup\$ – jonk Feb 10 at 0:12
  • \$\begingroup\$ Wasn't that wrong? \$\endgroup\$ – X J Feb 10 at 0:13
  • \$\begingroup\$ I think the OP has the initial conditions correctly stated (though I don't care because the lecturer's solution leaves that as a detail.) And no, I do think the lecturer was right, per the development I performed here. But, of course, only the OP knows if I understood the question well enough. There seems to be a lot of confusion over the details. So I won't know until the OP responds. But the fact I get the same solution as the lecturer is ... suggestive. \$\endgroup\$ – jonk Feb 10 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.