@almarg,
AC transmission using wire conductors.
Al,
Could you please explain in more detail the relationship of the electromagnetic wave, that travels in the space outside of the conductor, (At near the speed of light), and the "current" that travels very slowly slightly vibrating back and forth at 60Hz in the conductor. From what I understand the movement of the current in the conductor is quite slow.... Correct?
The electromagnetic wave is caused by the applied source voltage and the "current", "charge", in the conductor? (Amount of current in the closed circuit determined by the resistance of the connected load. I = E/R)..... Correct?
Am I correct in saying you can’t have the electromagnetic wave without having current? Install an on/off switch in series in the circuit. Close the switch the current passes through the switch contacts through the load and back to source.... Correct?
The bigger the load, the more current in the conductor. The more current in the conductor the larger the electromagnet wave.... Correct? And of course the conductor, wire, must have a current, ampere rating, to safely carry the current in the wire so the wire will not overheat.
IF the wire is too small to handle the amount of current in the wire is it the current that causes the wire to overheat or is it the energy of the electromagnetic wave? Please explain in detail.
.
Not to confuse things, if only a voltage, (potential), is present, an electromagnet field will exist outside of the conductor/s without there being current... Correct?
.
I know it is the energy, from the electromagnetic wave, that makes a heating element heat up and gives off its’ heat into the surrounding air around it. It is not the "current" directly causing the resistance element to heat up.... Correct?
I know the amount of energy consumed,(in watts), by the resistance element is determined by the source voltage and the resistance, in ohms, of the resistance element. E / R = I and we know the current..... Correct?
The Fuse.....
E x I = P
E = voltage
I = Current, amps
P = power, energy, measured in, watts, VA
A fuse rated at 2 amps with a maximum voltage rating of 250V. herman said it is the energy of the electromagnetic wave passing on the outside of the fuse element link that causes it to melt and blow open when the fuse is overloaded.
OK
Isn’t the size, (for lack of a better word), of the electromagnetic wave energy determined by the applied source voltage and the current in a closed circuit? E x I = P. Is not P the energy of the electromagnetic wave?
So say the load is 150 watts and a 2 amp 250V fuse is used to protect the load. The FLA of the 150 watt load is, 150W/120V = 1.25 amps.
Here is where I get hung up. As you know a 2 amp 250V fuse can be used for any voltage 250V or less. It could be used where the voltage is 24V. The ampere rating of the fuse is still 2 amps. So to me the current has to be some component that causes the fuse to blow when the current that passes through the fuse link and exceeds 2 amps in the time curve set by the fuse manufacture. NOTE I did not say current flow.
WOW,... I know,..... I sure have a lot of questions on my mind. Blame herman.
Very best regards,
Jim