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schematic

simulate this circuit – Schematic created using CircuitLab

So I was trying to solve problems in preparation for an exam regarding Laplace Transforms. I was able to solve some problems and then I came up with this circuit. I cannot go anywhere since I find it hard and not sure how to simplify the circuit at t = 0- and at t > 0.

Just wanna get an idea how to first simplify and solve the circuit at t = 0- and hopefully I'll be able to take it from there.

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  • \$\begingroup\$ The problem is that I1 is not subject to the action of the switch, and I3 is operational up to t=0 but we don't know when it started (t= -infinity?). You have no way of calculating the starting point values (at t=0) unless the circuit reaches some sort of equilibrium position you will need to simply define your initial values ( at t=0 ) \$\endgroup\$ Mar 25 '17 at 13:47
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The circuit as shown is impossible because the 2A current source (Item #13) is open circuited, thus producing an infinite voltage across the gap. There must be another switch which shorts this current source which then must be opened while the other (shown) switch changes position. If the #13 current source is intended to be connected across the inductance, then IL=2A at t=0- and t=0+. But I wouldn't waste any time trying to solve this bogus circuit until clarification is supplied.

A current source can NEVER be left floating, it must be shorted when not in use. The dual to a voltage which must be left open when not in use.

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  • \$\begingroup\$ It's actually not open, that is why I placed a note. At t > 0, it is dangling and should be disregarded but at t = 0 minus, the switch connects to the 2A current source and the circuit disregards the 2e^-t current source since it's the one dangling at that time. \$\endgroup\$
    – AndyMarty
    Mar 25 '17 at 15:15
  • \$\begingroup\$ In simpler terms, at t = 0 minus, the 2e^-t is dangling and disregarded. At t > 0, the 2A current source is dangling and disregarded. \$\endgroup\$
    – AndyMarty
    Mar 25 '17 at 15:16

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