Obnoxious creep.
<< The equation is simply I = C (dV/dt), where I is the current, V is the voltage,and C is the capacitance. If the input for V(t) is a sine wave, then the time derivative will convert the sine wave into a cosine wave, and the I(t) will thus be a cosine wave, regardless of any of the other details. The other details affect the magnitude of the current response, but the phase angle will always be the same, the angle implied by the shifting to a cosine wave. >>
Yes. Note also that dV(t) = (1/C) I(t) dt. Or, assuming no initial charge on the capacitor, upon integration : V(t) = (1/C) S I(t) dt, where S is the long-s integration sign. If the capacitor is driven by a constant current source, a function with w = zero, say I(t) = K, then V(t) = (K/C) t; a straight line function with time. If the current source continues to operate and some device is used to discharge the capacitor at some set voltage, then the voltage time function will be a sawtooth periodic time function whose frequency will depend on the circuit component values. In the old days, this kind of circuit, constructed with tubes or transistors, was used for sweeping functions.
<<Our rodent friend is engaging in tiresome sophistry by trying to apply a narrow definition of "airborne" (physicallly supported by air currents), >>
Indeed, even with his quibble, theobject couls stillbe *initially* airborne, which is all that is required for the question to make sense.
I'm pretty sure I encountered that rodent before. He isome sort of technician, and in that earlier conversation he asked about the relationship between voltage and current in a circuit with capacitance. I gave him the simple formula which involves a time derivative of the voltage, and he thought it was wrong, because he did not know what a derivative was, and was expecting me to list the phase shift and the amplitude elationship--both of which unbeknownst to him, come from the formula with the time derivative.
Also, I note that in the recent conversation he used the term EMF (Electromotive Force) and used "E" for it, neither of which would likely be done by a modern scientist or engineer.
Most of Mombosonny's messages this morning must be filled with loving hugs and kisses....this is the message I'm getting on about 1/2 his/her posts-----
"This message is hidden by your ratings filter."
<<I believe you are correct. For a pure sinusoidal waveform and assuming the phase angle of the voltage is zero, the phase angle of the current would be negative pi/2, regardless of the frequency >>
Yes.
The equation is simply I = C (dV/dt), where I is the current, V is the voltage,and C is the capacitaqnce. If the input for V(tt) is a sine wave, then the time derivative will convert the sine wave into a cosine wave, and the I(t) will thus be a cosine wave, regardless of any of the other details. The other details affectt he magnitude of the current response, but the phase angle will always be the same, the angle implied by the shifting to a cosine wave.
<< Where have you been? well, I hope>>
Thanks. I've ben busy with other stuff, and might be tied up for a few more weeks.
<< The current phase angle would not change. >>
I believe you are correct. For a pure sinusoidal waveform and assuming the phase angle of the voltage is zero, the phase angle of the current would be negative pi/2, regardless of the frequency (for w > 0, I quess). Where have you been? well, I hope.
<<You really should learn to control yourself. Your name calling is only a reflection of your lack of ability to carry on a debate and is indicative of your lack of self control and vocabulary.
>>
No, my name calling is because I am having a conversation with a complete idiot.
<<One more thing. I don't believe I said anything about doubling voltate. I said doubling the frequency, didn't I?>>
True, you did ask about doubling the frequency. It doesn't msatter though. Noi matter what you do to the voltage function (for a sinusoidal voltage), the phase angle will not vchange. There will always be a 90 difference. That is because the current goes as the time vdeerivative of the voltage.
You never answered when I asked you if you knew what a time derivative is. Why not?
<<Let me know if you still think a falling rock is airborne>>
Tell me precisely what you think "airborne" means.
<<Oh, I didn't say you thought E and I were in phase. >>
Yes, you did--you told me I was wrong, and then pproceded to tell me that the voltage (E) and the current (I) were not in phase. Here is your exact quote:
Absolutely wrong. Without telling you the amount, if any change in the relationship, I will tell you that voltage and current are not in phase and are, therefore, different at any given point in the circuit.
<<Congratulations, you did much better than last time>>
My answers were exactly the same both times, you pathetic piece of idiocy.
Show me a difference in my answers.