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QUESTION: Hello Steve,

I have here a signal generator, which is connected to a capacitor. To calculate the displacement current I need the di/dt value, can you help me ? The frequency is 500 kHz, the current is 140 mA......I am able to calculate the Q value in the first quarter period(between zero and Pi/2)Q = 7 exp(-8) coulomb, so, dq/dt, would be equal to i, which is again aqual to 140 mA, we need the value d2q/dt2......I used another equation id = C x (dV/dt), my XC is equal to 0,4681 ohm, C = 0,68 micro Farad, so, V would be equal to 0,065534 Volt, and id = 0,065534 x (4x500.000)x 0,68 exp (-6) = 0,0891 A = 89,12 mA, is this calculation correct ?

Yours Sincerely

Birol

ANSWER: No, don't go jumping into equation salad. When you say you know the current, is that the maximum current (Imax)? Because you mention a frequency, so is that the peak or the RMS value of your (presumably sinusoidal) wave (meaning you need to multiply by the sqrt(2) to get Imax)? If so, that would make the current given by i=Imax*sin(w*t), where w is the angular frequency = 2*pi*f. Then you take the derivative to get di/dt=Imax*w*cos(w*t) to give you the time-dependent di/dt. Why did you start by calculating Q in the first quarter period? Ignore that, do what I said above.

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QUESTION: Hello Steve,

Thank you very much for your answer, I like it very much :-)

Well, I was at my city electrician, we wanted to calculate the power of his wave (sinus) generator. We put a resistance of 1000 ohm across input and connected it to an oscillator. I was able to see nearly 10 Volts, this was the maximum (peak) value on the screen !

P = Vē/R = 100/1000 = 0,10 Watt.

I = V/R = 10/1000 = 0,01 A = 10 mA--> peak value, you are right !

Imax = 10 mA (Irms would be divided by sqrt(2) = 7,07 mA)

I will work at 500 kHz.

Hmmm, at 0 and pi, I would have the maximum di/dt value, lets calculate for cos (0) = 1,

w = 2 x 3,14 x 500.000 = 3.141.592 rad/s

di/dt = 0,01 x 3.141.592

di/dt = 31415,92 (A/s)

L = 10 mH (the inductance of the N1/N2 = 1, transformer)

EMK = - L di/dt

EMK = 0,01 x 31415,92

EMK = 314,159 Volt, is this the potential which is induced in the coil ?

Strange situation, I bought a transformer of 0,10 W, with dimensions of 0,50 x 0,50 x 0,50 inches, do you think it can tolerate 300 volts ?

Thank you for this beautiful discussion !

Yours Sincerely

Birol

ANSWER: No, you're mixing up self-inductance with mutual inductance in a transformer, but the math seems right. You're going to have seeeerious losses at 500 kHz. I have no idea what your transformer can handle, that's slightly outside my area of expertise and I have no idea what kind of wire gauge your transformer has. To thin, it won't handle the current.

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QUESTION: Hello Steve,

Thank you for your help. The mutual inductance coefficient would not be available, while N2/N1 is one......

Anyway, have a little problem in my project, I was not aware of it, now I can see (with the help of my electrician) it better. I have a capacitor of a certain capacitance (0,68 micro Farad). As you know it has 2 electrodes and a dielectric between these electrodes. I need to pulse one of this electrode, how ? Hmmm, I will ground the other electrode, so I can pull the charges from the ground or send the charges to the ground, it depends on the input signal. The input signal should vary with time, 400-500 kHz is my target, AC, when it is positive for example, the other electrode would get negative, by extracting charges from the ground, or vice versa......

The point is, which possibilities are there available to pulse the electrode periodically, in a sinus curve like configuration ?

I have this sinus generator as I told you before, when I ground the generator and connect the input to my capacitor, the circuit is open, I am not able to close the circuit, so, there would not flow any current through the system, how can I close the circuit or pulse the electrode of my capacitor ?

Another way is to connect a transformer across the wave generator, now I have a closed circuit, this is the primary winding. But, I have a secondary coil, the capacitor is connected to the one side, and the other side is grounded, as Tesla did, my electrician says, it would not work again, while there is a grounding on the secondary winding, open circuit, so what ?

Isn't here any possibility to pulse this capacitor with a sinus wave ?

This part is important, if you want we can do this step by step !

I am sending you a simple diagram !

Many thank,

Birol

What do you mean? Just connect the the generator to the capacitor. Tesla's notions of signal transmission didn't take into account the ridiculous inefficiency of such mechanisms...I hope you're not trying to replicate power transmission via radio waves, it's just a bad idea.

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I can answer most basic physics questions, physics questions about science fiction and everyday observations of physics, etc. I'm also usually good for science fair advice (I'm the regional science fair director). I do not answer homework problems. I will occasionally point out where a homework solution went wrong, though. I'm usually good at explaining odd observations that seem counterintuitive, energy science, nuclear physics, nuclear astrophysics, and alternative theories of physics are my specialties.

I was a physics professor at the University of Texas of the Permian Basin, research in nuclear technology and nuclear astrophysics. My travelling science show saw over 20,000 students of all ages. I taught physics, nuclear chemistry, radiation safety, vacuum technology, and answer tons of questions as I tour schools encouraging students to consider careers in science. I moved on to a non-academic job with more research just recently.**Education/Credentials**

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