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**Unformatted text preview: **) + 5 (-e-t + 11.25e-2.25t ) =-105e-t + 56.25e-2.25t d) Evaluate when t=0 v 1 (0) =-105e-t + 56.25e-2.25t =-105 + 56.25 =-48.75 V Notes: To solve the system, we needed to find the initial inductor current (the part on the left side of the switch), and the time constant (the part to the right). Once we have these, since it’s a first order system (there is only one component with a d/dt characteristic), we can apply the relevant exponential solution. To find the equivalent resistance connected to the inductor, after the switch is opened, we calculated the Thevinin resistance by applying a test current source. A test current source is easier than applying a voltage as we see the current divides into the 100 and 200 Ω resistors, then using i ∆ we work backwards to get v T , and R TH ....

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- Spring '08
- Judy
- Resistor, The Current, Electrical resistance, Exponential decay, test current source