How Does an Inductor Affect Current Change Upon Switch Closure?

[collinsmark]

Because there is no current through it

That logic would be true if the component was a resistor. But it doesn't hold for inductors (nor does it for capacitors).

An inductor can and often does have plenty of voltage drop across it, even though no current is flowing through it.

The voltage drop across an inductor only affects the rate of change of the current flowing through the inductor, not the magnitude of the current itself flowing through it.

 

[Sho Kano]

No! For a resistor, V = IR, so no current means no voltage drop. For an inductor, V = L dI/dt, so I can be zero, but dI/dt not zero, and there is a voltage drop with no current.

{ \phi }_{ B }\quad =\quad L(0)\\ V\quad =\quad \frac { d{ \phi }_{ B } }{ dt } \quad =\quad L\frac { d(0) }{ dt } \quad =\quad 0

 

[phyzguy]

I'm not sure exactly what you mean by this, but you seem to not understand that a function can have a value of zero and a first derivative which is not zero. Consider the function 1-e^{-t} . At t=0, f = 0, but df/dt = 1.

 

[collinsmark]

{ \phi }_{ B }\quad =\quad L(0)\\ V\quad =\quad \frac { d{ \phi }_{ B } }{ dt } \quad =\quad L\frac { d(0) }{ dt } \quad =\quad 0

But the inductor's \frac{d \Phi_B}{dt} is not zero immediately after the switch is closed.

Let me step back and use an analogy. Imagine dropping a ball, initially at rest. Earth's gravitational acceleration is 9.81 m/s^2. At time t = 0 you let go of the ball. Acceleration, a, can be defined as
a = \frac{dv}{dt}
where v is the ball's velocity.

The moment after you let go of the ball, the ball's velocity is still 0 m/s. But does that mean the ball's acceleration is 0?

 

[Sho Kano]

What is the voltage drop across a given resistor in terms of the current flowing through that particular resistor?[boldface mine]

What current is that?

0 Current?

 

[collinsmark]

0 Current?

Yes, that much is correct.

[Edit: and back to my earlier point: Just because the current flowing through a component is 0 does not necessarily mean that the rate of change of current flowing through that component is 0. They could be quite different.]

 

[Sho Kano]

Yes, that much is correct.

Ok, so because there is no current through the inductor, there is also no current through the resistor? I can't put one and two together.

 

[phyzguy]

Since there is no current through the inductor, how can there be current through the resistor? Where would the current go when it hit the inductor? It's like the water in a pipe. If the valve is closed, there is no flow anywhere in the pipe.

 

[collinsmark]

Ok, so because there is no current through the inductor, there is also no current through the resistor? I can't put one and two together.

Correct again!

This assumes we are talking about R_1 and L_1, btw.

For R_2 and L_2 [Edit: and in particular the voltage drop across R_2] you should be able to reach a similar conclusion, but only after invoking Kirchhoff's Current Law.

 

[Sho Kano]

Correct again!

This assumes we are talking about R_1 and L_1, btw.

For R_2 and L_2 [Edit: and in particular the voltage drop across R_2] you should be able to reach a similar conclusion, but only after invoking Kirchhoff's Current Law.

Okay, I used Kirchhoff's current rule. It seems to say exactly that. (if I'm using the rule correctly) Current loops around the left-ward loop, and has to be uniform everywhere. Since the current through inductor 1 is zero, then there must be no current at all.

 

[Sho Kano]

Both currents run through $R_2$ generating a voltage drop. The term in the second (and first) loop equations should be $(i_2 - i_1)R_2$, right?

Can someone clarify this? The way I was taught was just that the current splits off so that there are no more than 1 current through each resistor. You claim otherwise, but I would love to know if this way works too.

 

[collinsmark]

Okay, I used Kirchhoff's current rule. It seems to say exactly that. (if I'm using the rule correctly) Current loops around the left-ward loop, and has to be uniform everywhere. Since the current through inductor 1 is zero, then there must be no current at all.

I think you are on the right track, but I just want to be sure...

Using the same logic we have been discussing so far, what is the current through L_2? [Edit: immediately after the switch is closed, and assuming the switch had been open for a long time before it was closed.]

Invoking Kirchhoff's Current Law (KCL), what is the current flowing through R_2 in terms of the current flowing through L_1 and the current flowing through L_2?

(Hint: Just above R_2 there is a node that contains 3 paths connecting to it. One path (to the left) is the current flowing through L_1. Another path (to the right) is the current flowing through L_2. What does KCL say about how this relates the the third path, that path that goes through R_2 ?)

 

[Paul Colby]

Currents add. If two currents flow in a conductor then the total current is the sum of the two. Currents are the motion of charge which is never lost or created. So say I have $i_1$ it goes up through $R_1$ and $L_1$, down through $R_2$ across the switch (which is closed) through the battery. The current $i_1$ forms a closed loop and has some value. Now, imagine a second current $i_2$ going up through $R_2$ and down through $L_2$ forming a closed loop.

 

[collinsmark]

Can someone clarify this? The way I was taught was just that the current splits off so that there are no more than 1 current through each resistor. You claim otherwise, but I would love to know if this way works too.

You should be able to answer this using Kirchhoff's Current Law. If you have a node with three individual paths connected to it, and two paths each contain an individual current, what does KCL say about the third path*?

*(i.e., in terms of the currents through the other two paths)

 

[Paul Colby]

By choosing loop currents the current law is automatically imposed. Doing so means that your original equations have terms which should look like $(i_1-i_2)R_2$

 

[Sho Kano]

I think you are on the right track, but I just want to be sure...

Using the same logic we have been discussing so far, what is the current through L_2? [Edit: immediately after the switch is closed, and assuming the switch had been open for a long time before it was closed.]

Invoking Kirchhoff's Current Law (KCL), what is the current flowing through R_2 in terms of the current flowing through L_1 and the current flowing through L_2?

(Hint: Just above R_2 there is a node that contains 3 paths connecting to it. One path (to the left) is the current flowing through L_1. Another path (to the right) is the current flowing through L_2. What does KCL say about how this relates the the third path, that path that goes through R_2 ?)

The current through L2 just after the switch is closed is also zero.
If the initial current (through L1) is i, it will split up into two currents at the junction, i1 into R2, and i2 into L2.
i = i1 + i2
i1, the current through R2 is = i - i2

 

[Sho Kano]

You should be able to answer this using Kirchhoff's Current Law. If you have a node with three individual paths connected to it, and two paths each contain an individual current, what does KCL say about the third path*?

*(i.e., in terms of the currents through the other two paths)

Ah I see he's just expressing it in terms of the other currents right?

 

[collinsmark]

The current through L2 just after the switch is closed is also zero.
If the initial current (through L1) is i, it will split up into two currents at the junction, i1 into R2, and i2 into L2.
i = i1 + i2
i1, the current through R2 is = i - i2

Ah I see he's just expressing it in terms of the other currents right?

Right.

The "+" or "-" sign that goes in front of a particular current depends on how you set up and define your currents at the beginning.

Back in your Post #4, it seems that you used Kirchhoff's Voltage Law (KVL, as opposed to KCL) to define your currents in the form of current loops. You get to define these loops yourself. (There are multiple ways to define the loops in KVL. It's your job to choose one and be consistent from then on.)

You may have made a mistake in there somewhere when setting up your equations, btw (in post #4).

But using KVL (instead of KCL) to set up and define your current directions is perfectly okay. Once you define your currents, the important part after that is to maintain consistency.

Whether you use KCL or KVL to set up your system of equations, the final results should produce identical answers in the end, when evaluating the current through and voltage across any given component.

You are the one in charge of setting these things up. And you must maintain self consistency.

So yes, i = i_1 + i_2 is correct, or at least can be correct, depending on how you define i, i_1 and i_2.

In KVL it depends on which loops pass through a given component, and the direction of each of the loops (clockwise or counterclockwise).

In KCL it depends on which currents are directed into a node and which currents are directed out of a node.

 

[Sho Kano]

So to answer the question, since there is no current at all, the inductor must be "holding back" the battery's emf right? So EMF = Ldi/dt

 

[collinsmark]

So to answer the question, since there is no current at all, the inductor must be "holding back" the battery's emf right?

I've never heard it in those words before, but yes, I suppose that does sound like a qualitative way to put it.

So EMF = Ldi/dt

Yes, that certainly looks correct to me.

 

[Sho Kano]

I've never heard it in those words before, but yes, I suppose that does sound like a qualitative way to put it.

Yes, that certainly looks correct to me.

thank you :)

 

[lychette]

the current through an inductor may not change instantaneously and to talk about this is misleading!
When the switch is closed the current may be zero but it will be changing at a rate given by e = -L.di/dt so it ios possible to describe the current !

 

[phyzguy]

the current through an inductor may not change instantaneously and to talk about this is misleading!
When the switch is closed the current may be zero but it will be changing at a rate given by e = -L.di/dt so it ios possible to describe the current !

I think this is what we have all been saying. The current is zero immediately after the switch is closed, but dI/dt is not zero. What is your point?

 

[lychette]

just making the point that seems to have been lost searching for 'qualitative' explanations...inductor is 'holding back' the battery emf does not sound like a physics explanation. This aftert 49 posts!

 

[Sho Kano]

it

just making the point that seems to have been lost searching for 'qualitative' explanations...inductor is 'holding back' the battery emf does not sound like a physics explanation. This aftert 49 posts!

Please explain n a physics explanation then :)

 

[lychette]

An inductor is a device that produces an 'induced emf' whenever it experiences a changing magnetic fluk linkage.
In this example the changing magnetic flux linkage is caused by a changing current and therefore an induced emf occurs when the current changes.
The induced emf opposes the change in magnetic flux linkage, i.e.opposes the changing current. When the switch is closed the current tends to increase and therefore the induced emf opposes the increasing current i.e opposes the applied emf (I think it is bad to describe this as 'holding back' the battery emf )
When the switch is opened the current tends to decrease and now the induced emf tries to maintain the flow of current i.e it is in the same direction as the battery emf ! Trying to explain this complex problem qualitatively is extremely difficult and confusing.
The explanation is in mathematics.
The induced emf e = -Ldi/dt The minus sign conveys opposition to changing I (magnetic flux) and di/dt conveys the magnitude of the effect