Current Through LR Circuits

In summary: If the current cannot change instantaneously, then neither can the voltage across the inductor. This is due to the relationship between voltage and current in an inductor (V = L * dI/dt).In a series circuit, the current at t=0 would be zero, as the inductor would resist any change in current. This means that the voltage across the resistor would also be zero, since V = IR.In a parallel circuit, the current would flow through the path of least resistance, which would be the resistor. The inductor would still resist any change in current, but since there is another path for the current to flow through, the resistor would still
  • #1
garthenar
35
8
Homework Statement
Rank the circuits based on the current through the battery immediately after the switch is closed.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Relevant Equations
none given but i=(E/R(1-e^-(R/L)t))
lrCircuits.PNG

current through an lr circuit.PNG

is the equation I found in the book for the current through an LR-circuit. The problem is I can't find anything in the text I have that deals with resistors in series vs parallel, if a coil acts as a short, etc. So can I ask you...

1) is this equation valid for both resistors in parallel and series?

2) If I have a set up like the second and third from the left will the current go through the coil like a short circuit or does the coil have a resistance?

3) in a situation like the second from the right, I can use this equation with the total resistance of the branch and the inductance of the coil? Or does the order matter and if so why?

Are there any concepts I'm missing that I need to analyze an RC circuit?

Thank you for your help.
 

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  • #2
what book did you get these questions from?
 
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  • #3
garthenar said:
Homework Statement: Rank the circuits based on the current through the battery immediately after the switch is closed.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Homework Equations: none given but i=(E/R(1-e^-(R/L)t))

View attachment 252223
View attachment 252222
is the equation I found in the book for the current through an LR-circuit. The problem is I can't find anything in the text I have that deals with resistors in series vs parallel, if a coil acts as a short, etc. So can I ask you...

1) is this equation valid for both resistors in parallel and series?

2) If I have a set up like the second and third from the left will the current go through the coil like a short circuit or does the coil have a resistance?

3) in a situation like the second from the right, I can use this equation with the total resistance of the branch and the inductance of the coil? Or does the order matter and if so why?

Are there any concepts I'm missing that I need to analyze an RC circuit?

Thank you for your help.
Firstly, if the battery is treated as ideal, meaning it has zero internal resistance, then 2 out of the 6 circuits will lead to nonsensical results. This can be avoided by treating the battery as non-ideal, having a small, non-zero internal resistance.

I would not worry about the equation for this exercise. The key word is "immediately," in "immediately after the switch is closed."

One key factor in solving this problem is to answer these two questions:
(i) Is it possible to instantaneously change the voltage across the terminals of an inductor?
(ii) Is it possible to instantaneously change the current through an inductor?
 
  • #4
collinsmark said:
Firstly, if the battery is treated as ideal, meaning it has zero internal resistance, then 2 out of the 6 circuits will lead to nonsensical results.
HUH ? I don't see that at all.
 
  • #5
phinds said:
HUH ? I don't see that at all.
If all elements in the circuit are treated as ideal, including the battery, then in 2 out of the 6 circuits, the current through the battery will be blow up to infinity immediately.

If the battery is non-ideal (meaning it has some, albeit small, internal resistance), the current through the battery will rise quickly, but without blowing up to infinity.
 
  • #6
Anti Hydrogen said:
what book did you get this questions from?
It's not from the book its from our homework (I checked to see if there was a comparable problem before posting) We are using University Physics 14th edition by Young and Freedman and This homework corresponds to Chapter 30.
 
  • #7
collinsmark said:
Firstly, if the battery is treated as ideal, meaning it has zero internal resistance, then 2 out of the 6 circuits will lead to nonsensical results. This can be avoided by treating the battery as non-ideal, having a small, non-zero internal resistance.

I would not worry about the equation for this exercise. The key word is "immediately," in "immediately after the switch is closed."

One key factor in solving this problem is to answer these two questions:
(i) Is it possible to instantaneously change the voltage across the terminals of an inductor?
(ii) Is it possible to instantaneously change the current through an inductor?
I know that the current in the circuit isn't going to change instantaneously as the inductor is going to resist any change to magnetic flux due to self induction...? If the current can't change instantaneously then voltage can't change instantaneously right?

so an inductor resists rapid changes to current?
so with the series circuits the current at t=0 would be zero (and then increase at something like a time constant I saw with capactiors)?
if the current through the inductor won't change, won't the current go through the resistors in the parallel circuits (path of least resistance?)?
 
  • #8
garthenar said:
I know that the current in the circuit isn't going to change instantaneously as the inductor is going to resist any change to magnetic flux due to self induction...?
Correct. The current through an inductor cannot change instantaneously.

If the current can't change instantaneously then voltage can't change instantaneously right?
While the current through an inductor cannot change instantaneously, there is no such restriction to the voltage across an inductor.

The voltage across an ideal inductor will be whatever it takes to keep the current from changing instantaneously [Edit: "Discontinously" might be a better word here].

so an inductor resists rapid changes to current?

Correct. Inductors resist rapid changes to current.

so with the series circuits the current at t=0 would be zero

Correct again! :smile: However (and this is important), that assumes that the initial current flowing through the inductor is 0. For this problem, I think you are supposed to assume that the initial current is 0.

(and then increase at something like a time constant I saw with capactiors)?

Yes, it is similar in that respect. There are some differences though, but yes, the exponential time constant idea is similar.

While we're at it, you also might want to ask yourself these questions about capacitors:

(iii) Is it possible to instantaneously change the voltage across the terminals of an ideal capacitor?
(iv) Is it possible to instantaneously change the current through an ideal capacitor?

(It doesn't matter for this problem, but it might be useful for future problems involving capacitors.)

if the current through the inductor won't change, won't the current go through the resistors in the parallel circuits (path of least resistance?)?

Yes, both current and voltage can change instantaneously across the terminals of a resistor. Now, knowing what you know about the inductors, you can use the current involving the resistors to rank the circuits.

[Edit: when I say that "the current through an inductor cannot change instantaneously," what I mean is it cannot change by a non-zero, finite amount, instantaneously. In other words, there cannot be discontinuities in inductor current.]
 
  • #9
I'm going to be working on capacitors sometime tomorrow so you can expect a follow up on that. Thank you for all of your help.
 
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  • #10
collinsmark said:
If all elements in the circuit are treated as ideal, including the battery, then in 2 out of the 6 circuits, the current through the battery will be blow up to infinity immediately.

If the battery is non-ideal (meaning it has some, albeit small, internal resistance), the current through the battery will rise quickly, but without blowing up to infinity.
I just don't get it. Can you be more explicit please.
 
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  • #11
phinds said:
I just don't get it. Can you be more explicit please.
If you connect an ideal inductor across the terminals of an ideal battery, the current will blow up to infinity immediately because there is 0 series resistance.

Putting it another way, connecting an ideal inductor across the terminals of an ideal battery -- and leaving it at that -- is nonsense. It's like asking, "what happens when an unstoppable force encounters an unmovable object," or some-such nonsense.
 
  • #12
collinsmark said:
If you connect an ideal inductor across the terminals of an ideal battery, the current will blow up to infinity immediately because there is 0 series resistance.
No, it won't. The inductance will limit the current.
 
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  • #13
Solving the differential equation ##E=L\frac{dI}{dt}## that corresponds to the circuit of an ideal inductor attached to an ideal emf source we get the solution ##I=\frac{E}{L}t+c##, I don't see how that goes up to infinity immediately...
 
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  • #14
collinsmark said:
If you connect an ideal inductor across the terminals of an ideal battery, the current will blow up to infinity immediately because there is 0 series resistance.

Putting it another way, connecting an ideal inductor across the terminals of an ideal battery -- and leaving it at that -- is nonsense. It's like asking, "what happens when an unstoppable force encounters an unmovable object," or some-such nonsense.
Whew ! You really had me going there. I thought I was having a brain fart or something, but what it is, is that you are mistaken. When the switch is thrown, the inductor has INFINITE resistance to a sudden spike in voltage, so acts like an open circuit. Yes, it becomes a problem in the steady-state long run but not when the switch is thrown. That's undoubtedly part of WHY the question is being asked, to see if the student understands that fact, and I see from post #8 that you clearly DO understand that so I'm not clear why you would say what you did regarding the answer to the question.
 
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  • #15
@phinds, @vela, and @Delta2 are correct. The current will not blow up to infinity [well, not immediately], even if all components are considered ideal. My mistake. Sorry. My bad.
 
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1. What is a LR circuit?

A LR circuit is an electrical circuit that consists of an inductor (L) and a resistor (R). It is used to control the flow of current and can create a delay in the current flow.

2. How does current flow through a LR circuit?

In a LR circuit, current flows through the inductor and resistor in a series. The inductor initially resists the flow of current, causing a delay, and then gradually allows the current to flow through. The resistor then limits the amount of current that can flow through the circuit.

3. What is the time constant of a LR circuit?

The time constant of a LR circuit is the amount of time it takes for the current to reach 63.2% of its maximum value. It is calculated by multiplying the inductance (L) in henries by the resistance (R) in ohms.

4. How does the inductance affect the current in a LR circuit?

The inductance of an inductor in a LR circuit affects the current by producing a magnetic field that opposes the change in current flow. This results in a delay in the current flow and a gradual increase in the current over time.

5. What happens to the current in a LR circuit when the power source is turned off?

When the power source is turned off in a LR circuit, the current will continue to flow for a short period of time due to the inductor's magnetic field. However, as the magnetic field dissipates, the current will eventually decrease and reach zero.

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