General second order circuit problem - Find V(t) for t > 0

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  • #1
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Homework Statement



http://img396.imageshack.us/img396/2781/chapter8problem55oy6.jpg [Broken]

For the circuit above, find [itex]v(t)[/itex] for [itex]t\,>\,0[/itex].

Assume that [itex]v\left(0^+\right)\,=\,4\,V[/itex] and [itex]i\left(0^+\right)\,=\,2\,A[/itex].



Homework Equations



[tex]i_c\,=\,C\,\frac{d\,v_c}{dt}[/tex]



The Attempt at a Solution



I made a new diagram:

http://img507.imageshack.us/img507/4142/chapter8problem55part2zm4.jpg [Broken]

[tex]i_1\,=\,C_1\,\frac{d\,V_{C_1}}{dt}[/tex]

[tex]i_2\,=\,\frac{V_1\,-\,V_2}{2\,\Omega}[/tex]

[tex]i\,=\,C_2\,\frac{d\,V_{C_2}}{dt}[/tex]

KCL @ [itex]V_1[/itex]) [tex]i_1\,+\,\frac{i}{4}\,-\,i_2\,=\,0[/tex]


[tex]C_1\,\frac{d\,V_{C_1}}{dt}\,+\,\frac{C_2}{4}\,\frac{d\,V_{C_2}}{dt}\,-\,\frac{V_1\,-\,V_2}{2\,\Omega}\,=\,0[/tex]


[tex]0.2\,\frac{d\,V_{C_1}}{dt}\,+\,0.25\frac{d\,V_{C_2}}{dt}\,-\,V_1\,+\,V_2\,=\,0[/tex]



KCL @ [itex]V_2[/itex]) [tex]i\,+\,i_2\,=\,0[/tex]


[tex]C_2\,\frac{d\,V_{C_2}}{dt}\,+\,\frac{V_1\,-\,V_2}{2\,\Omega}\,=\,0[/tex]


[tex]\frac{d\,V_{C_2}}{dt}\,+\,V_1\,-\,V_2\,=\,0[/tex]


[tex]V_2\,=\,\frac{d\,V_{C_2}}{dt}\,+\,V_1[/tex]


Now substituting the KCL @ [itex]V_2[/itex] equation into the other equation:


[tex]0.2\,\frac{d\,V_{C_1}}{dt}\,+\,1.25\,\frac{d\,V_{C_2}}{dt}\,=\,0[/tex]


Here I am stuck. I don't know how to proceed, any hints? I know that I am supposed to get a second order differential equation for the circuit, but where from?
 
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Answers and Replies

  • #2
Ne0
12
0
Have you done circuits in the s-domain? That would make this problem much easier to solve for. It would become a simple two-mesh problem in which you could use Cramer's rule to solve for with the constraint equation for the dependant source. And I am pretty sure it is not going to be a second order differential equation since there is only one reactive element per mesh.
 
  • #3
489
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No, I have not done s-domain yet.
 
  • #4
berkeman
Mentor
59,039
9,141
I think you need to use your substitution to eliminate one of the unknowns, instead of getting an equation that still has both in it. And then you'll need to make an assumption about the form of the solution, differentiate it and plug it back in to solve, and then use the initial conditions for the full form of the final solution. By looking at the circuit, I'd guess the solution is a damped exponential, but it might have other terms....
 
  • #5
SGT

Homework Statement



http://img396.imageshack.us/img396/2781/chapter8problem55oy6.jpg [Broken]

For the circuit above, find [itex]v(t)[/itex] for [itex]t\,>\,0[/itex].

Assume that [itex]v\left(0^+\right)\,=\,4\,V[/itex] and [itex]i\left(0^+\right)\,=\,2\,A[/itex].



Homework Equations



[tex]i_c\,=\,C\,\frac{d\,v_c}{dt}[/tex]



The Attempt at a Solution



I made a new diagram:

http://img507.imageshack.us/img507/4142/chapter8problem55part2zm4.jpg [Broken]

[tex]i_1\,=\,C_1\,\frac{d\,V_{C_1}}{dt}[/tex]

[tex]i_2\,=\,\frac{V_1\,-\,V_2}{2\,\Omega}[/tex]

[tex]i\,=\,C_2\,\frac{d\,V_{C_2}}{dt}[/tex]

KCL @ [itex]V_1[/itex]) [tex]i_1\,+\,\frac{i}{4}\,-\,i_2\,=\,0[/tex]


[tex]C_1\,\frac{d\,V_{C_1}}{dt}\,+\,\frac{C_2}{4}\,\frac{d\,V_{C_2}}{dt}\,-\,\frac{V_1\,-\,V_2}{2\,\Omega}\,=\,0[/tex]


[tex]0.2\,\frac{d\,V_{C_1}}{dt}\,+\,0.25\frac{d\,V_{C_2}}{dt}\,-\,V_1\,+\,V_2\,=\,0[/tex]



KCL @ [itex]V_2[/itex]) [tex]i\,+\,i_2\,=\,0[/tex]


[tex]C_2\,\frac{d\,V_{C_2}}{dt}\,+\,\frac{V_1\,-\,V_2}{2\,\Omega}\,=\,0[/tex]


[tex]\frac{d\,V_{C_2}}{dt}\,+\,V_1\,-\,V_2\,=\,0[/tex]


[tex]V_2\,=\,\frac{d\,V_{C_2}}{dt}\,+\,V_1[/tex]


Now substituting the KCL @ [itex]V_2[/itex] equation into the other equation:


[tex]0.2\,\frac{d\,V_{C_1}}{dt}\,+\,1.25\,\frac{d\,V_{C_2}}{dt}\,=\,0[/tex]


Here I am stuck. I don't know how to proceed, any hints? I know that I am supposed to get a second order differential equation for the circuit, but where from?
With the reference senses you used, you have

[tex]i_1\,=\,-C_1\,\frac{d\,V_{C_1}}{dt}[/tex]

[tex]i\,=\,-C_2\,\frac{d\,V_{C_2}}{dt}[/tex]

and [tex]i_2 = -i[/tex]

You can use this with

KCL @ [itex]V_2[/itex]) [tex]i\,+\,i_2\,=\,0[/tex]

to eliminate [tex]i_1[/tex] and [tex]i_2[/tex]
 
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  • #6
84
0
Have you done circuits in the s-domain? That would make this problem much easier to solve for. It would become a simple two-mesh problem in which you could use Cramer's rule to solve for with the constraint equation for the dependant source. And I am pretty sure it is not going to be a second order differential equation since there is only one reactive element per mesh.
I'll second to that, s-domain (laplace) will make your life much more easier ;)
Neo, are you a mind-reader or something, was about to say about Cramer's rule as well :biggrin:
 
  • #7
489
0
Okay, I'll try s-domain.

http://img400.imageshack.us/img400/9269/problem55part3cs4.jpg [Broken]

[tex]i_c\,=\,\frac{V_1\,-\,0}{\frac{10}{s}}\,=\,\frac{V_1\,s}{10}[/tex]

[tex]i\,=\,\frac{V_1\,-\,0}{\frac{2}{s}\,+2}\,=\,\frac{V_1\,s}{2\,+\,2s}[/tex]

KCL @ [itex]V_1[/itex]:

[tex]-i_c\,-\,\frac{i}{4}\,-\,i\,=\,0[/tex]

[tex]-i_c\,-\,\frac{5}{4}\,i\,=\,0[/tex]

[tex]-\frac{V_1\,s}{10}\,-\,\frac{5}{4}\,\frac{V_1\,s}{2\,+\,2s}\,=\,0[/tex]

How do I proceed now?
 
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  • #8
SGT
Okay, I'll try s-domain.

http://img400.imageshack.us/img400/9269/problem55part3cs4.jpg [Broken]

[tex]i_c\,=\,\frac{V_1\,-\,0}{\frac{10}{s}}\,=\,\frac{V_1\,s}{10}[/tex]

[tex]i\,=\,\frac{V_1\,-\,0}{\frac{2}{s}\,+2}\,=\,\frac{V_1\,s}{2\,+\,2s}[/tex]

KCL @ [itex]V_1[/itex]:

[tex]-i_c\,-\,\frac{i}{4}\,-\,i\,=\,0[/tex]

[tex]-i_c\,-\,\frac{5}{4}\,i\,=\,0[/tex]

[tex]-\frac{V_1\,s}{10}\,-\,\frac{5}{4}\,\frac{V_1\,s}{2\,+\,2s}\,=\,0[/tex]

How do I proceed now?
Since you are looking for a zero input response, you must include the initial conditions in your equation.
Or you can use the hint I gave you in my previous post and solve the problem in the time domain.
 
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  • #9
489
0
Using your hint in a previous post, I obtained a new KCL @ [itex]V_1[/itex] equation:

[tex]i_1\,-\,\frac{5}{4}\,i\,=\,0[/tex]

[tex]-C_1\,\frac{dV_{c1}}{dt}\,-\,\frac{5}{4}\,C_2\,\frac{dV_{c2}}{dt}\,=\,0[/tex]

How do I get initial conditions and how to finally solve?
 
  • #10
SGT
Using your hint in a previous post, I obtained a new KCL @ [itex]V_1[/itex] equation:

[tex]i_1\,-\,\frac{5}{4}\,i\,=\,0[/tex]

[tex]-C_1\,\frac{dV_{c1}}{dt}\,-\,\frac{5}{4}\,C_2\,\frac{dV_{c2}}{dt}\,=\,0[/tex]

How do I get initial conditions and how to finally solve?
You must have only one variable.
Remember that
[tex]V_{c1} = V_{c2} - R i = V_{c2} + RC_2\frac{dV_{c2}}{dt}[/tex]
so,

[tex]\frac{dV_{c1}}{dt} = \frac{dV_{c2}}{dt} + RC_2\frac{d^2V_{c1}}{dt^2}[/tex]
For the initial conditions you must have
[tex]V_{c2}(0^+)[/tex] and [tex]\frac{dV_{c2}}{dt}(0^+)[/tex]

[tex]V_{c2} = V_{c1} + R i [/tex]
So, [tex]V_{c2}(0^+) = V_{c1}(0^+) + R i(0^+) = 4 + 2x2 = 8V [/tex]
[tex]\frac{dV_{c2}}{dt} = -\frac{i}{C_2}[/tex]
So, [tex]\frac{dV_{c2}}{dt}(0^+) = -\frac{i}(0^+){C_2} = \frac{2}{0.5} = 4V/s[/tex]
 
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