Hi Jim, thanks for your response.I know that an op-amp is a device that wants to make it's inputs equal. I just don't understand how it forces it's inputs to be equal, and that is what I want to understand.
Maybe my misunderstanding has got to do with me thinking that before the positive input...
Thanks for your reply.
As the output voltage passes through the positive input voltage there will be a voltage (appearing across the input resistor Rin) causing a new output voltage vo. If for example the positive voltage is 10mV and the gain is 1000 there will be 10V at the negative terminal...
Consider an op-amp in which there is a DC voltage of 100mV for example applied to the non-inverting (positive) terminal of the op-amp and the negative terminal of the op-amp is connected to the output directly. Also take the voltage supplied to power the op-amp, Vcc, to be 15V and the gain, A...
First of all thanks for all your replies.
@Hassan2 Yes you are right, in my model of electrons if their is more current passing through the same resistance then their will be more collisions with the ions leading to a higher voltage drop. I wasn't thinking that more I => more collisions => more...
Say a circuit has a 9V DC supply, a 2 ohm resistor in series with it and also a parallel combination of two 1 ohm resistors in series with it. So a simple series-parallel circuit.
What I don't understand is why the voltage drop across the 2 ohm resistor isn't 6V and the voltage drop across both...
Thanks so much for all your help.
I think I know what my conceptual difficulty was, it struck me when sophiecentaur said, why should it be transformed?
I am so use to textbook physics examples where for example you have an inclined plane, he has a certain GPE at the top - what is his KE at...
For 1.
If you have for example 60m/s and want to convert to km/hr you do the following:
(60*1000km)/(60*60hr)=16.67km/hr. (You convert the numerator to km, and the denominator to hours).
There is also a very nice conversion factor: 60/3.6=16.67km/hr, to go the other way simply multiply by 3.6...
Thanks for your help rcgldr, I really appreciate it.
Yeah, I see what your talking about, it just doesn't make sense to me that no energy is converted through conductive sections. In what way is a circuit different than if a wire was short circuited between the capacitor terminals? Would the...
It depends on the context actually. In terms of mechanical waves, for example a water wave ripple in a pond, more energy is carried by the wave for high amplitudes. Photons are another type of wave, but they carry energy in their frequency. A sound wave is a mechanical wave, and so the amplitude...
That's ok, that's understandable.
What is it that drives the electrons through the circuit if it isn't the electric field from the capacitor? I thought they felt an electric field due to the protons at the other end of the capacitor (or rather the lack of electrons). And hence why I think of...
Actually I am the one who is correct, the formula you have quoted is the formula for electric force - Coulomb's law. If you integrate force with respect to distance you arrive at the formula I stated.
I thought the electrons on the negative terminal of the capacitor experience the field of the...
But V=Ed and E is a constant. So U=qEd, that is analogous. That is the mechanism. You could similarly say U=mV. Where V is J/kg.
How is charge transfer different to mechanical transfer? Both occur due to a force created by a field.
Thanks for putting in the time to help me so far.
Ok I agree, but then the energy has to be converted into some form as the electron travels through the wire. What is the energy converted to then, it wouldn't be heat because there are almost zero collisions inside the conductor?
If this were not the case, as charges flow through a wire they convert electric potential energy to kinetic energy mainly through conductors. Why do you speak of hardly any energy? Electric potential energy is a function of distance (charge is constant in this circuit), and as such, as particles...