I'm stuck on this problem I'm working on:

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With the following questions:

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So my approach:

Well to start with there is two cases: t<0 and t>0, so I thought that I would start to find IL(0), the initial value.

And I redrew the circuit like this: enter image description here

I was thinking that the inductor conducts current and that it doesn't change which leads to the voltage over the inductor to be 0 V and that makes the inductor behave like a short circuit when the switch is closed.

Now what is a little bit hard for me to understand is the case with the R2 resistor. I don't understand if I should consider it in my calculations or not? Should I do a Rtotal = 100kohm x 1kohm / 100khom + 1kohm?


2 Answers 2


think that nature is lazy, so the current will allways take the shortest path, wich means the path with less resistance.

In this case Elliot is right, you can forget about R2 before t=0 cause the switch is closed, and you can forget L to calcul the current (that a DC analysis). Now if you are analysing this as a real circuit (AC and trasient analysis), once you will open the switch the energy stored in L will search to continue driving the same amount of current, that is the INERTIA of L, it goes againt changes in the circuit.

So in this case we have 10V/1000 ohm = 1mA, once you open the swich you will have 1mA going for a short amount of time, ( in a real life circuit is quite more difficult and unpredictable), and as crazy as it should sound 1mA over 100K will give you 100V over R2, for a short amount of time, it depends on the size of L. you can use the law V = L . di/dt(variation of current)

this is the reason why we put a diode, or a transil over a load in inductive circuits, to protect against overvoltages when the circuits opens or close fast. For exemple when using Relays

for the time of realease, if my memories are good, for RL circtuits is around 5 x L/R, and because is a logarithmic equation( the complicated one) with 1 x L/R you allready discarche 65% so at 5 you can say is all gone

  • \$\begingroup\$ Hello, and thank you for explaining so well! :) Just one little thing, I think you got 10/1000 wrong: should it not be 10 mA? and for the voltage over R2 = 1000 V? \$\endgroup\$
    – Vetenskap
    May 6, 2020 at 12:18
  • \$\begingroup\$ OHH yes you are right !!! \$\endgroup\$ May 6, 2020 at 12:27
  • \$\begingroup\$ Happy to help! have a good day \$\endgroup\$ May 6, 2020 at 12:27
  • \$\begingroup\$ Could I ask you to just explain the last part about the time of realease (the time constant tau). I didn't really catch that. How should I calculate the time constant? How did you get that it discharge 65 % at 5? \$\endgroup\$
    – Vetenskap
    May 6, 2020 at 12:38
  • \$\begingroup\$ yes, what we are watching is the current inside L. Imaging you put a square signal of 10 volt this time, the current will take 5 tau to get to 10V/1K = 10mA. at 1 tau it will allready be at 63%, because it's a logarithmic fonction \$\endgroup\$ May 6, 2020 at 13:17

Note that the switch is in parallel with R2. When the switch is closed it can be treated as a resistor of 0 ohms. So, replace the switch with a resistor of 0 ohms and then analyze the circuit using the usual techniques.

  • \$\begingroup\$ So you mean that I should not consider the inductor as a short circuit? If I consider the switch as 0 ohm and think of it as in parallel with R2 I get 0 ohm? I don't quite understand how it works out. \$\endgroup\$
    – Vetenskap
    May 6, 2020 at 11:41
  • \$\begingroup\$ I didn't say anything about the inductor. Your question was about how to handle R2. I gave you a hint, and you are correct that any resistance in parallel with 0 ohms gives an equivalent resistance of 0 ohms. So replace the parallel combination of the switch and R2 with a resistor of 0 ohms and continue the analysis. \$\endgroup\$ May 6, 2020 at 11:52
  • \$\begingroup\$ I get my IL(0) = VA/R1 = 10/1000 = 10 mA. So I get the inital value to 10 mA. However when the switch opens up at t = 0 s, t>0, then I want to think that the R2 resistor isn't "included" in the circuit and what is left i VA, R1 and L. Am I thinking correctly? \$\endgroup\$
    – Vetenskap
    May 6, 2020 at 12:04
  • \$\begingroup\$ I see no reason to think that R2 should not be included after the switch opens. \$\endgroup\$ May 6, 2020 at 12:29

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