a few circuit analysis questions.

[SpeedBird]

hey guys
i posted this in the electrical engineering section
but realised this may have been a better place to
post since i am studying this in university :-)
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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. im 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 dont 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

[ehild]

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. im 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?

[/QUOTE}

You are exactly right. :smile:

[Quote]
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.





So it is 0.5 A altogether downward. And the same to the left through the 10 ohm resistor and 1 A on the other 10 ohm resistor connected parallel to the capacitor, and there is 10 V across the capacitor.

ehild

[SpeedBird]

thanks very much!
one thing though..
how did you come to the conclusion that there is 10V across the capacitor?
cheers

Nik

[SpeedBird]

yes i know that, but where did the extra 5 volts come from?

[ehild]

yes i know that, but where did the extra 5 volts come from?

From the voltage across the 10 ohm resistor connected in series with the voltage source. There is 0.5 A current flowing through it, and it produces 5 V across it, according to Ohm's law. This adds to the 5 V of the voltage source.
And the 0.5 A current through the 20 ohm resistor produces 10 V and it is connected to the terminals of the capacitor. The current through the other 10 ohm resistor is 1 A, it also produces 10 V.
Those 10 ohm and 20 ohm resistors are connected parallel with each other, and with the capacitor, too. The same voltage has to be across all of them.

ehild