Question about Intuition of Inductor

[Master1022]

Homework Statement

I am trying to figure out how an inductor works in depth. It should be something very simple to find, but I have yet to find an explanation that goes through the process step-by-step in a non-circular way.

I can solve the inductor differential equations and do phasor analysis with inductors with little problem (I know AC is different), but have always struggled to gain a solid intuition.

My current thoughts:
(assume we have a DC voltage connected to an LR circuit)
This is what I think happens:
1. The voltage applied causes a rate of change of current as $V_{app} = L \frac{di}{dt}$
2. As the current increases, there will be a back emf created that opposes the applied voltage
3. Therefore, the applied voltage decreases and thus the rate of change of current decreases

This is where I cannot reconcile the beginning and end of the process as searches seem to suggest that the back-emf is proportional to the rate of change of current (as predicted by Lenz's law)

It would make sense if the back-emf was proportional to the current because:
- as the current increased, the back-emf increased
- voltage applied decreased and rate of change of current approaches 0
(much like a terminal velocity diagram), but I don't think this would lead me to the correct value of the final current

However, if it is proportional to the rate of change of current $V_{back} = k \frac{di}{dt}$ , I have problems understanding how the charging graph can arise.
Here are the two cases I can think of to do with the proportionality constant (ignored case where constant > L):

1. constant = L
- therefore, it would create a back-emf equal to the DC voltage, and thus the rate of change of current would = 0
- then would continually happen as the back-emf oscillates between said value and 0

2. constant < L
- doesn't really make the DC applied voltage would remain the same, but the applied voltage would never match the DC voltage and rate of change of current would always be greater than 0

I would appreciate any help in explaining which (if any) of the above explanations are correct and clarifying/correcting any wrong points I have made.

Thanks in advance.

 

[BvU]

assume we have a DC voltage connected to an LR circuit

Good. I would like to extend a little: disconnected for $t<0$, connected from $t=0$ onwards. And the voltage source is an ideal voltage source. Also, the inductor is an ideal inductor. Can you agree ?

With $V_{\rm back} = L\frac{di}{dt}$ for the inductor, you get a differental equation for the current:$$V_{\rm source} = i R + L {di\over dt} \Rightarrow \frac {V_{\rm source}}{R} = i + {L\over R} {di\over dt} $$ Inital condition $i_0 = 0$.
By the time things have come to rest, $\ {di\over dt} = 0 \$ and $\ i_\infty = \frac {V_{\rm source}}{R}$.

Solution straightforward for undergraduate level.

Next step : same stuff with $V_{\rm source} =V_0 \sin\omega t$.

--

[edit] had to correct the sign of $V_{\rm back}$

 

[BvU]

This is what I think happens:
1. The voltage applied causes a rate of change of current as $V_{app} = L \frac{di}{dt}$

Not so. An Inductor is defined by $V_{\rm \bf back } = L \frac{di}{dt}$

 

[Master1022]

Good. I would like to extend a little: disconnected for $t<0$, connected from $t=0$ onwards. And the voltage source is an ideal voltage source. Also, the inductor is an ideal inductor. Can you agree ?

With $V_{\rm back} = L\frac{di}{dt}$ for the inductor, you get a differental equation for the current:$$V_{\rm source} = i R + L {di\over dt} \Rightarrow \frac {V_{\rm source}}{R} = i + {L\over R} {di\over dt} $$ Inital condition $i_0 = 0$.
By the time things have come to rest, $\ {di\over dt} = 0 \$ and $\ i_\infty = \frac {V_{\rm source}}{R}$.

Solution straightforward for undergraduate level.

Next step : same stuff with $V_{\rm source} =V_0 \sin\omega t$.

--

[edit] had to correct the sign of $V_{\rm back}$

thanks for your reply and for the clarification around the back emf. I understand how to solve the ODEs, but would you be able to provide a qualitative explanation behind the process (i.e. an explanation of the basic underlying concepts). Thanks

 

[Merlin3189]

I think your problems arise at least partially from the slack terminology. You need to be clear what you mean by 'back emf' and 'applied voltage'.
There is only one emf, pd or voltage across the inductance. Whatever you call it, back pd, applied emf, applied back voltage, or whatever, it is all the same thing.
The rate of change of current is proportional to this value, be it voltage, pd, emf or whatever you designate it. (I'll call it V_L )

When you connect your 'DC Voltage' to the LC circuit, initially no current was flowing so there was no IR voltage across the resistor, V_R=0
Therefore V_L = V_DC and you get the corresponding maximum rate of change of current due to the $V_L=L \frac {dI} {dt}$ relation.

Since $L \frac {dI} {dt} > 0$ , then $I$ increases and $IR$ increase from zero
and now $V_L = V_{DC} - IR$ , which is less than the initial value.
That makes $\frac {dI} {dt}$ decrease, because $\frac {dI} {dt} = \frac {V_L}{L}$
$\frac {dI} {dt}$ is still positive, so the current still increases, just not as fast as it did initially.
So $V_R = IR$ is still increasing and $V_L = V_{DC} - IR$ is therefore decreasing.

And on and on it goes with current increasing at an ever smaller rate and $V_L$ decreasing towards zero.

It's all dead easy, just so long as you don't start inventing aliases for $V_L$ and changing one differently from another.

 

[Master1022]

I think your problems arise at least partially from the slack terminology. You need to be clear what you mean by 'back emf' and 'applied voltage'.
There is only one emf, pd or voltage across the inductance. Whatever you call it, back pd, applied emf, applied back voltage, or whatever, it is all the same thing.
The rate of change of current is proportional to this value, be it voltage, pd, emf or whatever you designate it. (I'll call it V_L )

When you connect your 'DC Voltage' to the LC circuit, initially no current was flowing so there was no IR voltage across the resistor, V_R=0
Therefore V_L = V_DC and you get the corresponding maximum rate of change of current due to the $V_L=L \frac {dI} {dt}$ relation.

Since $L \frac {dI} {dt} > 0$ , then $I$ increases and $IR$ increase from zero
and now $V_L = V_{DC} - IR$ , which is less than the initial value.
That makes $\frac {dI} {dt}$ decrease, because $\frac {dI} {dt} = \frac {V_L}{L}$
$\frac {dI} {dt}$ is still positive, so the current still increases, just not as fast as it did initially.
So $V_R = IR$ is still increasing and $V_L = V_{DC} - IR$ is therefore decreasing.

And on and on it goes with current increasing at an ever smaller rate and $V_L$ decreasing towards zero.

It's all dead easy, just so long as you don't start inventing aliases for $V_L$ and changing one differently from another.

Many thanks. This makes more sense now.