# Question about Inductors and Switches

1. Nov 8, 2014

### Bluestribute

1. The problem statement, all variables and given/known data
The switch has been open for a very long time, and then we close it. In the instant right after we close the switch, assuming that R1[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char3E.pngR2[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char3E.pngR3[PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmmi10/alpha/100/char3E.pngR4, [Broken] what is the total current through the circuit?

2. Relevant equations

3. The attempt at a solution
It's multiple choice, and the "relevant equation" is the correct answer, but why? So capacitors essentially act like wires when fully charged and breaks when charging (let me know if I have those backwards). So, the question is, what do inductors act like when a switch is open and closed?

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2. Nov 8, 2014

### Staff: Mentor

You need a pair of parentheses in your expression for I.

Inductors behave as though their current has inertia, it is reluctant to change. So the moment after the switch is closed, the current in the inductors is still what it was before, i.e. zero. So you can erase the inductors and their series elements to see what the circuit looks like at that first moment, t=0+

3. Nov 8, 2014

### Bluestribute

So when the switch is closed, does it slowly let current through? So like, maybe at t=0, current is nothing, but t=5 current might equal 10 A, and at t=10 current might equal 15 A? So basically, if instead of at t=0 it was t=2, what could I say about he inductor?

4. Nov 8, 2014

### Staff: Mentor

No, not this circuit. There is not an inductor in every branch. Only branches containing an inductor exhibit the characteristic "inertia".

Last edited: Nov 8, 2014
5. Nov 8, 2014

### Bluestribute

So what would this one look like after the switch has been closed for a while? Does the inductor take out that branch until the switch is open?

6. Nov 8, 2014

### Staff: Mentor

Analysis at t=0+ (the instant after the switch closes) and t=∞ ("a long time" after the switch closes) is much easier than analysis of the intermediate times. This is because inductors and capacitors can store energy, and they produce exponential or oscillatory effects when "hit" by sudden changes in circuit conditions (like closing a switch). Circuits containing these "reactive" components are described by differential equations.

At the instant after such a sudden "hit", capacitors show inertia for voltage and inductors show inertia for current. So an uncharged capacitor "looks like" a short circuit (a piece of wire), while an inductor without current "looks like" an open circuit. If the capacitor were charged to some voltage before the "hit" then it would "look like" a voltage source of that value in the instant after the switch closes. If the inductor were conducting some value of current when the switch closed, it would "look like" a current source of that value in the instant after the switch closed.

After a very long time when the circuit has settled to its steady state, capacitors "look like" open circuits and inductors "look like" short circuits.

See above. The inductors will "look like" wires and the capacitors like opens.

7. Nov 8, 2014

### Staff: Mentor

You can replace each inductor by a short piece of wire, but this is valid only once the currents have reached stable values.

inductor voltage = L. di/dt, so once di/dt becomes zero there is no voltage across the inductor yet there may be steady current in it. This latter characteristic is what you see in a short circuit: current flowing yet no potential difference, so under these steady conditions we can model the inductor as a short circuit.

8. Nov 10, 2014

### Bluestribute

Ok. So I see a problem I was getting at first off. Inductors "play with" current, conductors "play with" voltage.

So if I have an inductor before a resistor (let's just say it's in parallel, so it's its own "section" of the circuit), and the circuit has been going for a while, the "inductor-resistor" section is basically not part of the circuit anymore. At least, not until I open a switch and stop the current flow, in which case the "inductor-resistor" section will have a current going through it thanks to the inductor.

If it was a "capacitor-resistor" section, if it's been going for a while, is too not part of the circuit since the capacitor is charging. It remains that way until the switch opens up, blocking the current again. Then the capacitor can discharge and bring that section back to life . . . at least for a bit.

I hope I'm understanding this correctly!

9. Nov 10, 2014

### Staff: Mentor

No, at steady state ("after a long while") inductors look like a piece of wire. So at that point your inductor-resistor branch looks like just the resistor, conducting whatever current the resistor allows for that branch. At the instant the switch opens the branch continues to conduct that current. Then after another long while the branch will again look like just the resistor.

While the capacitor is charging current is flowing and its voltage is changing. After a long while the capacitor will have reached whatever potential difference is its final value for that circuit and current stops flowing. You can find that potential by removing the capacitor and finding what potential difference would be across the open terminals where it was connected. Since no current flows once the capacitor is fully charged to its final value, for purposes of analysis it "looks like" an open circuit at steady state.
You're getting there.

10. Nov 10, 2014

### Bluestribute

Ok, so after a while inductors just let current pass because there isn't a sudden change (because they resist that, right?). Glad I'm asking since some people use "like a wire" to mean a short and others use "like a wire" to mean like nothing is there. But this confuses me:

Doesn't that means the branch always looks like the resistor??

And the capacitor will charge while still letting current pass until it gets fully charged, in which case it'll stop the current until there's a break?

11. Nov 10, 2014

### Staff: Mentor

No, between steady states (the period from right after the switch changes things until "a long time" has passed) is what is known as the transient response of the circuit. Inductors and capacitors play an active role then. I'm sure you'll be covering this topic soon.

The current is due to the capacitor taking on or losing charge and changing its potential difference. The potential difference changes until it matches what the circuit can provide at the place where it's connected.

Perhaps an analogy will make it clear. If you take an object at some temperature and place it in an ambient environment at a different temperature, heat will flow (current), gradually changing the temperature of the object until it reaches the same temperature (potential) as the environment. This takes some time: The temperature of an object does not change instantaneously, it has what's called "thermal mass", and exhibits "thermal inertia". When the object approaches the same temperature as the environment ("after a while") the heat flow tapers off to zero.

The argument is the same for a capacitor. It's connected to a pair of nodes in a circuit and will accept or dish out charge until its potential difference matches the potential difference that you would see at the open terminals where the capacitor connects.

12. Nov 10, 2014

### Bluestribute

The sad part is we did cover it haha. All my classes just go at a really fast pace so I can't just stop and ask . . . or have free time to. I'm in between questions on a practice exam right now.

So, steady states are t=0 and t=∞ (in simple layman's terms)?

And that analogy is bomb. Definitely helps to understand. So capacitors will take in current (essentially acting as though they are just a normal wire going to, say, a resistor) until they reach the total potential. This total potential is the total potential of, say, the resistor the capacitor is in front of? Once the capacitor reaches the potential (can we say this is at t=∞, a steady state?), it "shorts" the branch. Is this correct (except for the bolded, which I'm not terribly sure of)?

13. Nov 10, 2014

### Staff: Mentor

Usually, but not necessarily. t=0 is just whenever the human decides to call start, and that is usually after conditions have been allowed to reach a steady state.

If all sources are DC, and switch positions remain fixed, then things will reach a steady state at time = infinity.

14. Nov 10, 2014

### Staff: Mentor

For this problem, yes.
No the capacitor doesn't look like a normal wire while charging. It behaves like the object in the heat analogy and changes its potential slowly depending upon its capacitance and the rate of flow allowed by the circuit around it. There's a sort of "inertia" to its potential difference, and it won't allow instantaneous change. The potentials across the resistor and capacitor will both change over time.

The instant after a circuit change is made (switching), the capacitor maintains the same potential difference as immediately prior to the change. In this way it resembles (in that instant) an ideal voltage source. If the capacitor happened to be initially uncharged so that its potential difference was zero, then it would look like a voltage source of zero volts, or effectively a short circuit.

After a long time it eventually reaches the potential difference that the terminals where the capacitor connects would show if the capacitor were removed. In effect, at steady state the capacitor "looks like" an open circuit, not a short.