Inductors and circuit problem

[Sho Kano]

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?

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.

 

[Paul Colby]

Can you write down a differential equation for $i(t)$?

 

[Paul Colby]

Actually, more basic question, what rules do you know about all circuits?

 

[Sho Kano]

Here is an attempt
Loop 1 (left side), Loop 2
$\varepsilon \quad -\quad i_{ 1 }{ R }_{ 1 }\quad -\quad L\frac { di_{ 1 } }{ dt } \quad -\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0\\ -L\frac { di_{ 2 } }{ dt } \quad +\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0$

 

[Paul Colby]

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?

 

[Sho Kano]

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

 

[Paul Colby]

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?

 

[Sho Kano]

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
$\varepsilon \quad -\quad i_{ 1 }{ R }_{ 1 }\quad -\quad { L }_{ 1 }\frac { di_{ 1 } }{ dt } \quad -\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0\\ -{ L }_{ 2 }\frac { di_{ 2 } }{ dt } \quad +\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0$

 

[Paul Colby]

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 $L_2$?

 

[Sho Kano]

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?

 

[Sho Kano]

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.

 

[Paul Colby]

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 :)

 

[Paul Colby]

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.

 

[Sho Kano]

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
$\varepsilon \quad -\quad i_{ 1 }{ R }_{ 1 }\quad -\quad { L }_{ 1 }\frac { di_{ 1 } }{ dt } \quad -\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0\\ -{ L }_{ 2 }\frac { di_{ 2 } }{ dt } \quad -\quad i_{ 2 }{ R }_{ 2 }\quad =\quad 0$

 

[phyzguy]

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 ]

 

[Sho Kano]

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?

 

[collinsmark]

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 $L_2$?

 

[Paul Colby]

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?

 

[Sho Kano]

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?

 

[collinsmark]

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).

 

[phyzguy]

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.

 

[Sho Kano]

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?

 

[collinsmark]

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?

 

[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?

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.

 

[phyzguy]

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?

 

[Sho Kano]

How do you conclude that?

Edit: what do you mean?

 

[collinsmark]

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.]

 

[phyzguy]

Edit: what do you mean?

As collinsmark said, why do you think there is no voltage drop across the inductor.

 

[Sho Kano]

The voltage across the inductor is equal to the battery voltage, so you can solve for di/dt from the equation

As collinsmark said, why do you think there is no voltage drop across the inductor.

Because there is no current through it

 

[phyzguy]

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.

 

[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?