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Well, I've the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

In the circuit I want to find the voltage \$\text{V}_{\text{Y}_1}\left(t\right)\$ as a function of time, after I open the switch (this switch was closed 'forever' theoretically seen). And the question is: is my derivation for the voltage across the inductor correctly?


My work:

When the switch opens, the direction of the current in the inductor does not change, so the current will go in the direction of the red arrow in the circuit I've drawn down here:

schematic

simulate this circuit

When the switch is open we can find the current in the circuit using Faraday's law:

$$\text{V}_{\text{R}_1}\left(t\right)+\text{V}_{\text{R}_2}\left(t\right)+\text{V}_\text{C}\left(t\right)=-\text{V}_\text{L}\left(t\right)\tag1$$

Using the voltage and current relations in a resistor, inductor and capacitor we can write equation \$(1)\$ in terms of the current, \$\text{I}\left(t\right)\$, that will run trough the circuit when the switch is open (this current is the same trough all the components because it is a series circuit):

$$\text{I}'\left(t\right)\cdot\text{R}_1+\text{I}'\left(t\right)\cdot\text{R}_2+\text{I}\left(t\right)\cdot\frac{1}{\text{C}}=-\text{I}''\left(t\right)\cdot\text{L}\tag2$$

In order to solve this DE, we need two initial conditions:

  1. The first one is pretty simple, when the switch is closed the maximum current that will run trough the coil is equal to \$\text{V}/\text{R}_1\$. So when I open the switch at \$t=0\$ that current will be running trough the inductor: $$\text{I}\left(0\right)=\frac{\text{V}}{\text{R}_1}\tag3$$
  2. At the point \$\text{I}'\left(0\right)\$ we get that: $$\text{I}'\left(0\right)=0\tag4$$

So, now we are able to solve that DE that is written in equation \$(2)\$. I solve it using Mathematica.

Now, in order to find the voltage across the coil we can use the voltage current relation in an inductor:

$$\text{V}_\text{L}\left(t\right)=\text{L}\cdot\text{I}_\text{L}'\left(t\right)\tag5$$

So, in the end we get that the voltage at the node \$\text{Y}_1\$ is given by:

$$\text{V}_{\text{Y}_1}\left(t\right)=\text{L}\cdot\text{I}'\left(t\right)\tag6$$

Where \$\text{I}'\left(t\right)\$ is the derivative of the solution to our DE out of equation \$(2)\$.


If I plot my solution, using the values that are given in the circuit I get the following graph:

enter image description here

QUESTION: is my method of derivation right or did I make a mistake?

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Your equation (4) is not valid.

At \$t=0^+\$ you can't assume \$I'=0\$.

Before the switch opened, everything was in equilibrium so that there was no voltage across the inductor to change its current. But after the switch opens, this equilibrium is disturbed, and you can't assume that \$I'=0\$ any longer.

Keep track of what your state variables are. They are the inductor current and the capacitor voltage. You should expect to use both state variables to find your initial conditions.

In this case, you need to use the initial condition for the capacitor voltage, along with the known loop current, to find an initial condition \$I'(0^+)\$.

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  • \$\begingroup\$ Okay, and how can I find the value of \$I'(0^+)\$? \$\endgroup\$ – Polik Tuyip May 21 at 16:26
  • \$\begingroup\$ @PolikTuyip, write the KVL equation for the loop at \$t=0^+\$, and you'll get the voltage across the inductor, which will give you \$I'\$. \$\endgroup\$ – The Photon May 21 at 16:28
  • \$\begingroup\$ Can you help me doing that, I do not completely see what you mean. Thanks for your help anyway \$\endgroup\$ – Polik Tuyip May 21 at 16:29
  • \$\begingroup\$ @PolikTuyip, you already found the loop current at \$t=0^+\$. So you know the voltage across the two resistors. What do you know about capacitors that will let you get the capacitor voltage at \$t=0^+\$? \$\endgroup\$ – The Photon May 21 at 16:30
  • \$\begingroup\$ The capacitor voltage at \$t=0\$ equals \$10V\$ because it charged up when the switch was closed. \$\endgroup\$ – Polik Tuyip May 21 at 16:31
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Sorry, it is incorrect, but you just asked a yes or no question.

short Answer

since at t=0, I=10V/100ohms, =0.1A and Vc=10V every part will have exactly 10V across it. Thus Y1=-10V. This should be intuitive with Ohm’s Law or KVL. (And the switch has 10V across it too at t=0 due to equal resistors.)

There is no instant change in current or Cap voltage. This is because the parts are ideal with no ESR or DCR.


For high values of series R into C. At t=0 The impedance of XL=XC at some frequency which just happens to be near 100 Ohms so the flyback is (almost) critically damped starting near or = -10V.

You can work it out by Laplace, impedance ratios, or time domain ( and other ways).

Don't forget the Ic(t)=CdVc(t)/dt and VL(t)=LdI /dt while I, dI/dt & dt are equal in all parts now, while 100+100 = 200 ohms in the loop.

enter image description here

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    \$\begingroup\$ I do not understand what you're saying in your answer, if I'm honest. Where does my calculations go wrong? \$\endgroup\$ – Polik Tuyip May 21 at 15:32
  • \$\begingroup\$ Your Eq. 2, 4 are incorrect. After, Vc(0)=0 & I(0)=I’(0)=0.1A, (no change) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 21 at 16:20
  • \$\begingroup\$ Why is equation 2 incorrect? And why does $I'(0)=0.1$? \$\endgroup\$ – Polik Tuyip May 21 at 16:23
  • \$\begingroup\$ \$I'\$ can't have units of amps. \$\endgroup\$ – The Photon May 21 at 16:27
  • \$\begingroup\$ (2) the cap voltage will not change instantly at switch opens thus Vc(0)=10. (Typo) And the inductor current will not change instantly, thus I(0) =I’(0) must be 10V/100 Ohms =0.1A just before and after switch opens \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 21 at 16:49

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