Engineering Circuit with capacitor, current source, voltagr source and a switch

AI Thread Summary
The circuit analysis involves determining the current across a 2-ohm and a 5-ohm resistor before and immediately after closing a switch at t=0. Before the switch is closed, the current through the 2-ohm resistor is 0.15 amps, while the current through the 5-ohm resistor is not specified. At t=0, the current through the 2-ohm resistor is 0 amps due to the capacitor being fully charged, resulting in no instantaneous change in voltage. Conversely, the current through the 5-ohm resistor is not equal to 0 at t=0 because it is influenced by the voltage divider effect when the switch is closed. The analysis emphasizes the importance of understanding circuit behavior at different time intervals, particularly in relation to capacitors and resistors.
fishys182
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Homework Statement


the switch in the circuit is closed at t=0 for a very long time.

find the current across the 2 ohm resistor before the switch is closed.
find the current across the 5 ohm resistor before the switch is closed.
find the current across the 2 ohm resistor RIGHT at t=0.
find the current across the 5 ohm resistor RIGHT at t=0.
explain why the current across the 2 ohm resistor at 0- is equal to 0+
explain why the current across the 5 ohm resistor at 0- is not equal to 0+

here is the circuit i made in LTSpice (and a little with paint):
http://img528.imageshack.us/img528/2843/724ds9.jpg

Homework Equations


when the switch is open, everything to the right of the switch isin parallel with the switch... everything to the right of the switch is shunted when the switch is closed. the circuit is really then, two separate circuits.

ohm;s law:
V=iR

relationship btwn current, voltage, and capacitance:
i=C(dV/dt)

KVL and KCL


The Attempt at a Solution



this is as far as i have gotten:

i used (for the left side of the circuit when the switch is closed):
V4=i(R4)
R4 = 1 volt / 20 ohms = 0.05 amps

so, when the switch was open, the initial current through R3====>
0.1amps + 0.05 amps = 0.15 amps

the right side of the circuit is a voltage divider when the switch is closed

While the switch is open, the capacitor fully charges up to 1 volt. right at t=0, there is no instantanious change in voltage, therefore dV/dt = 0 therefore current through the capacitor is also 0.

any ideas?
(btw is there a way to make this image smaller without having to re-upload it?)
 
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Analyze the circuit at t = -0, t= +0, t=0 separately. I saw that you kind of did it that way, but just label just label it so that its easier to follow your work.

And what's your question?
 

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