A brief introduction to RL circuits for students studying electromagnetic induction in calculus-based physics courses such as AP Physics C. For more information, please visit APlusPhysics.com.
At the start, does what you say imply that the inductor in the series circuit makes the E-field around it "NON" conservative??? I am very interested in this.
Why is the electric field 0 in the inductor? Also, in the textbook I'm using the form of the final equation is for a RL circuit where the circuit has been switched off of a battery into a loop and the form of the equation given is I=I(sub 0)*e^(-t/tao) where I(sub 0) was the current when the circuit was switched off of the battery. Are they both valid forms of the equation? Thanks
E field in the inductor is zero because all the energy is stored in the magnetic field. And yes, for an RL circuit, tau is L/R, so the equations are equivalent. :-)
In a way, didn't you end up using KVL for the first Current in RL Circuits example? You had IR-V=L(dI/dt), which I think I can find with Kirchoff's voltage law. Thanks for the explanations! They help a LOT. You're an awesome teacher :)
+SNHD33 It's not quite KVL since KVL doesn't work for a changing magnetic flux, but we can use Faraday's Law to analyze the circuit (KVL is an outcome of Faraday's Law for a fixed magnetic flux). So they're very close. :-) Thanks!
Hello Mr. Fullerton, thank you once again for another great video lecture! Why do you have a positive sign for the IR for the drop in voltage of the resistor in your first example and a negative sign for the change in voltage of the power supply? Should it not be in the opposite direction? Thanks.
There are different methods of doing this. I make a loop around the circuit (in this case a clockwise loop) and I record the sign I come to first at each of the circuit elements. I see the negative side of the power supply first, so I write that as negative. I see the positive side of the resistor first, so I write that as positive. Others take the opposite convention. As long as you're consistent, it'll work out the same either way. :-)
Ohh ok. Thanks for the explanation. My impression was that since I go clockwise and I get to the negative side of the battery and then to the positive, that my deltaV = Vf - Vi is posiitive and thus I would be adding a voltage instead of subtracting it. What do you think of that?
Mr Fullerton, if you do not mind, I have another question for you: when you were deriving an equation for the voltage of the RL circuit, you said that the electromotive force E = -L (dI/dt) and you said that it was across the inductor. This is confusing me as the equation that you derived for the current was based off the same expression but used for the whole circuit. How were you able to say that the expression is applicable for the circuit in one situation and an inductor in the other? Thanks in advance for the help, and thanks for a great video!
The current in a series circuit has to be the same everywhere in the circuit. So if that's the current through the inductor, then that's also the current through the entire circuit if there are no branches.
Thank you for a great explanation. Here is another question, if you do not mind. When you derived the equation for the voltage across the inductor you said that V(inductor) = L * dI/dt However, I thought that the expression of faraday's law was integral( E dot dl) = - L * dI/dt where the E dot dl integral referred to the whole current, not just the individual inductor? Then, how were you able to use L * dI/dt by making it refer specifically to the voltage drop across the inductor, not the circuit, and then not make it - L * dI/dt [that is negative] per faraday's law? Thanks in advance, and have a good day!
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Always nice to see this smoothly explained, this time with definite integrals.
Excellent explanation, thanks!
At the start, does what you say imply that the inductor in the series circuit makes the E-field around it "NON" conservative??? I am very interested in this.
nice explanation
Excellent lectures.
Thank you!
Why is the electric field 0 in the inductor? Also, in the textbook I'm using the form of the final equation is for a RL circuit where the circuit has been switched off of a battery into a loop and the form of the equation given is I=I(sub 0)*e^(-t/tao) where I(sub 0) was the current when the circuit was switched off of the battery. Are they both valid forms of the equation? Thanks
E field in the inductor is zero because all the energy is stored in the magnetic field. And yes, for an RL circuit, tau is L/R, so the equations are equivalent. :-)
In a way, didn't you end up using KVL for the first Current in RL Circuits example? You had IR-V=L(dI/dt), which I think I can find with Kirchoff's voltage law.
Thanks for the explanations! They help a LOT. You're an awesome teacher :)
+SNHD33 It's not quite KVL since KVL doesn't work for a changing magnetic flux, but we can use Faraday's Law to analyze the circuit (KVL is an outcome of Faraday's Law for a fixed magnetic flux). So they're very close. :-) Thanks!
Hello Mr. Fullerton, thank you once again for another great video lecture!
Why do you have a positive sign for the IR for the drop in voltage of the resistor in your first example and a negative sign for the change in voltage of the power supply? Should it not be in the opposite direction?
Thanks.
There are different methods of doing this. I make a loop around the circuit (in this case a clockwise loop) and I record the sign I come to first at each of the circuit elements. I see the negative side of the power supply first, so I write that as negative. I see the positive side of the resistor first, so I write that as positive. Others take the opposite convention. As long as you're consistent, it'll work out the same either way. :-)
Ohh ok. Thanks for the explanation.
My impression was that since I go clockwise and I get to the negative side of the battery and then to the positive, that my deltaV = Vf - Vi is posiitive and thus I would be adding a voltage instead of subtracting it.
What do you think of that?
Also, why is the electric field in the solenoid zero in the first example?
Absolutely, you can do it that way.
The inductor holds energy in a magnetic field, not an electric field.
You said Kirchoff's voltage law is not applicable if there is a changing magnetic flux. But why did you use KVL at 3:05?
thats if there is a change in magnetic flux throught the whole circuit such as in a lenz's law problem
Mr Fullerton, if you do not mind, I have another question for you:
when you were deriving an equation for the voltage of the RL circuit, you said that the electromotive force E = -L (dI/dt) and you said that it was across the inductor. This is confusing me as the equation that you derived for the current was based off the same expression but used for the whole circuit. How were you able to say that the expression is applicable for the circuit in one situation and an inductor in the other?
Thanks in advance for the help, and thanks for a great video!
The current in a series circuit has to be the same everywhere in the circuit. So if that's the current through the inductor, then that's also the current through the entire circuit if there are no branches.
Thank you for a great explanation.
Here is another question, if you do not mind.
When you derived the equation for the voltage across the inductor you said that V(inductor) = L * dI/dt
However, I thought that the expression of faraday's law was integral( E dot dl) = - L * dI/dt where the E dot dl integral referred to the whole current, not just the individual inductor?
Then, how were you able to use L * dI/dt by making it refer specifically to the voltage drop across the inductor, not the circuit, and then not make it - L * dI/dt [that is negative] per faraday's law?
Thanks in advance, and have a good day!
i appreciate very much your explanation, but i am trying to understand the difference between VL=L.di\dt and EL=--Ldi\dt