Sorry i got stuck in a Rl circuit problem

In summary: This is not the same as saying that the back EMF is in the opposite direction of the current. The back EMF will be in the same direction as the current, and it will be trying to prevent any changes in the current.Your calculations are correct. The EMF that gets generated will be in the same direction as the current and the original DC voltage that was removed. Meanwhile, if you turn on a voltage source with zero current, the EMF that gets generated will be a back EMF that resists any changes in current. This is not the same as saying that the back EMF is in the opposite direction of the current. The back EMF will be in the same
  • #1
garylau
70
3
Sorry

this is the first time i come to this forum.

Why the -Ldi/dt in the left side is minus but not plus sign

i got confused

Does everyone can help me in this problem

thank you
-------------------------------------------------------------
i know V-Ri-Ldi/dt=0 is true in the right hand side
however,it doesn't make sense that the "-L di/dt" term is still the same
when the di/dt is decreasing rather than increasing in the left hand side.
V=0
the direction of the induced emf is reverseing(so it should be Ldi/dt but not- Ldi/dt?)
the Ri is still the same when the i is in the same direction as before.
i think it should be L*di/dt-Ri=0

what's wrong with that?
 

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  • #2
garylau said:
Sorry

this is the first time i come to this forum.

Why the -Ldi/dt in the left side is minus but not plus sign

i got confused

Does everyone can help me in this problem

thank you
-------------------------------------------------------------
i know V-Ri-Ldi/dt=0 is true in the right hand side
however,it doesn't make sense that the "-L di/dt" term is still the same
when the di/dt is decreasing rather than increasing in the left hand side.
V=0
the direction of the induced emf is reverseing(so it should be Ldi/dt but not- Ldi/dt?)
the Ri is still the same when the i is in the same direction as before.
i think it should be L*di/dt-Ri=0

what's wrong with that?
The inductor causes a back EMF, so that the equation reads ## V-L \, dI \, /dt=IR ##. The reason it occurs in this direction is ## L=\Phi/I ## where ## \Phi ## is the magnetic flux. By Faraday's law, the EMF from the inductor is ## \varepsilon=-d \Phi/dt ##.
 
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  • #3
Charles Link said:
The inductor causes a back EMF, so that the equation reads ## V-L \, dI \, /dt=IR ##. The reason it occurs in this direction is ## L=\Phi/I ## where ## \Phi ## is the magnetic flux. By Faraday's law, the EMF from the inductor is ## \varepsilon=-d \Phi/dt ##.
But if V=0 than the di/dt will decreasing
than -L di/dt will become L di/dt ?
the sign reverse when the rate of current change is decreasing?
 
  • #4
i understand why V-L(di/dt)-Ri=0 is true

but when V=0 ,then (di/dt)will not increases but decreasing(so that (di/dt)will become (-di/dt))
therefore the whole terms will reverses its sign(from -L di/dt to L di/dt--> the direction of induced emf is in opposite way)?
but nothing happen? why?

the term (-Ldi/dt) in both side are the same even the direction of voltage supply(induced emf)is in opposite way
 
  • #5
garylau said:
But if V=0 than the di/dt will decreasing
than -L di/dt will become L di/dt ?
the sign reverse when the rate of current change is decreasing?
For V=0, the differential equation is ## L \, (dI /dt)+IR=0 ## and the solution is ## I(t)=Aexp(-(R/L)t) ##.When the current starts to decrease,e.g. if we shut off a DC voltage, there is a forward EMF in the inductor that attempts to keep the current from changing.
 
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  • #6
Charles Link said:
For V=0, the differential equation is ## L \, dI \, /dt+IR=0 ## and the solution is ## I(t)=Aexp(-(R/L)t) ##.When the current starts to decrease,e.g. if we shut off a DC voltage, there is a forward EMF in the inductor that attempts to keep the current from changing.
this is the correct solution in phyiscal sense.
yes you are right.
But i got no mathematical sense in this solution.
why we don't need to reverse its sign when there is forward EMF in the inductor to keep current changing(in the V=/0 case,it is clearly the EMF is backward but not forward)
if backward EMF= -L (di/dt)
then it is reasonable to think the forward EMF= +L(di/dt)

So: V-Ri-L(di/dt)=0

should becomes 0-Ri+L(di/dt)=0

clearly your solution is correct but mine is wrong
my reasoning got wrong when i try to think like this,why?
 
  • #7
garylau said:
this is the correct solution in phyiscal sense.
yes you are right.
But i got no mathematical sense in this solution.
why we don't need to reverse its sign when there is forward EMF in the inductor to keep current changing(in the V=/0 case,it is clearly the EMF is backward but not forward)
if backward EMF= -L (di/dt)
then it is reasonable to think the forward EMF= +L(di/dt)

So: V-Ri-L(di/dt)=0

should becomes 0-Ri+L(di/dt)=0

clearly your solution is correct but mine is wrong
my reasoning got wrong when i try to think like this,why?
When the current is decreasing, ## dI/dt ## is negative, so that ## \varepsilon=-L (dI/dt) ## is positive and thus a forward EMF.
 
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  • #8
Charles Link said:
When the current is decreasing ## dI/dt ## is negative, so that ## \varepsilon=-L (dI/dt) ## is positive and thus a forward EMF.
but by the same logic,

we can also write V-Ri-L(di/dt)=0 in this form
""V+Ri-L(di/dt)=0""


So that the i is<0 and Ri will be negative
and di/dt>0 and-L(di/dt) it will be positive

and "0+Ri-L(di/dt)=0
when (di/dt)<0 then L(di/dt)will be negative
and Ri will be negative when i <0
 
  • #9
garylau said:
but by the same logic,

we can also write V-Ri-L(di/dt)=0 in this form
""V+Ri-L(di/dt)=0""


So that the i is<0 and Ri will be negative
and di/dt>0 and-L(di/dt) it will be positive

and "0+Ri-L(di/dt)=0
when (di/dt)<0 then L(di/dt)will be negative
and Ri will be negative when i <0
Your calculations are correct. The EMF that gets generated will be in the same direction as the current and the original DC voltage that was removed. Meanwhile, if you turn on a voltage source with zero current, the EMF that gets generated will be a back EMF that resists any change.
 
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  • #10
Charles Link said:
The EMF that gets generated will be in the same direction as the current and the original DC voltage that was removed.
garylau said:
but by the same logic,

we can also write V-Ri-L(di/dt)=0 in this form
""V+Ri-L(di/dt)=0""


So that the i is<0 and Ri will be negative
and di/dt>0 and-L(di/dt) it will be positive

and "0+Ri-L(di/dt)=0
when (di/dt)<0 then L(di/dt)will be negative
and Ri will be negative when i <0
Just one additional comment: Your equation inside the quotes is incorrect. You can't arbitrarily change the sign of these terms. Your first equation above it is correct.
 
  • #11
Charles Link said:
When the current is decreasing, ## dI/dt ## is negative, so that ## \varepsilon=-L (dI/dt) ## is positive and thus a forward EMF.
Why should you do like this
you can put the minus sign outside directly?
for example:

when you write the "current is decreasing" and therefore "di/dt" is negative,then
you can write something like" -di/dt"

~ and "-di/dt" is increasing?
 
  • #12
garylau said:
Why should you do like this
you can put the minus sign outside directly?
for example:

when you write the "current is decreasing" and therefore "di/dt" is negative,then
you can write something like" -di/dt"

~ and "-di/dt" is increasing?
When you begin, you need to assign one direction, e.g. clockwise, as positive. In the solution ## I(t)=Aexp(-(R/L)t) ## when we say ## I ## is decreasing and ## dI/dt ## is negative, we are assuming ## A ## is positive. The same differential equation, ## V=IR+L(dI/dt) ## works equally well for negative (counterclockwise) currents and voltages.
 
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  • #13
Charles Link said:
When you begin, you need to assign one direction, e.g. clockwise, as positive. In the solution ## I(t)=Aexp(-(R/L)t) ## when we say ## I ## is decreasing and ## dI/dt ## is negative, we are assuming ## A ## is positive. The same differential equation, ## V=IR+L(dI/dt) ## works equally well for negative (counterclockwise) currents and voltages.
Oh i See
it sounds reasonable now
 
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  • #14
The Minus sign is just to follow Lenz's Law and it is the electrical equivalent of Newton's Third Law. Both statements imply that there is a reaction against change. Without the negative sign, the current would increase without limit or the trolley you push against would accelerate without an applied external force.
The Maths behind this just follows the normal conventions for how signs can be moved about inside formulae and equations.
 
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1. What is an RL circuit?

An RL circuit is a type of electrical circuit that contains both a resistor (R) and an inductor (L). The inductor is a coil of wire that creates a magnetic field when current flows through it, while the resistor helps to control the flow of current in the circuit.

2. What causes someone to get stuck in an RL circuit problem?

There can be several reasons why someone may get stuck in an RL circuit problem. These could include not understanding the principles of inductance and resistance, incorrect calculations or assumptions, or simply not having enough knowledge or experience with circuit analysis.

3. How do you solve an RL circuit problem?

To solve an RL circuit problem, you will need to use the principles of inductance and resistance to calculate the voltage, current, and other parameters in the circuit. This can be done through various techniques such as Kirchhoff's laws, Ohm's law, and the use of differential equations.

4. Are there any tips for understanding and solving RL circuit problems?

One helpful tip for understanding and solving RL circuit problems is to break down the problem into smaller, more manageable parts. This can help to identify the key components and their relationships within the circuit. Additionally, practicing with different types of RL circuits can improve understanding and problem-solving skills.

5. What are some real-life applications of RL circuits?

RL circuits have various applications in everyday life, such as in power supplies, motors, generators, and transformers. They are also used in electronic devices such as radios, televisions, and computers. Additionally, RL circuits are essential in industries such as telecommunications, automotive, and aerospace for various functions, including signal processing and power distribution.

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