Inductors and circuit problem

[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