# Solving simple RLC curcit

#### Luljj

Joined May 21, 2023
4
For t>=0 the switch comes from(1) to (2). E=10V R=2 (OHM) C=3F L=2H. I'm not sure how to find iL0 and uc0 and when are uc0 and iL0 equal to zero and when not. If you can explain that to me I'd be very greatful, Thanks in advance

#### Papabravo

Joined Feb 24, 2006
20,392
With the switch in the initial position the voltage source E charges capacitor C up to the voltage of the source E. It will approach this value asymptotically and NEVER actually get there. If we wait long enough the difference betwee E and the charge on the capacitor C will be measured in very small units. Perhaps femtovolts, if that is small enough for you at 15 orders of magnitude.

When the switch breaks from it's initial position, but before making contact with the other side of the switch, the voltage source E and the R on the left of the switch are out of the picture. When the switch makes contact with the other side, the Capacitor will discharge through the two resistors and the inductor.

Do you now have a handle on what is going on?

#### Luljj

Joined May 21, 2023
4
Yes, but if he says for t>=0, the switch goes to position (2). Does that mean iL0=0 and uC0=0 cause there is no time for C to charge. Or maybe I'm wrong

#### Papabravo

Joined Feb 24, 2006
20,392
I think the problem wants you assume that a steady state is established with the switch connecting the voltage source and the resistor to the capacitor C1. This all happens for t<=0. A real switch cannot change instantaneously. I have a generic simulation with no component values, that shows the three régimes.
1. For t < 0, the source E charges the capacitor C1 through R1.
2. For an arbitrarily small interval just before t=0, the switch interrupts the charging and nothing is happening on either side of the capacitor C1.
3. At t=0 the switch connects the charged capacitor C1 to the resistor R2, R3 and the inductor L1.

You still need to solve the problem for the explicit behavior of the discharge. I believe it is the case that at t=0 there are no voltages or currents in the resistors or the inductor to the right of the switch.

#### WBahn

Joined Mar 31, 2012
29,165
Yes, but if he says for t>=0, the switch goes to position (2). Does that mean iL0=0 and uC0=0 cause there is no time for C to charge. Or maybe I'm wrong
No. By saying that the switch is at position (2) for t >= 0, they are implying that it is NOT at position (2) for t < 0. In this simply scenario, if it is not at position (2), it must be at position (1).

The key to working these problems is not to try to memorize a bunch of special cases where the voltage and/or current is zero. The think to understand is that energy cannot be created or destroyed, just moved around or converted from one form to another. Since the energy stored in a capacitor is related to the voltage across it, this means that the voltage on a capacitor physically cannot change instantly. The current through it can, but the voltage cannot. So whenever you have something like a switch changing state, figure out what the voltages across all of the capacitors was just before the switch changed and this will be the same voltages across those capacitors immediately after the switch changes. For inductors, the energy is related to the current through the inductor, so the current cannot change instantly and must be continuous across changes in state, such as switches opening or closing.

#### RoofSheep

Joined Mar 7, 2023
36
For t<0, the circuits is in the steady-state. All the inductors become short circuits and all the capacitors are open circuits. That's how you calculate the initial conditions.

For t>0, you have a second order circuit. The response will not be exponential. Energy will be exchanged between the capacitor and inductor.

#### Papabravo

Joined Feb 24, 2006
20,392
For t<0, the circuits is in the steady-state. All the inductors become short circuits and all the capacitors are open circuits. That's how you calculate the initial conditions.

For t>0, you have a second order circuit. The response will not be exponential. Energy will be exchanged between the capacitor and inductor.
I do not think that is correct. I think the simulation demonstrates that there is no exchange of energy between the capacitor and the inductor. The presence of the resistors changes everything. I would agree that without them you would see 2nd order behavior of oscillation with a decaying exponential envelope.

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#### Danko

Joined Nov 22, 2017
1,737
Energy will be exchanged between the capacitor and inductor.
I do not think that is correct. I think the simulation demonstrates that there is no exchange of energy between the capacitor and the inductor.
Simulation shows exchange:

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#### Danko

Joined Nov 22, 2017
1,737
I do not think that is correct. I think the simulation demonstrates that there is no exchange of energy between the capacitor and the inductor. The presence of the resistors changes everything. I would agree that without them you would see 2nd order behavior of oscillation with a decaying exponential envelope.

View attachment 294715
On diagram in post #7 amplitude of oscillations can not be more than initial capacitor voltage:

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#### Papabravo

Joined Feb 24, 2006
20,392
Yes, the energy of the initial condition is dissipated, in the resistors and the inductor. By "excahnge", I meant "back and forth" as in 2rd order oscillatory behavior. Sorry if that was unclear.
On diagram in post #7 amplitude of oscillations can not be more than initial capacitor voltage:

View attachment 294731
That is odd. Did I write the initial condition statement wrong?
I did, I enforced an initial condition on the node "c", rather than on the capacitor, and the UIC condition to skip finding the initial operating point. My mistake.

That looks much better.

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#### Danko

Joined Nov 22, 2017
1,737
By "excahnge", I meant "back and forth" as in 2rd order oscillatory behavior.
It still "back and forth", only difference is in value of damping coefficient:

#### Papabravo

Joined Feb 24, 2006
20,392
I see the light. It depends entirely on the values that you select for R, L, and C. The system for t>0 is 2nd order, but it can be under damped, critically damped, or over damped. The values I chose did fool me.

#### Luljj

Joined May 21, 2023
4
For t<0, the circuits is in the steady-state. All the inductors become short circuits and all the capacitors are open circuits. That's how you calculate the initial conditions.

For t>0, you have a second order circuit. The response will not be exponential. Energy will be exchanged between the capacitor and inductor.
So are initial conditions for this example iL0=0 and uc0=I*R +E, am I right?

#### WBahn

Joined Mar 31, 2012
29,165
So are initial conditions for this example iL0=0 and uc0=I*R +E, am I right?
What is I in your equation? Where does it come from?

#### Luljj

Joined May 21, 2023
4
What is I in your equation? Where does it come from?
I don't have I, but I'd put I=E/R and then find uc0 cause it's paralel to that brench. If I am not right, could you please write the correct way cause I feel like I understand it, but I'm not sure. And when the switch goes to position (2) will my Uc be same as uc0

#### WBahn

Joined Mar 31, 2012
29,165
I don't have I, but I'd put I=E/R and then find uc0 cause it's paralel to that brench. If I am not right, could you please write the correct way cause I feel like I understand it, but I'm not sure. And when the switch goes to position (2) will my Uc be same as uc0
This is the I that I was referring to:

So are initial conditions for this example iL0=0 and uc0=I*R +E, am I right?
If you try to put I=E/R, then you claiming that E is the voltage across that resistor (and I'm only guessing which resistor you are referring to).

If the switch has been in position (1) for a long time, what is the current in the left-most resistor?

Given that current, what is the voltage across it?

Given that voltage across it, what is the voltage across the capacitor at that time?