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RLC Circuit, Initial Conditions! - Noob Question

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fouadalnoor

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Hello guys,

Can you please tell me how I know the initial conditions of this circuit (attached).

It says the switch is left open for a "long" time and then at t=0 its closed. I want to find the current through the 3ohm resistor for t<0.

What I know up to now is that:

the current through the 3ohm resistor should be given by@

i(t) = Ae-t + Be-2t <--- check that this is correct...

but I need two initial conditions to find A and B and thus know how the current changes.

Thanks for your help.

Fouad
 

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Oops, I didn't look closly at the diagram. "For a long time" means the inductor current is zero and capacitor voltage is 20V.
 
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yeah, but wouldnt it be 10v (since there is a potential divider?)

Also, I cant seem to work out A and B from my initial conditions...

I know that at t=0, then i=io since current will just start to flow from the cap as it starts to discharge?

I dont know actually... im confused about what happens when the switch is closed ...
 
I think you realize initial charge on cap is 10 vdc.

Have learned differential equations yet?

Think dampened resonance. (energy flows back and forth between cap and inductor with power dissipation due to loop current through 3 ohm resistor)
 
I'll show you the answer without giving you the equation:

The switch in the simulation closes at T=1 (not 0) so that the initial conditions can be established. The switch has an Off resistance of 100GΩ and an On resistance of 1µΩ. The switch is closed when V(On) > 1, so V2 is used to control the switch for the purpose of the simulation.

As you would expect, the initial conditions are that C1 is charged to 10V, and I(L1)=0. After the switch closes, the simulation shows a classic RLC response where the transient is highly damped. I'll let you write the equation...
 

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Hmm I dont get what the V(a) is for. Is that the voltage across R1?

What I do get is that at t<0 (T=0 in your simulation) the switch is left open for a long time and the capacitor has charged up to 10v because of the potential divider. This means there is no potential difference between the RLC circuit and the 5ohm resistor, so the curent through R1 is 0A (as shown on the simulation). Now when the switch is closed at t=0 (t=1 in your case) then the current is increasing and then decaying to 0A (and V(a) is 0 v but still not sure what that is pointing to..)

so really..its

t<0 i=0

t=0 i = ? - some function

t>0 i = 0 (as t tends to infinity)

I cant really get anywhere with these values with my equation: (t) = Ae-t + Be-2t as I can only get one constant..
 
fouadalnoor,

i(t) = Ae-t + Be-2t <--- check that this is correct...

No, your characteristic equation should be i(t) = Ae^-t + Be^-2t . (eq. 1)

You are right, you need two initial conditions. Both conditions are at t=0 .

The first condition is applied for eq. 1, 0 = A + B

Taking the derivative we get d(i(t))dt = -Ae^-t - 2Be^-2t . (eq. 2)

The current through the inductance is continuous and zero at t=0 , but the voltage across the inductance is equal and opposite of the voltage across the capacitance at t=0.

Therefore Ld(i(t))/dt = 1*d(i(t))/dt =-10 , so substituting into eq. 2 at t=0 we get -10 = -A -2B . Solving 0 = A + B and -A -2B = -10 , we easily get A=-10, and B=10 .

Therefore the complete equation is i(t) = 10e^-2t - 10e^-t

Ratch
 
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RCinFLA,

Think dampened resonance. (energy flows back and forth between cap and inductor with power dissipation due to loop current through 3 ohm resistor)

Not in this circuit. The dampening factor is too high for it to oscillate.

Ratch
 
Hmm I dont get what the V(a) is for. Is that the voltage across R1?...

No, V(a) in the simulation is the voltage at the node labeled A with respect to the ground symbol, which is effectively the voltage across the switch. At the instant before the switch is closed, the voltage at that node is 10V, which is also the voltage at the capacitor. When the switch is closed, the voltage instantly goes to zero, because the switch resistance is effectively zero. It is the voltage at node A that establishes the initial conditions for the circuit just as the switch closes. The capacitor discharges through the inductor (whose current was zero) and the 3Ω resistor. The equation you are looking for describes the current through the inductor vs time, starting from the initial conditions described above.
 
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Hi there fouad,


When you solve for initial conditions you focus on what *does not* change and try to use that to figure out the conditions on what does change.

For this circuit, we start with the switch open at t=0- and it's been open since t=-inf, so the inductor goes to a short circuit and the capacitor goes to an open circuit so we have at t=0- these conditions:
iL=0 amps
vC=10 volts
The current is 0 amps because the capacitor is open after a long time, and the two five ohm resistors causes a voltage divider effect which divides the 20v source in half so the cap voltage is 10 volts. This is all before the switch is closed.

Next the switch is closed and we consider that to be at the time t=0+, or just between t=0- and t=0+.
At this time the circuit is divided into two parts, the 20v source and two 5 ohm resistors, and the 3 ohm resistor and L and C. We can break off the source and 5 ohm resistor now, and we are left with the right hand side 3 ohm and L and C.
Since at t=0- the cap voltage was 10 volts, that becomes the driving force now. Everything that happens next is due to that cap voltage because the inductor current is zero so we wont get any energy from there. Before t=0- all of the inductor energy if any would have dumped into the cap or resistor.

Now we first focus on what can not change in that tiny tiny short interval between t=0- and t=0+. That's considered to be no time at all so that means any small time period like delta t or dt is really zero, and this is what makes finding the initial conditions possible.
We know that the inductor current can not change in zero time, and the cap voltage can not change in zero time (and remember we only care about the time between t=0- and t=0+ which is zero time) so the cap can be considered a voltage source of 10 volts and the inductor a current source of zero amps. The derivatives however may in fact be non zero as you know because the derivative is defined for dt=0 (ie we let dt approach zero to get the derivative).

Lets solve for di/dt at t=0+.
At that time the cap voltage is 10 volts, the inductor current is 0 amps, the drop across the resistor is zero, so we can use:
v=L*di/dt
so solve for di/dt.
We know the cap voltage is 10 volts and that cant change in zero time, so that makes v=-10 volts (it's negative because of the orientation in the circuit).
We know L=1 because that is given, so we get:
v=L*di/dt
-10=1*di/dt
and solving for di/dt we get:
di/dt=-10

So now you know at t=0+ that
i=0
and
di/dt=-10
So there you have two initial conditions.

Does that help you find the solution now?

Sometimes it helps to list all the conditions at t=0- and all at t=0+ too like so:
vR(0-)=0
vL(0-)=0
vC(0-)=10
iR(0-)=0
iL(0-)=0
iC(0-)=0

vR(0+)=0
vL(0+)=-10
vC(0+)=10
iR(0+)=0
iL(0+)=0
iC(0+)=0

Notice what changed at t=0+ was the inductor voltage, but we were able to use the fact that the cap voltage did NOT change to get that inductor voltage. From there we were able to get the required di/dt.
 
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...
We know the cap voltage is 10 volts and that cant change in zero time, so that makes v=-10 volts (it's negative because of the orientation in the circuit).
....

Huh? Look at V(c) green line in the plot. I(R1) = I(L1) = blue line in the plot.
 

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Huh? Look at V(c) green line in the plot. I(R1) = I(L1) = blue line in the plot.

Hi Mike, you are correct that V(c) is +10V. However, for the purpose of the initial contidions analysis, the current that is being solved for, I, has been defined to flow towards the capacitor. When the switch is closed, capacitor current flows out, opposite to the way I was defined. So you're both right, but the explanation could be a little better. Hope that helps.
 
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Here is the simplified circuit that exists immediately after the switch is closed, showing a consistent view of voltage polarity, current direction and the initial conditions.
 

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...
Therefore the complete equation is i(t) = 10e^2t - 10e-t

Ratch

No, it cannot be that because it blows up. It looks to me that the correct equation is as shown below:
 

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Here is the simplified circuit that exists immediately after the switch is closed, showing a consistent view of voltage polarity, current direction and the initial conditions.

Hi. Again, you are correct. But this is an acedemic question, so the answer must be given in terms of the orignal definition, or else it's points off. The teacher is the 'customer' in all cases, so his definition always the correct one :) But thanks for your input and insight.
 
MikeMl,

You are correct in that I made a typo by neglecting to write a minus sign in one of the terms. It should have been I=10e^-2t - 10e^-t . My current value is opposite with respect to yours because I used a clockwise current direction. The coefficients should be correct, however.

Ratch
 
**broken link removed** Originally Posted by MrAl **broken link removed**
...
We know the cap voltage is 10 volts and that cant change in zero time, so that makes v=-10 volts (it's negative because of the orientation in the circuit).
....


Huh? Look at V(c) green line in the plot. I(R1) = I(L1) = blue line in the plot.

Hello Mike,

You'll note that you took that line out of context just a little, as others have noted when we solve for the current in the inductor we have to stay consistent with the current direction that was originally going through the inductor at some time before t=0. That would have made the current through the inductor oriented clockwise in that schematic (as usual we use conventional current flow, positive to negative). The lower case 'v' is the voltage with its polarity the way it would appear just after the switch is closed. This is overdamped, but it would work the very same way in the underdamped case if we made the cap value equal to say 0.005 F instead of 0.5 F.
If we didnt stay consistent in this way then the circuit would not work properly if we opened and closed the switch repeatedly as in PWM.

Using the definition for the inductor:
v=L*di/dt

that makes v=-10 volts.

If you still dont believe this, try modulating the switch using PWM at say 50 percent duty cycle and see what result you get.

Alternately we can assume a different current direction, but if we do chose to do that then we have to stay consistent with that direction too. That would make the capacitor voltage be negative at the start, and it would in fact stay negative, but for the purposes of evaluating the current v becomes the opposite polarity.

Maybe that is what caused the confusion. It's not that Vc changes itself, it's just that the polarity of v has to change for the purpose of evaluating the current through the inductor after the switch has just been closed (in the equation v=L*di/dt). Vc stays positive using conventional current flow.

Its good to discuss this because polarity in circuits like these is very important.
At the very least we should stay consistent with the direction of the current. If we start out clockwise, then we should finish clockwise or else note that we switched direction. Note that initially with the 20v source the current is oriented clockwise until it falls to zero, but then even a tiny leakage current in the cap 1e-32 amps would keep it clockwise. Thus it calculates out clockwise and then changes polarity, but you could say that it kept its polarity but changed direction if you really wanted to, as a positive current in the positive direction is the same as a negative current in the negative direction, and a positive current in the negative direction is the same as a negative current in the positive direction.
Interesting, if we look at the circuit in a circuit analyzer and allow time for the 20v source to pump up the cap, the current is positive in the resistor (and inductor) until the cap charges up, then when the switch is closed the current will dip negative. Try it.
 
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The voltage at node C (see my circuit) is always positive with respect to the bottom node (GND) That is not negotiable!!!

Fixing your (IMHO: poor) choice that current flows clockwise (instead of the actual direction of counter clockwise) can only be fixed by putting a - sign in the current term; not by arbitrarily flipping the capacitor voltage to -10V!
 
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Hi again Mike,


As i was saying earlier, it's a good idea to discuss this because this is something that comes up in circuits like these and it can become confusing.

If you read my post over, you'll see that i did not say that the cap voltage 'flipped'.

As Brownout was nice enough to point out, this is a semi academic issue here, and the ultimate answer depends on what the instructor is looking for. He may be satisfied with a definition of current in either direction as long as that direction is pointed out.

I, on the other hand, come from the real world, where in the real world i usually base my results on the sum off *all* parts of a problem, not on the sum of *some* of the parts. I see this problem has having more than one part but dont restrict my analysis to only one part of the circuit, but include the *entire* circuit in my analysis. That means that for this circuit i want to include as much as possible into the result, and that in turn means that i have to consider what happens at times before we close the switch. It's true that the solution called for is for what happens *after* the switch is closed, but there are other parts shown too and the only way we can solve for the initial capacitor voltage is to include these other parts of the circuit and take it from there. Having done that, i dont want to just forget that these other parts were once in the circuit too and that they caused the initial response that i have to start out with in order to get the required solution that was asked for.
Note also that we can not even solve for the cap voltage without considering what happened at times before the switch is closed, so we end up with an inductor current direction from that to start with.

To illustrate this point a bit further, i've included a drawing. Refer to this drawing for the following discussion.

First analyze the circuit in Fig. A for the inductor current after the switch is closed at t=0. The initial cap voltage is 10 volts with the positive on top and negative on bottom (ground). That puts +10v at the left side of the inductor. The initial inductor current is zero. Express the inductor current as an equation.

Second, analyze the circuit in Fig. B for the capacitor voltage if the switch has been open for a very long time so steady state has been reached, then close the switch at t=0 and again analyze the current through the inductor solving for the current with time. Express the inductor current as an equation.

Next compare the inductor current equations found above from Fig A and from Fig B.
Now explain what role the extra parts in Fig B play and why they might be included in that circuit.

If this doesnt seem to help, then repeat the second analysis, but this time starting at t=0 the switch is open and there is no initial energy in any of the elements so the cap voltage and inductor current are both zero to start with at t=0, then analyze the inductor current until t=10 seconds, then at t=10 seconds turn on the switch and again analyze the inductor current. Express the current before t=10 seconds as an equation, and express the current after 10 seconds as another equation. Compare the two equations.
Now ask the question, "What happened to the direction of the current in the inductor after the switch closed?"
 

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