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I need to find the inductor current for t > 0 and initial conditions as in example below. I think it's enough to divide V1 by 5*s and take the inverse Laplace Transform but my answer 3*e^-2*t - 7*e^-t + 4 is not the same as the book answer. It has +3 instead of my +4. It seems to be that book answer adds -1A initial current to my answer but I can't fully understand why it's so. It's clear that the original nodal equation was created with taking initial values of capacitor voltage and inductor current into account and it's not need to make any manipulations to current result, because the nodal difference V1 to ground already takes into account the existence of the inductor's initial current.

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My solution is:

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Multisim part of question is placed in another post: Multisim setup transient analysis with initial conditions

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  • \$\begingroup\$ Please provide an attempt at a solution, if your posting homework \$\endgroup\$ – Voltage Spike Dec 12 '17 at 17:44
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    \$\begingroup\$ @laptop2d My solution is simple and was added to post \$\endgroup\$ – MaxMil Dec 12 '17 at 17:52
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Your textbook is correct. The inductor is connected in between the Node V1 and ground. It is modeled as a constant current source i(0)/s in parallel with corresponding inductor reactance. Now, Look at the original circuit. The current through inductor = the current flowing between node V1 and ground. In the below laplace model, it is nothing but the sum of two currents: i(0)/s and V1/5s (Kirchoff's Current Law). V1/5s will give you only the transient part of current through the inductor, while their sum will be the total current through inductor at any time t > 0.

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  • \$\begingroup\$ Why you say: "sum will be the total current through inductor at any time"? The textbook say that the "total current" must be "pransient part" + "steady state part". "steady state" - means value of current at infinite time, but not at zero conditions. What about v1(t) in that situation - it's only transient part too? \$\endgroup\$ – MaxMil Dec 13 '17 at 7:29
  • \$\begingroup\$ In the laplace model, its obvious that the current is split as i(0)/s and v1/5s. So you should add both to get total current flowing from that node to ground. V1(t) is the effective voltage at the node, anytime t > 0. if t --> tends to infinity it will be nothing but steady state voltage. \$\endgroup\$ – Mitu Raj Dec 13 '17 at 12:33
  • \$\begingroup\$ Ya at any time. Say for eg. If you put t = 0, you have to get the current through the inductor at t=0 , as -1 A. In your solution you will get it as 0 A. So its wrong. \$\endgroup\$ – Mitu Raj Dec 13 '17 at 12:40
  • \$\begingroup\$ The task question says that I need to find voltage across the capacitor but V1 isn't a voltage across the capacitor - it's the voltage between V1 and ground! Why it's interpreted as the capacitor voltage? \$\endgroup\$ – MaxMil Dec 13 '17 at 15:10
  • \$\begingroup\$ V1 is ofcourse the voltage across the capacitor here. What makes you think, it is not ? v(0)/s has to be included while calculating capacitor voltage at any time t. \$\endgroup\$ – Mitu Raj Dec 13 '17 at 15:16

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