For example, a 30 watt bulb versus a 100 watt bulb. Is the filament thicker? Thinner? Longer?
Since the voltage is the same, to get higher power you need a lower resistance.For example, a 30 watt bulb versus a 100 watt bulb. Is the filament thicker? Thinner? Longer?
I don't follow the reasoning here. If the smaller thickness resulted in greater resistance, then for the same voltage applied across it the power dissipated will go down. Now, if you used a constant current source, the power would go up, but that is not the usual case.I recall grinding a 1/2-Watt rated carbon resistor and the resistance went up. The cross section (thickness) was reduced
When the resistance goes up the current goes down. So, I think the smaller thickness (cross sectional area) is the higher Watt.
If the circuit is sourced with 12 Vdc and has a 1 ampere load, then the resistance is neither 10 Ω nor 1000 Ω, but rather 12 Ω.In a circuit sourced with 12Vdc and 1 Amper load you could compare heat given off of a 10 Ohm and a 1000 Ohm.
I still don't know what the 1 Amp is all about. How do you put a 1000 Ω resistor in series with a load and get 12 Vdc at 1 A?Further explanation about using resistors to replace a light bulb in showing heat difference.
I should have said 10 Ohm resistor in series with a load can be compared to a 1000 Ohm resistor in series with a load will show that the heat given off is greater for the 10 Ohm resistor. The values 12Vdc at 1 Amp should demonstrate heat but I cannot recommend using 120Vac because of a safety concern. Be careful. I am sure there is a better and safer demonstration where resistors are used instead of a light bulb. It's about showing warmth and not having flames.