- #1
SpeedBird
- 28
- 0
hey guys
i posted this in the electrical engineering section
but realized this may have been a better place to
post since i am studying this in university :-)
*************************************
*************************************
hello there,
i'm wondering how to go about solving this problem
A diagram of the circuit can be seen here
http://www.compsoc.nuigalway.ie/~filter/images/circuit.gif
(i threw it together in microcap) the pulse sourse is actually
meant to be a 2Amp current source. the capacitor
has a value of 1micro-Farad.
By using superposition i have to find the steady state DC
current flowing in the 20 Ohm resistor. i could usually do
this easy but the capacitor is throwing me off. I am assuming
that once a steady state has been reached, the capacitor
will be fully charged and no current will actually be flowing
into it right? if this is the case.. i decided that i could ignore
its effects.. because it should really have any. and this
appears to work. i don't know if its a fluke or not though.
the answers are 0.4A flows in the 20 Ohm resistor due to
the current source and 0.1A flows in the 20 Ohm resistor
due to the voltage source.
any ideas on how to go about this would be welcome.
i need my mind to be put at rest :-) cheers, Nik
i posted this in the electrical engineering section
but realized this may have been a better place to
post since i am studying this in university :-)
*************************************
*************************************
hello there,
i'm wondering how to go about solving this problem
A diagram of the circuit can be seen here
http://www.compsoc.nuigalway.ie/~filter/images/circuit.gif
(i threw it together in microcap) the pulse sourse is actually
meant to be a 2Amp current source. the capacitor
has a value of 1micro-Farad.
By using superposition i have to find the steady state DC
current flowing in the 20 Ohm resistor. i could usually do
this easy but the capacitor is throwing me off. I am assuming
that once a steady state has been reached, the capacitor
will be fully charged and no current will actually be flowing
into it right? if this is the case.. i decided that i could ignore
its effects.. because it should really have any. and this
appears to work. i don't know if its a fluke or not though.
the answers are 0.4A flows in the 20 Ohm resistor due to
the current source and 0.1A flows in the 20 Ohm resistor
due to the voltage source.
any ideas on how to go about this would be welcome.
i need my mind to be put at rest :-) cheers, Nik
Last edited by a moderator: