Conceptual question regarding inductors and capacitors

In summary: If the new steady state conditions are different from the current state, then there will be some temporary currents and voltages that arise as the circuit adjusts to the new conditions. You can treat these as a kind of "initial conditions", since they will last only for a short time.So, the capacitor's current will change, but only for a short time. The initial value of that current is what you're trying to find out.As you can see, the current in the capacitor is not really "steady" because it's still changing to get to the new steady state condition. The capacitor's current will eventually end up at 0, but it will take some time to get there.
  • #1
NewtonianAlch
453
0

Homework Statement



http://img207.imageshack.us/img207/3756/24134024.jpg

N.B: (0+) means just after switch is open, and (0-) means just before switch is open.


I have the solutions for this, but there are some problems I have with understanding how inductors and capacitors work in circuits like this.


1) Before that switch is open (t < 0), the inductor is essentially a short, and the capacitor is an open circuit

2) When that switch is open, the currennt I[itex]_{R}[/itex] at time (0+) is going to be different to (0-) because resistors unlike capacitors and inductors respond quickly to changes in current.

3) The current through the inductor i[itex]_{L}[/itex] at time (0+) is considered to be the same as (0-) because inductors oppose sudden changes in current.

What I do not understand is when the switch is open, current i[itex]_{C}[/itex] at time (0+) - the current through the capacitor has a value of 0.278mA, I thought the capacitor is acting like an open circuit and like the inductor doesn't respond suddenly, how is there a current flow through it?

The solutions also say V[itex]_{C}[/itex](0+) = V[itex]_{C}[/itex](0-) = 0

I understand that just before the switch is open, the capacitor is acting like a short circuit, and there is no current through it, but how can there be 0 voltage at time (0+) and have a current i[itex]^{C}[/itex](0+) = 0.278mA?

i = Cdv/dt, there is no change in voltage, so current should be zero.
 
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  • #2
Hi NewtonianAlch! :smile:
NewtonianAlch said:
… when the switch is open, current i[itex]_{C}[/itex] at time (0+) - the current through the capacitor has a value of 0.278mA, I thought the capacitor is acting like an open circuit and like the inductor doesn't respond suddenly, how is there a current flow through it?

as SGT said seven years ago (https://www.physicsforums.com/showthread.php?t=89729) …
The current in a capacitor is
i = Cdv/dt​
and the voltage in an inductor is
v = Ldi/dt​

So, for an instantaneous change of voltage (dt = 0) in a capacitor, you need an infinite current. In the same way, for an instantaneous change of current in an inductor, you need an infinite voltage.

Notice that there is an infinite mean power, but a finite energy, since dt = 0.​
… i = Cdv/dt, there is no change in voltage, so current should be zero.

no, it's like when you let go of something …

its acceleration is g, but its speed is 0

here, there's no instantaneous change in voltage (v), but there is an instantaneous change in dv/dt :wink:
 
  • #3
Instantaneous change in [itex]\frac{dv}{dt}[/itex]? You mean [itex]\frac{d}{dt}[/itex]([itex]\frac{dv}{dt}[/itex])?

What do you mean by an instantaneous change in [itex]\frac{dv}{dt}[/itex]?

Although if the capacitor is still acting as an open-circuit at t = (0+), how is there going to be a current through it anyway?

P.S - Nearly 20,000 posts, congrats!
 
  • #4
NewtonianAlch said:
Instantaneous change in [itex]\frac{dv}{dt}[/itex]? You mean [itex]\frac{d}{dt}[/itex]([itex]\frac{dv}{dt}[/itex])?

What do you mean by an instantaneous change in [itex]\frac{dv}{dt}[/itex]?
No, an instantaneous change in V. That is, some substantial dV while dt = 0. For a capacitor that would require an infinite current.
Although if the capacitor is still acting as an open-circuit at t = (0+), how is there going to be a current through it anyway?
The capacitor behaves as an open circuit simply because it has matched the potential being presented to its terminals -- there is no potential difference available to drive current one way or the other, so the current is zero just as for an open circuit. If you change the conditions though (such as by closing a switch and presenting the capacitor with a new 'equilibrium target') it will take in or release charge accordingly to reach a new equilibrium condition.
P.S - Nearly 20,000 posts, congrats!
Agreed! :smile:
 
  • #5
20,000th post!

NewtonianAlch said:
… when the switch is open, current i[itex]_{C}[/itex] at time (0+) - the current through the capacitor has a value of 0.278mA, I thought the capacitor is acting like an open circuit and like the inductor doesn't respond suddenly, how is there a current flow through it?

i = Cdv/dt, there is no change in voltage, so current should be zero.
NewtonianAlch said:
Instantaneous change in [itex]\frac{dv}{dt}[/itex]? You mean [itex]\frac{d}{dt}[/itex]([itex]\frac{dv}{dt}[/itex])?

What do you mean by an instantaneous change in [itex]\frac{dv}{dt}[/itex]?

P.S - Nearly 20,000 posts, congrats!

oooh, this is my 20,000th post! thanks for pointing it out! :smile:

in a graph of v against t, an instantaneous change in v (from 0) would look like a flat-line to the left of 0, then a discontinuity, a jump up the vertical (v) axis, and then a continuous curve to the right

but an instantaneous change in dv/dt (from 0) would look like a flat-line to the left of 0, then a "corner" as the graph starts rising to the right …

not a smooth change at 0, but a sharp corner :wink:

for a capacitor, the former (a jump) isn't allowed, but the latter (a corner) is allowed
 
  • #6
I was just doing this problem again and ran into the same kind of issue, although a bit different this time.

I still do not understand how ic(0+) can be non-zero because iL(0-) = iL(0+) has a value, because it's essentially a short-circuit, so if in the inductor is short-circuiting the capacitor at 0+, why is there a current through the capacitor at 0+ ?
 
  • #7
The short-circuit and open-circuit behaviors of inductors and capacitors apply to their DC steady state conditions. That is, when things have settled down and the voltages and currents in the circuit stop changing.

When you create a change (by, say, opening or closing a switch), then the circuit is no longer in steady state -- A change in voltage is presented to the inductor or capacitor, or a previous current path for the inductor's current is altered, which sets new "target" currents and voltages for a new steady state.

To find out why there might be a given current or voltage, determine what the new steady state conditions might be. Do they differ from the current state? If yes, how will the circuit get from here to there?
 
  • #8
So iL(0+) has the same current as iL(0-) because the inductor doesn't respond immediately to changes in current, BUT at 0+ the inductor is no longer a short-circuit but now starting to function as an inductor is that what this means?
 
  • #9
NewtonianAlch said:
So iL(0+) has the same current as iL(0-) because the inductor doesn't respond immediately to changes in current, BUT at 0+ the inductor is no longer a short-circuit but now starting to function as an inductor is that what this means?

In truth, inductors always behave as inductors and capacitors always behave as capacitors. They're not schizophrenic components :smile: We say that an inductor "behaves like" a short circuit at steady state because at that time all the transient AC components have died away and a steady current flows through it as though it were nothing more than a piece of wire. BUT! add some new transient voltages or currents and you quickly find out that that inductor is still an inductor after all.

The same goes for the capacitor. At steady state its potential difference matches the potential presented to it from the surrounding circuit. With no differences, no current flows. No current flowing even though there's a potential difference across its terminals makes it "look like" an open circuit. BUT! Change the circuit a bit so that there's a new current flowing into the capacitor and voila! You find out it's still a capacitor.
 
  • #10
Thanks gneill, you've always been a big help. Much appreciated!
 

Related to Conceptual question regarding inductors and capacitors

What is an inductor and how does it work?

An inductor is an electronic component that stores energy in the form of a magnetic field. It consists of a wire wound into a coil, which creates a magnetic field when current flows through it. The strength of the magnetic field depends on the number of turns in the coil and the current flowing through it. When the current stops flowing, the magnetic field collapses and induces a voltage in the inductor, which opposes the change in current flow.

What is a capacitor and how does it work?

A capacitor is an electronic component that stores energy in the form of an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied to the capacitor, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. The strength of the electric field depends on the surface area of the plates and the distance between them. The capacitor can store this energy until it is discharged.

What is the difference between an inductor and a capacitor?

The main difference between an inductor and a capacitor is the type of energy they store. An inductor stores energy in the form of a magnetic field, while a capacitor stores energy in the form of an electric field. Additionally, the way they react to changes in current or voltage is opposite. An inductor opposes changes in current, while a capacitor opposes changes in voltage.

What is the relationship between inductance and capacitance?

Inductance and capacitance are two fundamental properties of an electrical circuit. Inductance is the ability of a circuit to store energy in a magnetic field, while capacitance is the ability to store energy in an electric field. They are inversely proportional to each other, meaning that as one increases, the other decreases. This relationship is important in many applications, such as in filter circuits where inductors and capacitors are used together to block certain frequencies.

How are inductors and capacitors used in electronic circuits?

Inductors and capacitors are used in a variety of electronic circuits for different purposes. Inductors are commonly used in filters, oscillators, and power supplies, while capacitors are used in filters, timing circuits, and power supplies. They can also be used together in resonant circuits, which are used in radio and television receivers. Additionally, inductors and capacitors are also used for energy storage in renewable energy systems, such as solar panels and wind turbines.

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