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step response of RLC circuit, two sources, IE7.11

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PG1995

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Hi

Please have a look on the attachment. It has many questions there. Please help me with them. It would really kind of you. Thank you.

Regards
PG

PS: I think I understand what I have to do with the two source. I think it's: Vs1 - Vs2. In this case, it would be: (4V - 12V)/LC.
 

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Hi there PG,

You probably know some of this already but i'll mention everything for completeness.

Part 1, switch at pos 1:
To start with, the switch is at position 1 so you just have an inductor in series with two resistors excited by two voltage sources. You need to calculate iL for t=+int to get the iL(0) you'll need for the time when the switch moves to position 2.
To simplify this a little, you can combine the two voltage sources into one.

Part 2, switch moves to pos 2:
Once the switch moves to pos 2, we have an RLC circuit with two voltage sources and there is some initial current through the inductor as found above. It's just a plain old RLC circuit again with some initial current in the inductor and the two sources can be reduced to one by combining them (subtraction). Note however that if the solution this time is the output voltage, we'll have to remember to adjust the output to the correct voltage later. In other words, the source voltage (looking at the bottom of the 18 ohm resistor as ground) will be -12+4=-8 volts now, but later once we have the output calculated we have to add that 12v back. So for example say we got 2.3v output that way, we'd have to add 12v back to get 14.3v later as the actual value of the output voltage.
When the capacitor is drawn with no initial conditions shown, it usually means that there is no initial voltage so that Vc(0)=0. If there is an initial voltage then it would be shown as something like ic=5.2v or something like that, or just Vc(0)=5 or similar. There is also a chance that the associated text declares an initial voltage but that's not the case here either, so we take Vc(0) to be 0v. If there had been an initial voltage there, we would have had to add that to the 4v supply before we start, and also then make Vc(0)=0v again. So for example if they declared Vc(0)=2v we would then have to change the 4v supply there 6v and then make Vc(0)=0 again. So it's pretty simple really, in that the 4v supply there could either be a separate supply or they wanted Vc(0) to be 4v and indicated that with an external supply.

To verify your results at least partly, the solution for the output voltage at time t=0.1 seconds after the switch is moved to pos 2 is very close to +10.60309 volts, as calculated and verified by simulation.

Have you done an RLC circuit before that had some initial current in the inductor?
 
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