# I Sorry i got stuck in a Rl circuit problem

1. Nov 11, 2016

### garylau

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
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i know V-Ri-Ldi/dt=0 is true in the right hand side
however,it doesnt 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. Nov 11, 2016

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 $\varepsilon=-d \Phi/dt$.

3. Nov 11, 2016

### garylau

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. Nov 11, 2016

### garylau

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. Nov 11, 2016

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.

6. Nov 11, 2016

### garylau

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. Nov 11, 2016

When the current is decreasing, $dI/dt$ is negative, so that $\varepsilon=-L (dI/dt)$ is positive and thus a forward EMF.

8. Nov 11, 2016

### garylau

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. Nov 11, 2016

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.

10. Nov 11, 2016

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. Nov 11, 2016

### garylau

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. Nov 11, 2016

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.

13. Nov 11, 2016

### garylau

Oh i See
it sounds reasonable now

14. Nov 12, 2016

### sophiecentaur

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.