1
\$\begingroup\$

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.

\$\endgroup\$
7
  • 1
    \$\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
    Commented Jul 2, 2018 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\$
    – D.A.S.
    Commented Jul 3, 2018 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
    Commented Jul 3, 2018 at 0:32
  • 2
    \$\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
    Commented Jul 3, 2018 at 0:41
  • 1
    \$\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
    Commented Jul 3, 2018 at 1:02

2 Answers 2

Reset to default
1
\$\begingroup\$

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+.

\$\endgroup\$
3
  • \$\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
    Commented Jul 3, 2018 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
    Commented Jul 3, 2018 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
    Commented Jul 3, 2018 at 16:27
0
\$\begingroup\$

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.

\$\endgroup\$
15
  • \$\begingroup\$ The OP just needs the initial conditions, I don't think much more than algebra is required. \$\endgroup\$
    – Elliot Alderson
    Commented Jul 3, 2018 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
    Commented Jul 3, 2018 at 2:18
  • \$\begingroup\$ @Jaden we shouldn't give complete answers to homework. \$\endgroup\$
    – Chu
    Commented Jul 3, 2018 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
    Commented Jul 3, 2018 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
    Commented Jul 3, 2018 at 11:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.