***Updated***Circuit experts: Last question.

[Bluga]

Last question:

How can i find Vx in terms of Vs? Circuit

Thanks alot.

 

[blahblah99]

when you close the switch, you are grounding that node, so all current on left side is going into ground, and all energy stored on the right side is going to ground also. Hence, there is a 'zero-input response' because there are no inputs (ie, no current flowing into the components).

when the switch is opened after being closed for a long time, the components are being charged up (current is flowing) so there is an input to the system.

 

[RossGr]

The way I see it is that after the switch has been open for a long period of time the cap. will have charged to the battery voltage and total circiut current will be 0. So your zero input responce is 0 current.

Things will start to happen when you close the switch. Batter current will flow through the switch and the resister nearest the battery, the cap will discharge and the inductor will fight that current change, not sure what the current decay graph will look like, but depending on the capasitance and inductance it could be underdamped and oscilate or overdamped which would mean a rapid decay. Just the right pairing of inductance and capasitance will create critically damped system which has a different pattern of current decay.

Edit, just reread the last part of your question.

A long period of time after the switch is closed, current will continue to flow through the battery and resister but the capasitor will have have discharged so there will be zero current in the right hand side. Since the battery/ resister circiut is still conducting you do not have the zero input response of no current flow and you will be reading zero volts across the cap and/or inductor. (The inductor will always read 0 volts in a 0 current situation.)

 

[Bluga]

Thanks for the clear explanations!

Also is this circuit a zero-state response as well?

 

[rimshaker]

It's just what it sounds like... a system (a circuit in this case) which needs no external excitation for it to start doing something. With the switch off, nothing is happening except that the capacitor is charged to whatever value the DC battery is at, there is no current flowing whatsoever anywhere. With the switch on, you now have 2 circuit loops with current coming from the battery side and the capactor side. After a few time constants have passed, the cap eventually discharges completely... so you're left with a simple battery/resistor circuit.

Yea, that last circuit you asked about is considered zero-input response as well. After both switches activate, you'll be left with a simple decaying RLC circuit (the middle part).

 

[Bluga]

i just updated a question, please help me out, thank you.

 

[blahblah99]

use the final value and initial value theorem to find the voltage across the capacitor to find voltages at t=infinity and t=0.

if you want a general equation for voltage across that cap after the switch is closed, redraw the circuit with the closed switch and take the laplace of all components, use kirchoff's voltage law (sum of voltage around a loop is zero), and inverse-transform the result. i think that's correct... by the way, what class is this? i remember taking a basic circuit analysis class a few years ago and had to do this.

 

[Krakerjak]

Thats what im doing in my Math class. Im in first year engineering.
Probably the samy type oc class bluga is in as well

 

[Bluga]

thanks for the reply. It's an introductory course on linear circuits.



<< Thats what im doing in my Math class. Im in first year engineering. >>



In Math class?

 

[Bluga]

One last question

 

[blahblah99]

you change those v sources into current sources using norton's equivalent theorem? (i dont remember), but Vs in series with R1+R2 gives you

Vs/(R1+R2) in parallel with R1+R2.. do the same with the 200Vx and R3 and from there, add the currents together, find the equivalent R and re-transform it back.

 

[Bluga]

thanks alot!

I'm all set for final tomorrow