# solving the circuit after initial conditions have been determined

Thank you to everyone who helped out with my previous post concerning the singular circuit.
Now, what is the best way for solving this circuit, assuming I know my
initial conditions for t=0+ ? http://ca.pg.photos.yahoo.com/ph/enginquiry/detail?.dir Ä4b&.dnm1...
Do I incorporate the initial conditions as such? : http://ca.pg.photos.yahoo.com/ph/enginquiry/detail?.dir Ä4b&.dnmo9...
Assuming I want to solve the above circuit for V3 using superposition: I have to sum the contribution to V3 from the 3 DC sources, one at a time right? So, let's say I am finding the output at V3 due the initial charge on C1 (call this contribution V3-dueto-1). To solve for
V3-dueto1, do I "shut-off" all the DC sources except V1(0+) by replacing them with a short circuit? Then, still solving for V3-due-to-1, I expect to obtain a differential equation with "unknown constants". To figure out what these unknown constants are, I express my differential equation for time t=0, and equate it to the DC output at V3 when all the capacitors are shorted. Then, I express my differential equation for time=infinity and equate it to the DC output voltage at V3 when all the capacitors are open-circuited.
Can someone confirm the validity of this approah ?
Is there another way ? Here is the Laplace circuit: http://ca.pg.photos.yahoo.com/ph/enginquiry/detail?.dir Ä4b&.dnmf9...
This would be simpler I supposed because there wouldn't be any "unknown
constants" to solve for, correct?
Any suggestions / corrections ? Am I on the right track here ? I get ridiculously long expressions when I solve the Laplace circuit by superposition ...
Thanks, Dan
âœ–
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Thank you to everyone who helped out with my previous post concerning the singular circuit.
Now, what is the best way for solving this circuit, assuming I know my initial conditions for t=0+ ? http://ca.pg.photos.yahoo.com/ph/enginquiry/detail?.dir Ä4b&.dnmÔf7re2.jpg&.src=ph
Do I incorporate the initial conditions as such? : http://ca.pg.photos.yahoo.com/ph/enginquiry/detail?.dir Ä4b&.dnmo97re2.jpg&.src=ph
Assuming I want to solve the above circuit for V3 using superposition: I have to sum the contribution to V3 from the 3 DC sources, one at a time right? So, let's say I am finding the output at V3 due the initial charge on C1 (call this contribution V3-dueto-1). To solve for
V3-dueto1, do I "shut-off" all the DC sources except V1(0+) by replacing them with a short circuit? Then, still solving for V3-due-to-1, I expect to obtain a differential equation with "unknown constants". To figure out what these unknown constants are, I express my differential equation for time t=0, and equate it to the DC output at V3 when all the capacitors are shorted. Then, I express my differential equation for time=infinity and equate it to the DC output voltage at V3 when all the capacitors are open-circuited.
Can someone confirm the validity of this approah ?
Is there another way ? Here is the Laplace circuit: http://ca.pg.photos.yahoo.com/ph/enginquiry/detail?.dir Ä4b&.dnmf96re2.jpg&.src=ph
This would be simpler I supposed because there wouldn't be any "unknown
constants" to solve for, correct?
Any suggestions / corrections ? Am I on the right track here ? I get ridiculously long expressions when I solve the Laplace circuit by superposition ...
Thanks, Dan
âœ–
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On 2/2/06 4:53 PM, in article snipped-for-privacy@g49g2000cwa.googlegroups.com, " snipped-for-privacy@yahoo.ca"

Go to a circuit book that tells you how to use Laplace transforms. That method is ideal for incorporating initial conditions. I did not check your work.
Remember, however. if you can solve a problem at all, there are others methods to do so as well. Use whatever tricks make it easier for you.
Bill
-- Ferme le Bush
âœ–
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I cannot access your files- permission needed. In any case- the circuit can be simplified and likely a nodal approach would be the best -convert Laplace "voltage" sources to current sources and you should have little trouble.
--

Don Kelly @shawcross.ca