Solving Inductor Circuit Homework Problems

In summary: I get that but like what's the correlation with the 70 ohm resistor and the 15 mH?The 70 ohm resistor is connected between the 15.0 mH inductor and the switch.
  • #1
superslow991
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1

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.
php0Fgk3k.png

(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.
 
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  • #2
Sorry, there is no figure to look at.
 
  • #3
kuruman said:
Sorry, there is no figure to look at.
Really? you try copy and pasting the gyazo link?
 
  • #4
superslow991 said:
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).
 
  • #5
gneill said:
There is no image. Please upload your image to Physics Forums as an attachment (use UPLOAD).
Fixed thanks
 
  • #6
Thank you for fixing the figure.
superslow991 said:
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.
 
  • #7
kuruman said:
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?
 
  • #8
superslow991 said:
(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 ... ?
 
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  • #9
kuruman said:
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.
 
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  • #10
superslow991 said:
the same as the 60 ohm resistor?
Yes.
superslow991 said:
Also I am not sure why no current goes through the 30 ohm resistor.
Only at t=0, because
kuruman said:
A charging inductor acts like an open circuit at t = 0
 
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  • #11
cnh1995 said:
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?
 
  • #12
superslow991 said:
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.
 
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  • #13
cnh1995 said:
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?
 
  • #14
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?
 
  • #15
superslow991 said:
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).
 
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  • #16
kuruman said:
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?
 
  • #17
superslow991 said:
Parallel?
Yes, so how does it compare with the voltage across the 60 ohm?
 
  • #18
kuruman said:
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?
 
  • #19
superslow991 said:
they should have the same resistance **voltage**
 
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  • #20
Oh ok thanks a lot for the help
 
  • #21
superslow991 said:
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? :wink:
 
  • #22
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?
 
  • #23
gneill said:
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? :wink:
I'm not sure lol is it going to short the 60 ohm resistor?
 
  • #24
kuruman said:
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?
 
  • #25
superslow991 said:
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.
 
  • #26
superslow991 said:
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.
 
  • #27
kuruman said:
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?
 
  • #28
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.
 
  • #29
kuruman said:
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?
 
  • #30
gneill said:
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
 
  • #31
superslow991 said:
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.
 
  • #32
kuruman said:
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.
kyMFM.&token=655e3af7-c1c6-4850-84dd-9aa220c02e68&owa=outlook.live.com&isc=1&isImagePreview=True.jpg
 
  • #33
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?
 
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  • #34
kuruman said:
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
 
  • #35
superslow991 said:
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 ... ?
 

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