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Inductors and circuit problem

  • #1
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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


[itex]{ \varepsilon }_{ induced }\quad =\quad L\frac { di }{ dt } \\ V\quad =\quad IR[/itex]

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.
 

Answers and Replies

  • #2
Paul Colby
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Can you write down a differential equation for ##i(t)##?
 
  • #3
Paul Colby
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Actually, more basic question, what rules do you know about all circuits?
 
  • #4
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Here is an attempt
Loop 1 (left side), Loop 2
[itex]\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[/itex]
 
  • #5
Paul Colby
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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?
 
  • #6
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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
 
  • #7
Paul Colby
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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?
 
  • #8
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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
[itex]\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[/itex]
 
  • #9
Paul Colby
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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
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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
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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
Paul Colby
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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
Paul Colby
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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
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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
[itex]\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[/itex]
 
  • #15
phyzguy
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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 ]
 
  • #16
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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?
 
Last edited by a moderator:
  • #17
collinsmark
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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 [itex] R_2 [/itex]? Thus what is the voltage across [itex] R_2 [/itex]? Thus what is the voltage across [itex] L_2 [/itex]? :wink:
 
  • #18
Paul Colby
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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
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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
collinsmark
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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
phyzguy
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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
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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
collinsmark
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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
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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
phyzguy
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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?
 

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