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I am trying to solve this question:


problem


However, I am having difficulties solving as I only have come across circuits which are initially closed then opened, not vice versa as is the case here. Specifically, I am trying to construct the simplified circuit for t<0. Usually, for the latter case, the inductor is a short circuit for the 20 Ohm resistor if t<0. However, I am unsure how it acts as this circuit starts open and then becomes closed.

So far, all I have been able to do is to a source transformation, making the current source a 150 V voltage source and combining the 10 and 5 Ohm resistors.

Any help would be appreciated, thanks!

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for t<0 you have this circuit :

schematic

simulate this circuit – Schematic created using CircuitLab

first, you should calculate IL for t<0: enter image description here

When SW1 move from a to b we have this circuit:

schematic

simulate this circuit

enter image description here

schematic

simulate this circuit

the equivalent resistor of 5 and 20 ohm is 4 So you have :

schematic

simulate this circuit

enter image description here

to obtain voltage equation we derivate the current equation:

enter image description here

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  • \$\begingroup\$ Thanks! From my understanding, in the second last equation for V = L di/dt, you used I = 10e^-80.2*t. Where did this come from? I assumed it came from the line above, however I(t) = 10e^-0.2 * e^-80*0.01 doesn't equal such. \$\endgroup\$ – Richard Robinson Jul 19 '18 at 20:11
  • \$\begingroup\$ I just calculate C1 int that line, not i. \$\endgroup\$ – Rashid Jul 20 '18 at 10:54
  • \$\begingroup\$ Where did you get the i in di/dt from then? \$\endgroup\$ – Richard Robinson Jul 20 '18 at 16:22
  • \$\begingroup\$ In the second circuit we have i=c1*e^(-80t) where c1 is 10e^(-0.2) . To obtain voltage we need to calculate Ldi/dt . \$\endgroup\$ – Rashid Jul 20 '18 at 16:40
  • \$\begingroup\$ But I = c1*e^(-80t) = 10e^(-0.2) doesn't equal 10e^(-80.2t) \$\endgroup\$ – Richard Robinson Jul 20 '18 at 18:50

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