# LINKS CORRECTED: How to solve now that I know initial conditions

E

#### [email protected]

Jan 1, 1970
0
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=c44b&.dnm=d4f7re2.jpg&.src=ph

Do I incorporate the initial conditions as such? :
http://ca.pg.photos.yahoo.com/ph/enginquiry/detail?.dir=c44b&.dnm=6f97re2.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=c44b&.dnm=6696re2.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

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