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I read a paper and simulate its schematic recently,and my simulation is not the same as authors,so i want to ask something.I hope i can know the answer from theory,and not from the simulation now.

Q1: enter image description here when current source(positive half wave) charge \$C1\$ and \$BAT1\$ flow through the \$L1\$,the current direction should be like this

enter image description here After charge \$C1\$ to a V,and flow through \$L1\$ to b V,b is higher than a.The SW1 close and SW2 opened.At this moment,what is the current direction? Theoretically shoule be "A" direction,because the current direction should be high voltage to low voltage.But my classmate told me the current in the inductor can't change suddenly,and the paper said the direction is "B".Why?i am confused now.

Q2:

enter image description here

At the beginning,A point have aV,and B point have bV,b is higher than a.

enter image description here

Then,the capacitor energy will flow to the inductor to let the inductor's energy flow to the capacitor,so the A point voltage will become from a to -b,the B point will become from b to a.It is like there is a force from A point to push to B point to let the energy from B flow to A.Now if i add a vpulse at B point,and at this moment,the vpulse voltage is 0.that is,the B point voltage is 0 now,will one of these two force be disappeared to let A point won't become -b and B point won't become a ?

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  • \$\begingroup\$ You have to clarify your concepts. A capacitor is charged, that is right. A coil is not "charged". A coil has a magnetic field which depends upon its CURRENT, and will oppose any change in its CURRENT. That means that the voltage on it can have sudden changes. Specifically, when the switch SW2 is opened, the voltage on the coil will have a sudden change in polarity while the coil is "trying" to keep its current flowing constant. \$\endgroup\$
    – Claudio Avi Chami
    Commented Apr 18, 2018 at 6:36
  • \$\begingroup\$ OK,I will modify "charge" in the inductor.But i am not understand for "the switch SW2 is opened, the voltage on the coil will have a sudden change in polarity while the coil is "trying" to keep its current flowing constant." so what is its current direction? \$\endgroup\$
    – XM551
    Commented Apr 18, 2018 at 7:40
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    \$\begingroup\$ @ClaudioAviChami I think it's acceptable to use the word charge for an inductor acquiring energy from the flow of electrons (rate of change of charge). Mathematically it's debateable but most folk will understand what is meant. \$\endgroup\$
    – Andy aka
    Commented Apr 18, 2018 at 7:53

1 Answer 1

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At this moment,what is the current direction?

An inductor cannot change the current in an infinitely small time else an infinite voltage is produced and nobody wants that.

V = L\$\dfrac{di}{dt}\$ is from Faraday and encapsulates what I've said

So, at the moment SW1 and SW2 swap over, the current flow through the inductor remains the same and it has to find a path to continue this current. The only path is through C1 in direction B.

Theoretically shoule be "A" direction

No it is not.

Theoretically shoule be "A" direction, because the current direction should be high voltage to low voltage

Only the capacitor has acquired a voltage that can be sustained. The voltage across the inductor will be the voltage acquired on the capacitor prior to the switches reversing their positions.

For a capacitor, Q = CV hence

\$\dfrac{dQ}{dt} = C \dfrac{dV}{dt}\$ = current

So, the change in current from being charged by I1 to receiving a different current from L1 results in a capacitor dV/dt change hence, the voltage rises or falls depending on the difference in those currents.

It rises or falls from the value it attained prior to the switches operating.

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  • \$\begingroup\$ @Thankyou Andy aka.By the way,do you know my last question answer?" B point voltage is 0 now,will one of these two force be disappeared to let A point won't become -b and B point won't become a ?" \$\endgroup\$
    – XM551
    Commented Apr 18, 2018 at 8:56
  • \$\begingroup\$ With b voltage at zero and knowing the capacitor formula in my answer, infinite current will flow because you are discharging the capacitor in an infinitely short time. \$\endgroup\$
    – Andy aka
    Commented Apr 18, 2018 at 10:11
  • \$\begingroup\$ you mean the formula \$c \frac{dV}{dt}=\$current,when \$V=0\$?current\$=0?\$ \$\endgroup\$
    – XM551
    Commented Apr 18, 2018 at 11:59
  • \$\begingroup\$ When V changes from "something" to "something else" in zero time, \$\frac{dV}{dt}\$ is infinite hence current is infinite. The inductor and capacitor formula underpin everything you need to understand about them. Those formulas are the definition of what they do. \$\endgroup\$
    – Andy aka
    Commented Apr 18, 2018 at 13:30

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