Timeline for How do I calculate the current in this circuit using Laplace transforms, with a time shift?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2018 at 22:51 | comment | added | Pedro | I said that when t<0.01s the switch has been connected to node 1 for a long time, therefore, v1 has been connected for a long time | |
| Jun 24, 2018 at 22:46 | comment | added | Chu | Your question does not say that v1 has been applied for a long time. | |
| Jun 24, 2018 at 22:41 | comment | added | Pedro | How? The switch only occurs at t=0.01s and at t=0s the first cosine wave is already applied | |
| Jun 24, 2018 at 22:15 | comment | added | Pedro | The question is, assuming that at t<0.01 s the switch is connected to node 1 for a long time, and so we can assume it has reached a steady state. Calculate i(t) for t>0.01s. | |
| Jun 24, 2018 at 21:50 | comment | added | Chu | Can you translate the full question - an initial condition does not make any sense. | |
| Jun 24, 2018 at 20:41 | comment | added | Pedro | But nothing happens at t=0, and since for t<0.01s it is an open circuit the voltage across the capacitor is v1(0.01). I don't think I understan where you are getting at | |
| Jun 24, 2018 at 18:59 | comment | added | Chu | \$\small 10 \sqrt {2} cos (100t) \$ is applied at t=0, hence there’s a transient. What would you have done if a step were applied at t=0 instead of a cosine? | |
| Jun 24, 2018 at 16:32 | comment | added | Pedro | Why isn't the system at steady state? | |
| Jun 24, 2018 at 9:04 | history | edited | Chu | CC BY-SA 4.0 |
added 28 characters in body
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| Jun 24, 2018 at 8:45 | history | answered | Chu | CC BY-SA 4.0 |