Measured Power: Dissipated or Undissipated?

Current does not dissipate. Current is the same through the entire circuit. Imagine a pipe with water flowing through it: The amount of water flowing into the pipe will be the same as the amount flowing out, as well as at any point along its path.

Forget about reactive power and imaginary numbers for now.

Power is current multiplied by voltage. If the voltage remains constant, less resistance means more current, and therefore more power.

The measured current is the current the flows in at one end and out the other. There is only one current.

It depends. If you assume a constant current, then yes. You would need a higher voltage to drive the current through the resistance, and since power is current multiplied by voltage, there would be more power.

If, however, you assume a constant voltage, the current will drop as resistance increases, resulting in less power.

The infinity concept is not as easy to understand as you think. In fact, it is a theoretical, almost philosophical, concept.

If the resistance is truly infinite, you would need infinite voltage to drive current through it. But since the resistance is infinite, there is no way to drive current through it. We end up with a paradox.

As we all know, anything multiplied by zero is zero, and anything multiplied by infinity is infinity. Since power is voltage multiplied by current, you would be multiplying infinity by zero, which, of course, is the same paradox.

As mentioned, it is not possible to say if the increased resistance results in more or less power unless you also specify what happens with either current or voltage. The three are tightly linked by Ohm's law.

Only if you also declare that current remains constant.

When you turn your electric heater to a higher (more power) setting, you decrease its resistance. Your house is supplied by a (relatively) constant voltage, so reduced resistance gives increased current, which gives increased power.

Because there is no such thing as infinite voltage.

Of course, there is no infinite resistance or infinite current either.

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RoRo

is measured power "true power" or "imaginary power" (at least in AC circuit s)?

Yes, true. But if it is measurable after the resistor then it is like the w ater coming out of a hose. If a large current remains then the power is not dissipated like when we close a valve so that the water becomes a trickle or a drip, but instead is capable of doing significant work.

Why? This becomes important for understanding what is truly happening?

100 watt incandescent bulb. More current flows through a 100 watt bulb tha n through a 60 watt bulb. That is why the 100 watt bulb has a higher power rating.

More available power perhaps but less power dissipated as heat, right?

is measured current the current dissipated by the resistance or is it the u ndissipated current?

Yes, I know. A flow valve on a faucet is like a variable resistor. If you c lose the valve no current flows. Slowly open the valve and the resistance d ecreases to allow at first a drip then a trickle then a significant current of water to flow.

r the resistance the greater the dissipated power.

Less dissipated power. Don't believe me? Imagine a superconducting coil (an ideal coil with only reactive power). What is the power dissipated by such a superconducting coil?

able) current flowing through an infinite resistance is zero amps.

if an increase of resistance results in an increase of dissipated power th en an infinite resistance should dissipate 100% of the power flowing throug h it.

Yes, true. Sorry, my original post was perhaps poorly worded. I meant to sa y that is with a constant voltage power supply.

s, more power is dissipated.

My understanding is that the increased heat isn't due to the increased curr ent alone but rather the property of the material used for the heating elem ent which responds to the 60Hz frequency by emitting higher harmonics all t he way up to the infra-red range. Or perhaps the resistance increases (expo nentially?) as it heats up so that it starts out with low resistance but in creases? Has anyone measured the resistance of these heating coils while in operation?

Think about this all before responding. I understand that I am challenging the conventional understanding but am I really? look at what is taught with ideal coils and capacitors with no resistance. Is any power dissipated? Wh at happens when we add a resistance? Power is suddenly dissipated (as "true power"), right? Doesn't it make sense to conclude that as resistance incre ases (while using the same power supply with the same voltage) that the amo unt of power dissipated as "true power" should also increase?

all (or so it is taught).

You missed the point that was just made: There is no "remaining" current. The current out of a circuit is the same as the current into a circuit.

If, say, a 10A current feeds a circuit, a 10A current will pass all the way through the circuit whether there is a resistor in the circuit or not.

In your example, if the water out of a valve is a trickle, the water into the valve is also a trickle.

To carry the water example a little further, if a turbine is driven by water, the quantity of water out of the turbine is the same as the quantity of water into the turbine no matter how much power is delivered by the turbine.

You need to understand the fundamentals of current flow before moving on to more complicated concepts.

You're trying to make things unnecessarily complicated.

The truth is simple: P = IE.

Increasing the current with a fixed voltage increases the power dissipation and, hence, the radiated heat. The fact that the power and heat increase has nothing to do with the heating element's material, or 60 Hz, or harmonics. The simple fact is that the heating element gets hotter and emits more infrared energy (heat) whether you power it with 60 Hz, DC, or anything else.