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Capacitor and inductor problem

  • #1
Dear All

I am struggling with 2 different problems with the basic concept of inductance and capacitance.

1) Firstly, I am not sure if I am understanding correctly how (hypothetically) an initially "charged" inductor can sustain, when short circuited, a constant current.

I am right in saying that, when the inductor is initially "charged" and then short circuited, and the initial current source is removed, the rate of change of current is negative (current decreasing w.r.t the initial current source), and so the inductor reacts by inducing an emf and hence a current proportional to this negative rate of change of current but in the OPPOSITE direction to the original current source?

If this is correct so far, I am at a loss as to what happens next? Surely the induced current starts decreasing since it is depleting the inductor's finite store of energy? Then what would cause an increase in current to keep the current constant, as I can't see when the inductor is being "charged" again to become a current source once again? My reasoning would seem to imply an oscillating AC current being produced, but I think it should be DC?!

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2) My other question is as follows:

Starting with circuit with an inductor and a capacitor connected together both "uncharged", if a battery is connected briefly in parallel with them, the capacitor should rapidly charge and the inductor should resist the sudden change in current by dropping a voltage and as a result be barely "charged". Then if the battery is removed, apparently, assuming no resistance, the circuit should indefinitely oscillate at its resonant frequency. I am stuck as to a qualitive explanation of what happens. One site says that the current increases through the circuit as the capacitor voltage decreases, but I thought the current decayed exponentially from a discharging capacitor?!
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Many thanks in advance

Paul
 

Answers and Replies

  • #2
collinsmark
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Okay, everything I'm about to say involves ideal components. In other words, inductors and capacitors that have zero internal resistances, zero parasitics, etc. Also when possible, I'll try to keep things in terms of lumped parameter models.
1) Firstly, I am not sure if I am understanding correctly how (hypothetically) an initially "charged" inductor can sustain, when short circuited, a constant current.

I am right in saying that, when the inductor is initially "charged" and then short circuited, and the initial current source is removed, the rate of change of current is negative (current decreasing w.r.t the initial current source),
No, here is where you're going awry. The rate of change of current is not negative, it's zero!

Let's step back a bit. When the inductor is being charged for a long time, eventually the current moving through the inductor is constant.

Let's step back a bit more. Most of the time, the charging circuit is represented by an ideal voltage source and a resistor (although you could do the same thing with an ideal current source and a resistor in parallel). Initially, when first connected to the charging circuit, the inductor resists any change in current going through it, and has a voltage drop across it equal to the voltage source's voltage. But after time, current starts flowing through the inductor and the voltage drop across its terminals decreases. After a very long time, the inductor essentially becomes a short circuit itself, having a 0 V voltage drop across its terminals, the current though it is constant, and the current is simply V/R.

Then the switch is flipped , and there is a short placed across the terminals of the inductor (also, essentially removing the voltage source and the resistor from the circuit), forcing the voltage drop across the inductor to be zero. So just before the switch is flipped, the voltage across the inductor's terminals was zero, and after the switch is flipped it is forced to be zero.

Note that:

[tex] v_L = L\frac{di_L}{dt} [/tex]

And we've forced vL = 0, essentially forcing diL/dt = 0.

diL/dt is the rate of change of the current. And since the rate of change of the current is 0, it means that the current flowing through the conductor does not change. In other words, the current flowing through the inductor is a constant. A DC constant that does not change.
and so the inductor reacts by inducing an emf and hence a current proportional to this negative rate of change of current but in the OPPOSITE direction to the original current source?
By flipping the switch, short circuiting the terminals of the inductor, you have essentially forced the emf to be zero. There is no emf. Thus there is no change in current. Thus the current in the inductor is constant.
If this is correct so far, I am at a loss as to what happens next? Surely the induced current starts decreasing since it is depleting the inductor's finite store of energy? Then what would cause an increase in current to keep the current constant, as I can't see when the inductor is being "charged" again to become a current source once again? My reasoning would seem to imply an oscillating AC current being produced, but I think it should be DC?!
Okay, let's talk energy. Inductors and capacitors never dissipate energy (i.e. change it to heat). Resistors (or anything with resistance) are the things that change energy to heat.

So if your circuit has zero resistance, and only inductors and capacitors, energy is conserved. It's like older physics problems involving frictionless roller-coasters or something. Energy can change from kinetic to potential energy, but besides that it's always conserved.

Resistors are like friction. Resistors can dissipate the amount of energy stored in the system. but if your circuit has only capacitors and inductors, but no resistors, there is no way to dissipate any energy.
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2) My other question is as follows:

Starting with circuit with an inductor and a capacitor connected together both "uncharged", if a battery is connected briefly in parallel with them, the capacitor should rapidly charge and the inductor should resist the sudden change in current by dropping a voltage and as a result be barely "charged". Then if the battery is removed, apparently, assuming no resistance, the circuit should indefinitely oscillate at its resonant frequency. I am stuck as to a qualitive explanation of what happens. One site says that the current increases through the circuit as the capacitor voltage decreases, but I thought the current decayed exponentially from a discharging capacitor?!
Current decays exponentially from a discharging capacitor only when a resistor is present. The current through a capacitor is;

[tex] i_C = C\frac{dv_C}{dt} [/tex]

And when a charged capacitor is connected to a resistor, this becomes (invoking Ohm's law dV = Rdi, and rearranging a bit),

[tex] i_C = RC\frac{di_C}{dt} [/tex]

meaning that the rate of change of current is directly proportional to the current. This sort of first order, linear, ordinary differential equation happens often in nature. Another example is radioactivity. The radioactivity, based on the rate of change from one radioactive element to some other element, is directly proportional to the amount of radioactive material present. The radioactive material has a half-life. Similarly with a charged capacitor and a resistor, the current flowing through the capacitor also has a half-life. The math involved is essentially identical in this case.

Replacing the resistor with an inductor gives a totally different result though. The voltage across an inductor is not simply v = iR, but rather v = Ldi/dt. And the current in the circuit is i = Cdv/dt. Combining these two simultaneous, first order, linear, ordinary differential equations leads to a second order, linear, ordinary differential equation, which leads to the oscillations.

Thinking about it qualitatively, when the capacitor is charged to its maximum, no current is flowing. But at this time, there is a voltage across both the capacitor and inductor. since there is a voltage across the inductor, current will gradually start flowing though it, and the terminal voltage (across the terminals of both components) will start to decrease. Eventually the voltage across the terminals is zero. At this time the current flowing is at its maximum. A negative voltage then develops across the terminals as the current begins to decrease. Eventually the voltage across the terminals is at its most negative value, and once again the current is zero. Then the current reverses direction, and increases to a point of maximum magnitude (where the voltage is now zero once again), but then keeps on going as the voltage gradually approaches the original voltage as the current magnitude falls back to zero. Then the whole process repeats.
 
  • #3
Thanks for the reply collinsmark. You have definitely clarified my issue with the first problem. I was thinking that short circuiting the inductor would suddenly decrease the current in the inductor for some reason, whereas it actually would remain constant.

I still have some issues understanding the explanation for the second question:
When the capacitor is charged to its maximum, no current is flowing. But at this time, there is a voltage across both the capacitor and inductor. since there is a voltage across the inductor, current will gradually start flowing though it, and the terminal voltage (across the terminals of both components) will start to decrease.
Are you assuming that the battery has now been removed, and the capacitor causes the voltage across the inductor here?

My understanding based on my basic knowledge and what has been said is that the capacitor emf, effectively caused by the seperation of the charge on the plates, pushes on the charge carriers with an initially high force/emf, causing a large acceleration of the charge carriers and hence a large di/dt term. However, the inductor is going to try and oppose this by creating an opposing and equal emf (e=Ldi/dt). So my question now is, if the inductor opposes this change in current, then why is di/dt not zero?
 
  • #4
gneill
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My understanding based on my basic knowledge and what has been said is that the capacitor emf, effectively caused by the seperation of the charge on the plates, pushes on the charge carriers with an initially high force/emf, causing a large acceleration of the charge carriers and hence a large di/dt term. However, the inductor is going to try and oppose this by creating an opposing and equal emf (e=Ldi/dt). So my question now is, if the inductor opposes this change in current, then why is di/dt not zero?
An analogy:
A force pushes on a mass so you'd expect a large dv/dt, but the mass is going to try and oppose this by creating an opposing and equal reaction force (Inertia + Newton 3, f = m*dv/dt). So why is dv/dt not zero?
 
  • #5
Thanks for the reply gneill. I understand that analogy; There is a large dv/dt for the mass since the two forces are acting on different objects (NIII law pair). However, I don't see how that analogy can be applied here since surely both the capacitor's emf and the inductor's "back emf" (which are initially equal and opposite) are acting on the same objects -the electrons?
 
  • #6
gneill
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Suppose the force on the mass is being applied via a stick. The reaction force is thus being applied to the stick, so the total force that the stick "has left" is zero. There's woolly thinking going on somewhere, since we know that the mass accelerates.

I think that the thing to take away from the analogy is that the inductance "behaves" like inertial mass for current when the applied voltage is analogous to force. The larger the inductance, the more slowly the current magnitude increases.
 
  • #7
In the analogy, the mass does accelerate as I said in my last post (dv/dt is large) since the pushing force and the reaction force are on different bodies. For the example of the stick, whilst there is a reaction force on the stick, this does not cancel the fact that that the stick is applying a force on the mass, so the mass still accelerates.

I am still very confused. Why does current of increasing magnitude (di/dt not zero) flow through an inductor if it produces an equal back emf to counter this increase. It seems counter-intuitive to me. Or perhaps I'm just being stupid.

If I simplify my original problem to the case of a battery source emf connected through an initially open switch to an ideal inductor. There is NO RESISTANCE in the circuit either as part of the inductor or a seperate resistor. If the switch is closed what happens? Is there any current flow, if so why? Surely by KVL, the voltage drop and hence back emf must be equal to the source emf, therefore they cancel and there is no current flow...
 
  • #8
gneill
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In the analogy, the mass does accelerate as I said in my last post (dv/dt is large) since the pushing force and the reaction force are on different bodies. For the example of the stick, whilst there is a reaction force on the stick, this does not cancel the fact that that the stick is applying a force on the mass, so the mass still accelerates.
The reaction force (back emf) is on the stick (electrons). I don't see why you can claim a problem in one case and not the other.

I am still very confused. Why does current of increasing magnitude (di/dt not zero) flow through an inductor if it produces an equal back emf to counter this increase. It seems counter-intuitive to me. Or perhaps I'm just being stupid.

If I simplify my original problem to the case of a battery source emf connected through an initially open switch to an ideal inductor. There is NO RESISTANCE in the circuit either as part of the inductor or a seperate resistor. If the switch is closed what happens? Is there any current flow, if so why? Surely by KVL, the voltage drop and hence back emf must be equal to the source emf, therefore they cancel and there is no current flow...
If the battery is up to it (in terms of its ability to supply current), there will be a constantly increasing current in the inductor. v = L*di/dt. So di/dt = v/L. i = (v/L)*t.

This is analogous to the mass problem, where somehow one manages to apply a constant force to a mass (despite the fact that the mass is going faster and faster, and should eventually outrun whatever mechanism is trying to apply the force). f = m(dv/dt),
so dv/dt = f/m, and v = (f/m)*t.

In the case of the mass and force, there is kinetic energy constantly being put into the inertial mass. For the inductor, the energy is being pushed into its magnetic field.
 
  • #9
collinsmark
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Going back to an earlier question...
I still have some issues understanding the explanation for the second question:

Are you assuming that the battery has now been removed, and the capacitor causes the voltage across the inductor here?
Yes, in my explanation of 2), particularly the qualitative explanation, I am describing it after the battery has been removed (along with any resistance), such that the only components remaining in the circuit are the inductor and capacitor, and they are connected together. One of the terminals of the inductor is connected to a terminal of the capacitor. Likewise, the remaining terminals (one of the capacitor, the other the inductor) are also attached. There is nothing else in the circuit. The capacitor has an initial charge on it.
My understanding based on my basic knowledge and what has been said is that the capacitor emf, effectively caused by the seperation of the charge on the plates, pushes on the charge carriers with an initially high force/emf, causing a large acceleration of the charge carriers and hence a large di/dt term. However, the inductor is going to try and oppose this by creating an opposing and equal emf (e=Ldi/dt). So my question now is, if the inductor opposes this change in current, then why is di/dt not zero?
diL/dt is not zero because there is a voltage (emf) across the terminals of the inductor. So what is diL/dt then? Well, recall that:

[tex] v_L = L\frac{di_L}{dt} [/tex]

So,

[tex] \frac{di_L}{dt} = \frac{v_L}{L}[/tex]

which obviously isn't zero, so long as there is a non-zero voltage across the inductor's terminals.
 

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