How Does an Inductor Affect Current Change Upon Switch Closure?

AI Thread Summary
The discussion centers on the behavior of inductors in a circuit immediately after a switch is closed. It emphasizes that the current through an inductor cannot change instantaneously, meaning that just after the switch closure, the current remains at zero. Participants explore the implications of this behavior on voltage drops across resistors and inductors, clarifying that while current is zero, the voltage across the inductor can still be non-zero due to its inductance. The conversation also touches on the importance of Kirchhoff's laws and the conservation of charge in circuit analysis. Ultimately, the key takeaway is that inductors resist changes in current, which affects circuit dynamics immediately after a switch is closed.
Sho Kano
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Homework Statement


In the following figure, R1= 3.1 Ω, R2= 4.9 Ω, L1= 0.155 H, L2= 0.2 H and the ideal battery has ε = 5.6 V. Just after switch S is closed, at what rate is the current in inductor 1 changing?
HW10Q16.png


Homework Equations


{ \varepsilon }_{ induced }\quad =\quad L\frac { di }{ dt } \\ V\quad =\quad IR

The Attempt at a Solution


Don't know where to start. I know that the inductor will induce an emf in the opposite direction of the battery's. Right after the switch is closed, I guess it has two resistors in parallel, so the current can be calculated from Ohm's law. But that doesn't seem right since there will be a "resistance" from the inductors.
 
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Can you write down a differential equation for ##i(t)##?
 
Actually, more basic question, what rules do you know about all circuits?
 
Here is an attempt
Loop 1 (left side), Loop 2
\varepsilon \quad -\quad i_{ 1 }{ R }_{ 1 }\quad -\quad L\frac { di_{ 1 } }{ dt } \quad -\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0\\ -L\frac { di_{ 2 } }{ dt } \quad +\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0
 
Looks close. I'm having a hard time understanding your definition of ##i_1## and ##i_2##? What happens to ##i_2## when it goes out of the inductor?
 
Paul Colby said:
Looks close. I'm having a hard time understanding your definition of ##i_1## and ##i_2##? What happens to ##i_2## when it goes out of the inductor?
I just have two currents because there are two loops. I guess in this case, i2 doesn't matter because the problem only asks for regarding the leftmost loop.

So only the first equation.

Edit: My bad, I think your confusion comes from the fact that I forgot to add subscripts to the L in the second equation
 
The reason I ask about ##i_2## is that your equations imply that ##i_2## just disappears. It's only in the second inductor. Doesn't that violate charge conservation?
 
Paul Colby said:
The reason I ask about ##i_2## is that your equations imply that ##i_2## just disappears. It's only in the second inductor. Doesn't that violate charge conservation?
Here, I edited them
\varepsilon \quad -\quad i_{ 1 }{ R }_{ 1 }\quad -\quad { L }_{ 1 }\frac { di_{ 1 } }{ dt } \quad -\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0\\ -{ L }_{ 2 }\frac { di_{ 2 } }{ dt } \quad +\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0
 
The thing about choosing loop currents as variables is they are automatically conserved because they form closed loops. If ##i_2## runs around the second loop shouldn't it be flowing through ##R_2## as well as ##L_2##?
 
  • #10
Paul Colby said:
The thing about choosing loop currents as variables is they are automatically conserved because they form closed loops. If ##i_2## runs around the second loop shouldn't it be flowing through ##R_2## as well as ##L_2##?
By that, equation #1 needs a i2R2 term as well correct?
 
  • #11
Conventional current flows from the positive of the terminal through inductor 1, then splits off into i1 downwards, and i2 rightwards. i2 then joins with i1 at the other junction. It doesn't make sense for i2 to flow upwards.
 
  • #12
Well, ##R_2## is part of both loops so both ##i_1## and ##i_2## have to contribute to the voltage drop across ##R_2##. Be careful with the signs :)
 
  • #13
I find it helpful to draw the two loop currents in their respective loops. Choose some loop direction for each loop and stick with it.
 
  • #14
Paul Colby said:
I find it helpful to draw the two loop currents in their respective loops. Choose some loop direction for each loop and stick with it.
okay, I have i1 for loop the left loop, and l2 for the right one
\varepsilon \quad -\quad i_{ 1 }{ R }_{ 1 }\quad -\quad { L }_{ 1 }\frac { di_{ 1 } }{ dt } \quad -\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0\\ -{ L }_{ 2 }\frac { di_{ 2 } }{ dt } \quad -\quad i_{ 2 }{ R }_{ 2 }\quad =\quad 0
 
  • #15
You're making it much too hard. The current through an inductor cannot change instantaneously. So immediately after the switch is closed, the current through the inductors is zero. So there is no voltage drop across the resistors, since there is no current.
[ moderator's note: message abridged, too much help provided ]
 
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  • #16
phyzguy said:
You're making it much too hard. The current through an inductor cannot change instantaneously. So immediately after the switch is closed, the current through the inductors is zero.
What about L2, doesn't that get factored in somehow too?
 
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  • #17
Sho Kano said:
What about L2, doesn't that get factored in somehow too?
By the same logic that @phyzguy just mentioned, and also invoking Kirchhoff's current law, what is the current through R_2? Thus what is the voltage across R_2? Thus what is the voltage across L_2? :wink:
 
  • #18
Sho Kano said:
okay, I have i1 for loop the left loop, and l2 for the right one
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?
 
  • #19
phyzguy said:
You're making it much too hard. The current through an inductor cannot change instantaneously. So immediately after the switch is closed, the current through the inductors is zero. So there is no voltage drop across the resistors, since there is no current.
[ moderator's note: message abridged, too much help provided ]
Is it because of the resistance in the inductor to the change in current, that in time = 0, there is no current through it?
 
  • #20
Sho Kano said:
Is it because of the resistance in the inductor to the change in current, that in time = 0, there is no current through it?
When working with ideal, lumped parameter components, and assuming that there are no voltage/current sources that can generate an infinite amount of voltage/current (even for a very short period of time, such as an ideal impulse), then the following holds:
  • The current through an inductor cannot jump from one value to a different value instantaneously.
  • The voltage across a capacitor cannot jump from one value to a different value instantaneously.
The current through an inductor can change gradually. That much is allowed. But it cannot jump from one value to another instantaneously. (The same is true for the voltage across a capacitor.)

For example, if the moment before the switch is closed, an inductor has 0 A flowing through it, then the current flowing through it immediately after the switch is closed must be infinitesimally close to 0 A (in other words, 0 A).
 
  • #21
Sho Kano said:
Is it because of the resistance in the inductor to the change in current, that in time = 0, there is no current through it?

Yes. Inductors resist changes in current. Since there is no current through it before the switch is closed, there is no current immediately after the switch is closed. To change the current from zero to some non-zero value instantaneously would require an infinite voltage. Similarly, capacitors resist changes in voltage, so the voltage across a capacitor cannot change instantaneously.
 
  • #22
phyzguy said:
Yes. Inductors resist changes in current. Since there is no current through it before the switch is closed, there is no current immediately after the switch is closed. To change the current from zero to some non-zero value instantaneously would require an infinite voltage. Similarly, capacitors resist changes in voltage, so the voltage across a capacitor cannot change instantaneously.
Okay, that helped a lot! Another question regarding post #15: How is there no voltage drop across the resistor that came before inductor 1? Current passes through that resistor, then encounters opposite emf from the inductor right?
 
  • #23
Sho Kano said:
Okay, that helped a lot! Another question regarding post #15: How is there no voltage drop across the resistor that came before inductor 1?

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

Current passes through that resistor, then encounters opposite emf from the inductor right?
[boldface mine]

What current is that? :wink:
 
  • #24
collinsmark said:
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? :wink:
Okay, so because there is no voltage drop across the inductor, there should no current flowing at all, between the inductor and the battery. So that's why my professor said that at time = 0, inductors can be modeled as open switches.
 
  • #25
Sho Kano said:
Okay, so because there is no voltage drop across the inductor, there should no current flowing at all, between the inductor and the battery.

How do you conclude that?
 
  • #26
phyzguy said:
How do you conclude that?
Edit: what do you mean?
 
  • #27
Sho Kano said:
Okay, so because there is no voltage drop across the inductor, there should no current flowing at all, between the inductor and the battery. So that's why my professor said that at time = 0, inductors can be modeled as open switches.
Wait, I think you misunderstood.

  • It is acceptable to change the voltage across an inductor instantaneously. That's totally OK.
  • It is not acceptable to change the current through an inductor instantaneously. That's forbidden (without a source that can produce infinite voltage).

  • Similarly, it is acceptable to change the current through a capacitor instantaneously. That's totally OK.
  • It is not acceptable to change the voltage across a capacitor instantaneously. That's forbidden (without a source that can produce infinite current).
So your statement about the voltage drop across the inductor being 0 is not correct. It's fine to change the voltage drop across an inductor instantaneously. [Edit: just as it is acceptable to instantaneously change the voltage drop across an open switch: a switch which blocks current from flowing.]
 
  • #28
Sho Kano said:
Edit: what do you mean?

As collinsmark said, why do you think there is no voltage drop across the inductor.
 
  • #29
Sho Kano said:
The voltage across the inductor is equal to the battery voltage, so you can solve for di/dt from the equation
phyzguy said:
As collinsmark said, why do you think there is no voltage drop across the inductor.
Because there is no current through it
 
  • #30
Sho Kano said:
Because there is no current through it

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.
 
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  • #31
Sho Kano said:
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.
 
  • #32
phyzguy said:
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
 
  • #33
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.
 
  • #34
Sho Kano said:
{ \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?
 
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  • #35
collinsmark said:
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? :wink:
0 Current?
 
  • #36
Sho Kano said:
0 Current?
Yes, that much is correct. :smile:

[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.]
 
  • #37
collinsmark said:
Yes, that much is correct. :smile:
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.
 
  • #38
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.
 
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  • #39
Sho Kano said:
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! :approve:

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. :wink:
 
  • #40
collinsmark said:
Correct again! :approve:

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. :wink:
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.
 
  • #41
Paul Colby said:
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.
 
  • #42
Sho Kano said:
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 ?)
 
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  • #43
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.
 
  • #44
Sho Kano said:
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)
 
  • #45
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##
 
  • #46
collinsmark said:
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
 
  • #47
collinsmark said:
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?
 
  • #48
Sho Kano said:
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 said:
Ah I see he's just expressing it in terms of the other currents right?
Right. :smile:

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.
 
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  • #49
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
 
  • #50
Sho Kano said:
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. :approve:
 
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