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I am studying for the FE Mechanical exam and am working my way through the electricity & magnetism section. Attached is a practice problem statement I found online and am having a bit of trouble working through.

My thought is that after the switch is closed the inductor will still act as a short while the capacitor will continue to act as an open circuit (with no current through that branch) since they are still connected to the voltage source. To that end I have redrawn a simplified circuit and applied KCL to solve for the "new" current through the 4K resistor. I end up with a result of 6.67mA. I understand that these exams often don't give the exact answer as a solution so my thought it that the correct answer would be 10mA. However, I am not confident that I have solved this question correctly.

Can anyone review this work and let me know of any missteps that I may have made? Any feedback is appreciated. This is my first post on the site so if something is unclear please let me know! Thank you.

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The inductor will oppose the change in steady state, hence, at the instant the switch is closed, the current through the inductor will be the same as the steady state value. Therefore \$\small I= \:... \$

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  • \$\begingroup\$ Hi, thank you for the response! I have in my initial post that I my calculations are based on my knowledge that the inductor should still act as short immediately after the switch is closed. However, this doesn't tell me whether my other assumption were correct or if I solved the for the new current through the 4K resistor correctly. \$\endgroup\$ Jun 15, 2021 at 12:45
  • \$\begingroup\$ Your analysis is not correct, and the assumption that the inductor is a short circuit after the switch is closed is also not correct. The key to this particular problem is to recognise that an inductor will always strive to maintain status quo in regard to the current flowing through it. You could do a transient analysis, but in this case there is no need as you are only required to find the current value at one, very brief, period of time. \$\endgroup\$
    – Chu
    Jun 15, 2021 at 13:22
  • \$\begingroup\$ I think I understand you point. If I assume my initial sketch is correct, then prior to the switch closing (t=0-) the inductor and the three resistors should be in series with the same current (I =10 mA) through all of them. Immediately after the switch closing (t=0+) current will flow to the second 4K resistor. Since the inductor will resist an instantaneous change in current it's value should not change (i.e I_inductor ( @ t=0+) = 10mA). Using KCL at the lower junction the current will split evenly since the parallel R values are the same. So I (@ t=0+) = 5mA \$\endgroup\$ Jun 15, 2021 at 13:29
  • \$\begingroup\$ Yes. You can safely ignore all the components to the left of the inductor - put them in a 'black box' if you wish. Your analysis is now correct, \$\small I=5\:\text{mA}\$ \$\endgroup\$
    – Chu
    Jun 15, 2021 at 14:05
  • \$\begingroup\$ Thank you for your help on this. Your points on how to think about the inductor are very useful. \$\endgroup\$ Jun 15, 2021 at 16:13

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