Solving Inductor Circuit Homework Problems

[superslow991]

Homework Statement

For the circuit shown in the figure, the inductors have no appreciable resistance and the switch has been open for a very long time.

(Image of circuit)

(a) The instant after closing the switch, what is the current through the 60.0-Ω resistor?
(b) The instant after closing the switch, what is the potential difference across the 15.0-mH inductor?
(c) After the switch has been closed and left closed for a very long time, what is the potential drop across the 60.0-Ω resistor?

Homework Equations

V=IR
I=V/R

The Attempt at a Solution

Part A- i assume since we are finding the current through the 60 ohm resister the 10 and 60 ohm resistors will be in series so for the current i just use I=V/R? I ignore both inductors and the 30 ohm resistor right? if so why? I also believe the circuit will act like an open circuit but I am at a loss for that.

Part B- Apparently the potential difference between the 15 mH inductor is equivalent to the potential difference between the 60 ohm resistor but why is that so?

Part C- I realize after some time has passed when the switch is closed the current will not fluctuate as much so there would be no change in current and the inductor behaved like a short circuit where no voltage is induced so the voltage would be 0? still not sure on this.

 

[kuruman]

Sorry, there is no figure to look at.

 

[superslow991]

Sorry, there is no figure to look at.

Really? you try copy and pasting the gyazo link?

 

[gneill]

Really? you try copy and pasting the gyazo link?

There is no image. Please upload your image to Physics Forums as an attachment (use UPLOAD).

 

[superslow991]

There is no image. Please upload your image to Physics Forums as an attachment (use UPLOAD).

Fixed thanks

 

[kuruman]

Thank you for fixing the figure.

I also believe the circuit will act like an open circuit but I am at a loss for that.

(a) What is the expression for the current through a charging inductor? What happens at t = 0?
(b) What is the voltage across the 30 Ω resistor? Hint: How much current is flowing through it?
(c) If dI/dt is zero, the voltage across an inductor is zero. That means you can replace it with a short.

 

[superslow991]

Thank you for fixing the figure.

(a) What is the expression for the current through a charging inductor? What happens at t = 0?
(b) What is the voltage across the 30 Ω resistor? Hint: How much current is flowing through it?
(c) If dI/dt is zero, the voltage across an inductor is zero. That means you can replace it with a short.

(a)- I figure I = V/R and if t=0 doesn't that mean the circuit is open?
(b)- I think no current goes through the 30 ohm resistor so there potential difference is 0?

 

[kuruman]

(a)- I figure I = V/R and if t=0 doesn't that mean the circuit is open?
(b)- I think no current goes through the 30 ohm resistor so there potential difference is 0?

Correct on both counts. A charging inductor acts like an open circuit at t = 0 and like a short a long time later.

On edit: To clarify, in (b) the potential difference across the 30 Ω resistor is zero. If that's the case, the potential difference across the 15.0 mH inductor is the same as ... ?

 

[superslow991]

Correct on both counts. A charging inductor acts like an open circuit at t = 0 and like a short a long time later.

On edit: To clarify, in (b) the potential difference across the 30 Ω resistor is zero. If that's the case, the potential difference across the 15.0 mH inductor is the same as ... ?

the same as the 60 ohm resistor?

Also I am not sure why no current goes through the 30 ohm resistor.

 

[cnh1995]

the same as the 60 ohm resistor?

Yes.

Also I am not sure why no current goes through the 30 ohm resistor.

Only at t=0, because

A charging inductor acts like an open circuit at t = 0

 

[superslow991]

Yes.

Only at t=0, because

ok Thanks but what about for part b why would the potential difference across the 15 mH be the same for the 60 ohm resistor?

 

[cnh1995]

ok Thanks but what about for part b why would the potential difference across the 15 mH be the same for the 60 ohm resistor?

Because the voltage across the 30 ohm resistor is zero. It can be treated as a wire with zero resistance.

 

[superslow991]

Because the voltage across the 30 ohm resistor is zero. It can be treated as a wire with zero resistance.

I get that but like what's the correlation with the 70 ohm resistor and the 15 mH?

 

[kuruman]

You mean the 60 ohm resistor. Look at the circuit imagining that the 30 ohm resistor is replaced with a straight wire. How is the voltage across the inductor related to the voltage across the 60 ohm resistor?

 

[cnh1995]

I get that but like what's the correlation with the 70 ohm resistor and the 15 mH?

Replace the inductors with open switches and apply KVL to the rightmost loop containing the 60 ohm and 30 ohm resistances and the open switch (15mH inductance).

 

[superslow991]

You mean the 60 ohm resistor. Look at the circuit imagining that the 30 ohm resistor is replaced with a straight wire. How is the voltage across the inductor related to the voltage across the 60 ohm resistor?

Parallel?

 

[kuruman]

Parallel?

Yes, so how does it compare with the voltage across the 60 ohm?

 

[superslow991]

Yes, so how does it compare with the voltage across the 60 ohm?

So if it's parallel they should have the same resistance so it would just tact on the 60 ohm resistor?

 

[cnh1995]

they should have the same resistance **voltage**

 

[superslow991]

Oh ok thanks a lot for the help

 

[gneill]

Could someone please help me with a detailed explanation as to why the 60 ohm resistor would have 0 voltage. I told my teacher that after some time the switch is closed the inductors should act like an short circuit and the 15mH inductor would short the 60 ohm resistor. But this was not enough. If anyone can help me some type of detailed explanation

Replacing the 15 mH inductor with a short would still leave the 30 Ω resistor in parallel with the 60 Ω resistor. So that's not a valid reason for the potential there to be zero.

Look for another reason. What's the other inductor doing?

 

[kuruman]

A long time later, the inductor carries maximum current. How do I know? Because that is how "long time" is defined. Maximum current means that the current through the inductor is not changing. What is the potential difference across an inductor when the current through it is not changing?

 

[superslow991]

Replacing the 15 mH inductor with a short would still leave the 30 Ω resistor in parallel with the 60 Ω resistor. So that's not a valid reason for the potential there to be zero.

Look for another reason. What's the other inductor doing?

I'm not sure lol is it going to short the 60 ohm resistor?

 

[superslow991]

A long time later, the inductor carries maximum current. How do I know? Because that is how "long time" is defined. Maximum current means that the current through the inductor is not changing. What is the potential difference across an inductor when the current through it is not changing?

Potential difference should be 0 right? The inductor would just be a wire?

 

[gneill]

I'm not sure lol is it going to short the 60 ohm resistor?

See if you can't make a logical deduction from the known behavior of an inductor.

 

[kuruman]

I'm not sure lol is it going to short the 60 ohm resistor?

Redraw the circuit with the inductors replaced with straight wires. Analyze it.

 

[superslow991]

Redraw the circuit with the inductors replaced with straight wires. Analyze

Wait so they would be in series correct total resistance would just be 100?

 

[kuruman]

I would redraw the circuit on a piece of paper and then study it. The three resistors are not in series if you replace the inductors with straight wires.

 

[superslow991]

I would redraw the circuit on a piece of paper and then study it. The three resistors are not in series if you replace the inductors with straight wires.

How so? Wouldn't 30 and 60 be in series? Also 10 and 30?

 

[superslow991]

See if you can't make a logical deduction from the known behavior of an inductor.

I mean inductor just induced an emf to oppose the changing current

 

[kuruman]

How so? Wouldn't 30 and 60 be in series? Also 10 and 30?

Please post the redrawn circuit as I indicated in post #28 and I will explain to you what is and what is not in series and why.

 

[superslow991]

Please post the redrawn circuit as I indicated in post #28 and I will explain to you what is and what is not in series and why.

 

[kuruman]

Right. Now remember that when two resistors are in series, the current that flows through one resistor must flow through the other. That's what "in series" means. Now imagine current I starting at the battery moving to the right through the 10 Ω. It reaches the junction where the straight wire is. What happens next? Does the current go through the straight wire or does it split itself in two?

 

[superslow991]

Right. Now remember that when two resistors are in series, the current that flows through one resistor must flow through the other. That's what "in series" means. Now imagine current I starting at the battery moving to the right through the 10 Ω. It reaches the junction where the straight wire is. What happens next? Does the current go through the straight wire or does it split itself in two?

Split itself

 

[kuruman]

Split itself

Think again. If some of the current went into the right branch, then some of that current would have to go through the 60 Ω. Now Ohm's law says V = IR which means that there must be a potential difference across the 60 Ω if current flows through it. But that can't be because the potential difference across the straight wire (which the same as the potential difference across the 60 Ω) is zero. Therefore ... ?

 

[superslow991]

Think again. If some of the current went into the right branch, then some of that current would have to go through the 60 Ω. Now Ohm's law says V = IR which means that there must be a potential difference across the 60 Ω if current flows through it. But that can't be because the potential difference across the straight wire (which the same as the potential difference across the 60 Ω) is zero. Therefore ... ?

See that's where I'm lost. I understand that the potential difference would be zero but what is the straight side doing that makes the 60 ohm resistornhave the potential difference of zero? And what about the 30 ohm or 10 ohm would they have no potential difference?

 

[kuruman]

Any straight segment in a drawn circuit is by definition and convention an equipotential. This means that the potential difference between any two points on a segment connecting circuit elements is zero. For example, the potential at the positive end of the battery is the same along the wire all the way to the 10 Ω.

 

[superslow991]

Any straight segment in a drawn circuit is by definition and convention an equipotential. This means that the potential difference between any two points on a segment connecting circuit elements is zero. For example, the potential at the positive end of the battery is the same along the wire all the way to the 10 Ω.

Okay so your saying since the current would travel down where the the straight wire would be. There would be 0 potential difference and since that straight wire is parallel to the 60 ohm resistor the 60 ohm resistor will have 0 potential difference?

 

[kuruman]

Okay so your saying since the current would travel down where the the straight wire would be. There would be 0 potential difference and since that straight wire is parallel to the 60 ohm resistor the 60 ohm resistor will have 0 potential difference?

Yes. The straight wire takes all the current so the equivalent circuit is just the battery and the 10 Ω. The 30 Ω and 60 Ω are in parallel with the straight wire but they draw no current.

 

[superslow991]

Yes. The straight wire takes all the current so the equivalent circuit is just the battery and the 10 Ω. The 30 Ω and 60 Ω are in parallel with the straight wire but they draw no current.

Hmm ok thanks a lot. Very appreciative. Still kinda iffy on the 30 and 60 ohm resistors being parallel to the straight wire

 

[superslow991]

Edit*

 

[cnh1995]

Hmm ok thanks a lot. Very appreciative. Still kinda iffy on the 30 and 60 ohm resistors being parallel to the straight wire

Assume for a moment, that there is no wire. What would be the equivalent of the 30 and 60 ohm resistors?

 

[superslow991]

Assume for a moment, that there is no wire. What would be the equivalent of the 30 and 60 ohm resistors?

Hmm so if I'm reading right the 30 and 60 would be in series so are you trying to say where the 60 ohms are at it would be replaced by 90 ohms and thus it would be parallel to that straight wire to the left?

 

[cnh1995]

the 30 and 60 would be in series

No. They would be in parallel. And

it would be parallel to that straight wire to the left

 

[superslow991]

No. They would be in parallel. And

Ok could you please help me understand how the 30 and 60 ohms would be parallel cause I'm not seeing it

 

[cnh1995]

Ok could you please help me understand how the 30 and 60 ohms would be parallel cause I'm not seeing it

Replace the 15mH inductor with a wire. See the ends of the 30 ohm and 60 ohm resistance. Are they connected between the same two points?

 

[superslow991]

Replace the 15mH inductor with a wire. See the ends of the 30 ohm and 60 ohm resistance. Are they connected between the same two points?

By ends are you talking about the bottom point of the 60 ohm resistor and the edge of the 30 ohm resistor? if so then no they are not connected.

But what about when the current reaches up until the point where it can travel to the 30 ohm resistor or go down to the 60 ohm?

 

[cnh1995]

Replace the 15mH with a wire first.

By ends are you talking about the bottom point of the 60 ohm resistor and the edge of the 30 ohm resistor?

By ends, I mean the "terminals" of the resistors. For parallel connection, the four terminals ( two for each resistance) should be connected between the "same" two points.

 

[superslow991]

Replace the 15mH with a wire first.By ends, I mean the "terminals" of the resistors. For parallel connection, the four terminals ( two for each resistance) should be connected between the "same" two points.

ok so if I am reading this right. The 60 ohms terminal goes up and down. the 60 ohms terminal that's above is at the same point of the 30 ohms terminal to the left?? now the 30 ohms terminal to the left is in the same point of the 60 ohm terminal above? now the 30 ohms terminal to the right go down then connects at the same point of the 60 ohms terminal down bottom since the 15 mH inductor will just be a straight wire?

so essentially the current can go through 60 ohms or 30 ohms but not both?

 

[cnh1995]

ok so if I am reading this right. The 60 ohms terminal goes up and down. the 60 ohms terminal that's above is at the same point of the 30 ohms terminal to the left?? now the 30 ohms terminal to the left is in the same point of the 60 ohm terminal above? now the 30 ohms terminal to the right go down then connects at the same point of the 60 ohms terminal down bottom since the 15 mH inductor will just be a straight wire?

Yes.

so essentially the current can go through 60 ohms or 30 ohms but not both?

I don't know what you mean by that. If the middle wire were not there, the current would flow through both the 30 and 60 ohm reistors. But since they are in parallel with the middle wire, no current flows in either of them. Look up 'current divider'.