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

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SUMMARY

The discussion focuses on analyzing a circuit with a capacitor, current source, voltage source, and a switch using LTSpice. Key findings include that the current across the 2-ohm resistor is 0 A immediately before and after the switch is closed, while the current across the 5-ohm resistor is not equal to 0 A at t=0. The analysis utilizes Ohm's Law, the relationship between current, voltage, and capacitance, as well as Kirchhoff's Voltage and Current Laws to derive these conclusions.

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  • Familiarity with Ohm's Law and its application in electrical circuits.
  • Knowledge of capacitor behavior in DC circuits, particularly during switching events.
  • Experience with LTSpice for circuit simulation and analysis.
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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?)
 
Last edited by a moderator:
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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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