Physics/What is the relation between C and dV/dt in a RLC circuit ?
QUESTION: Hello Stephen,
I have a simple question for you !
In a RLC circuit, we calculate the inductive voltage with the equation VL = i x XL, am I right ? But we can also define the voltage through “L” which means, VL = L x (di/dt), the change velocity of the current plays an important role, hence VL = f(L, (di/dt))
I try the same work for a capacitor (Capacitive voltage, XC), I used the change of the voltage, dv/dt, where V = VM x sin(wt), I had to add some other parameters, I add the C, but it was not complete, I also add the Xc value, which gives: Xc x C x (dV/dt), the multiplication, Xc x C yields to (1/w), w = 2 x Pi x f, so we get: Vc = (1/w) x (dV/dt), this is equall to Xc x i = Vc.
This equation works, if we just have a “C” (Capacitor) in the circuit, but not when we have a R and/or L, then it does not work, what can you say about this ?
How can I define an analogous equation like VL = L x (di/dt) and XL x i, as far as I can see, when we use just a “C” in the circuit, the potential does not depend on “C” when we use alternate current, it depends on w and dV/dt, am I right, where is C and why does it not work when we also have a R and/or L added ?
b- In the Physics book of Halliday & Resnick the displacement current is defined with the equation Vd = C x (dV/dt), which is correct, is this current an additional amount to the circuit current which is defined with: i = Vrms/Z, Z is the impedance.
When should I take the displacement current into account and when can I ignore it, does this play an important role in a RLC circuit ?
Wish you a beautiful Sunday,
ANSWER: Birol, you never ask me a simple question. They all come in multiple paragraphs like this. Please take a circuits course at a university, I'm really getting exhausted just reading these. You need a full, formal course. You're smart, after a formal course you won't need me.
First, you're treating these like they're all at maximum voltage, all the time. They're not, they're varying voltages that are different over time. At any once instance in time, all the voltages in an RLC loop (including whatever is driving it) will be zero when you add them up. The capacitor and inductor will be pi radians out of phase, the difference will be made up by the resistor. Study your phasor diagrams, this will become apparent.
Example: if I have a current that is 1A*sin(wt), then at some point the current will be 0.1A and increasing. At that moment, the factor of 0.1=sin(wt). Therefore that is the phase of the circuit. The voltage across the resistor will be 0.1A*R, as usual. The voltage across the inductor (since the inductor will resist the increase from zero) 1A*cos(wt)*wL. The voltage across the capacitor will be -1A*cos(wt)/(wC). The voltage around the loop will add up to whatever is driving the loop, which will be out of phase by the usual term (arctan(R/(XL-XC))).
The displacement current is not a real current. It is a mathematical device that physicists used to conceptualize the changing field inside a capacitor in relation to its creation of a magnetic field as if there was just a wire connecting the two plates. In general, you can ignore it.
[an error occurred while processing this directive]---------- FOLLOW-UP ----------
QUESTION: Hello Dr. Stephen Nelson,
Thank you very much for your recent answer.
Well, I have been studying the book from Halliday & Resnick, chapter 33, electromagnetic oscillations and alternating current. I have always this book in my hand, especially when I construct new ideas or analyse some extraordinary questions......
I know that the potential across the coil and capacitor is changing with time, and that there is a phase difference of pi. I just wanted to discuss with you, I am able to write the displacement current as i = C x (dv/dt), while V = VM x sin(wt), so, i = C x w x VM cos(wt).
The potential across the capacitor VC = q/C, q = i x t, so q = C x w x VM cos(wt) x (1/w)
q = C x VM x cos(wt), VC = VM x cos (wt) or VC = (1/w) x dV/dt.
Anyway, these are just some theoretical games, about formulating the potential across the capacitor, instead of writing Xc x i.
Theory is easy, experimenting is more difficult. Have you ever studied the wireless power concept of Nikola Tesla ? I want to conduct an experiment on this subject.
My question is about the transmitting device of Nikola Tesla, please check the photo which is attached to this mail. Consider the secondary circuit of a Tesla coil, here we have a grounded circuit, a coil between the ground and aerial capacitor (a ball shaped antenna).
We have a function generator and want to activate this “Tesla Antenna” with a sinus wave as defined above, my question is, how is that possible, is a primary circuit needed (oscillator) which activates the antenna circuit, according the equation: L1 x C1 = L2 x C2, as in the wireless concept of Nikola tesla, or could there be a possible way (trick) just to activate a Tesla Type antenna, with high efficiency, without needing a primary resonating circuit ?
What happens when I connect the lower end of the coil to the function generator, and the other cable of the function generator to a grounded plate, can I oscillate by this way the charges between ground and ball shaped antenna as in a conventional Tesla transformer ?
Can you discuss the concept from different perspectives please ?
Later we may analyse the other steps of this project study.
Many thanks for your help,
Woah, woah, you started in an oscillating current with q=i*t, then replaced a static I with a variable current instead of an INTEGRATED current! That's wrong from the get-go, try again and I'll try the rest of this for your math because I can't follow the rest.
Unfortunately, the drawing you included also doesn't contain enough detail for me to answer this question's second part. Perhaps you have your own drawing?