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The switch in the following circuit was in position 'a' for long time , before passing to position b at t=0.

Find the initial current.

Find the initial Voltages of the capacitor , the inductor and the resistors.

enter image description here

What i tried

At t=0- , the circuit will look like this :

enter image description here

Since the switch was closed for a longtime , the current going through the capacitor will become 0 , while the voltage of the inductor will become 0.

Applying KVL : 28 +14I +20=0 , Hence I = -3A = I0

The voltage through the capacitor will be equal to the voltage through the 4K resistor + 28v

vc(t=0) = 4(-(-3))-28=-16V

and finally VL(t=0) = 0 since the inductor is short circuited.

However when the switch passes to the position b , the KVL equation doesn't apply anymore: -16+8(-3)+0+20 is different than zero.

Thanks in advance.

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    \$\begingroup\$ @Misunderstood At t=0 the switch is sort of both open and closed. This is a fictional switch that models the unit step function \$u(t)\$ which has a discontinuity at t=0. Sometimes we talk about t=0- as the infinitesimal instant before the switch moves and t=0+ as the infinitesimal instant after the switch moves. \$\endgroup\$ – Elliot Alderson Jul 2 '18 at 23:39
  • \$\begingroup\$ Since there was no cap current before, a switch break before make causes no change, thus is not relevant here. \$\endgroup\$ – Sunnyskyguy EE75 Jul 3 '18 at 0:26
  • \$\begingroup\$ @Misunderstood We don't really talk about the impedance of the switch at t=0. At t=0- the switch has zero impedance to point a and infinite impedance to point b. At t=0+ the switch has infinite impedance to point a and zero impedance to point b. It takes zero time for the switch to move. We really don't care what happens at the infinitesimally small instant when t is exactly zero. \$\endgroup\$ – Elliot Alderson Jul 3 '18 at 0:32
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    \$\begingroup\$ @TonyStewartolderthandirt I don't understand what point you are trying to make. Capacitor current and inductor voltage can be discontinuous; they are never relevant at a switching event. Capacitor voltage and inductor current are relevant, even if their value is zero. \$\endgroup\$ – Elliot Alderson Jul 3 '18 at 0:41
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    \$\begingroup\$ @ElliotAlderson sorry, I misunderstood. I have deleted my comment. PS, I never understand Tony, be careful or he will take you down a rabbit hole. \$\endgroup\$ – Misunderstood Jul 3 '18 at 1:02
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No, KVL still holds. In the instant when the switch moves, it is possible that the capacitor current can change and/or the inductor voltage can change. The thing to remember is that the capacitor voltage cannot change instantaneously and the inductor current cannot change instantanely...these must have the same value at t=0- and t=0+.

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  • \$\begingroup\$ Yes, VL is not 0. The "before" current is I=E/R = 48/16K = 3ma. The "before" voltages across the 4K & 12K resisters are E=IR = .003*4K=12V & .003*12K=36V. The voltage across the cap is -28+12 = +20-36 = -16V. The "after" resistance of the 12K & 24K is 8K. So the "after" voltage across the resistors is E=IR = .003*8K=24V. Then 20-VL-24+16=0 so VL = 12V. But not for long. ( 2nd order differential equation ?!) \$\endgroup\$ – dcromley Jul 3 '18 at 3:29
  • \$\begingroup\$ This is what confused me I thought that the voltage across the inductor has to be continuous hence if it’s zero before closing the switch , it has to be zero the moment we close it. \$\endgroup\$ – Raku Jul 3 '18 at 4:21
  • \$\begingroup\$ Yes, as Elliot says, C can change I; L can change V; but C cannot change V; L cannot change I; [instantly]. \$\endgroup\$ – dcromley Jul 3 '18 at 16:27
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At \$\small t=0\$ the circuit comprises the series connection of: the C, with an initial voltage of \$ \small -16\: V\$ on the top plate relative to the bottom plate; a resistance of \$\small 8\: k\Omega\$; the L with an initial current of \$\small 3\: mA\$ anticlockwise; and the \$\small 20 \:V \$ source.

Let \$\small I \$ be the anticlockwise current flowing from \$\small t=0\$, then derive the 2nd order differential equation in \$\small I\$ and solve, taking the above initial conditions into account.

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  • \$\begingroup\$ The OP just needs the initial conditions, I don't think much more than algebra is required. \$\endgroup\$ – Elliot Alderson Jul 3 '18 at 1:56
  • \$\begingroup\$ This isn't the whole answer, but it is part of it. Still need to do some KVL/KCL to find the other voltages (and currents if that is indeed asked). \$\endgroup\$ – Jaden Jul 3 '18 at 2:18
  • \$\begingroup\$ @Jaden we shouldn't give complete answers to homework. \$\endgroup\$ – Chu Jul 3 '18 at 6:39
  • \$\begingroup\$ @ElliotAlderson 1st paragraph of answer gives initial current; initial inductor voltage can then be determined, but homework criterion applies so can't give the full answer. \$\endgroup\$ – Chu Jul 3 '18 at 6:54
  • \$\begingroup\$ My point was that your suggestion to derive a differential equation is unnecessary. I didn't want the OP to be confused by that. \$\endgroup\$ – Elliot Alderson Jul 3 '18 at 11:01

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