Measured Power: Dissipated or Undissipated?

Is measured current dissipated or undissipated current? In other words, is measured power "true power" or "imaginary power" (at least in AC circuits)?

For example, a 60 watt incandescent lightbulb has more resistance than a 100 watt incandescent bulb. More current flows through a 100 watt bulb than through a 60 watt bulb. That is why the 100 watt bulb has a higher power rating.

So, since a 100 watt lightbulb has less resistance than a 60 watt bulb, is measured current the current dissipated by the resistance or is it the undissipated current?

Here's another (related) paradox: Resistance dissipates power. The higher the resistance the greater the dissipated power. But what is the power dissipated by an infinite resistance? The (measurable) current flowing through an infinite resistance is zero amps.

So, it is said that an infinite resistance acts like an open switch. But if an increase of resistance results in an increase of dissipated power then an infinite resistance should dissipate 100% of the power flowing through it.

But that is NOT what is taught. It is taught that as resistance increases, more power is dissipated. Yet, for some reason, at infinite resistance...no power is dissipated at all (or so it is taught).

 

[Robert Roland]

On Tue, 28 Oct 2014 15:02:20 -0700 (PDT), rb.rb642@gmail.com wrote:

>Is measured current dissipated or undissipated current? In other words, is measured power "true power" or "imaginary power" (at least in AC circuits)?

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.

>For example, a 60 watt incandescent lightbulb has more resistance than a 100 watt incandescent bulb. More current flows through a 100 watt bulb than through a 60 watt bulb. That is why the 100 watt bulb has a higher power rating.

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

>So, since a 100 watt lightbulb has less resistance than a 60 watt bulb, is measured current the current dissipated by the resistance or is it the undissipated current?

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

>Here's another (related) paradox: Resistance dissipates power. The higher the resistance the greater the dissipated power.

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.

> But what is the power dissipated by an infinite resistance? The (measurable) current flowing through an infinite resistance is zero amps.

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.

>So, it is said that an infinite resistance acts like an open switch. But if an increase of resistance results in an increase of dissipated power then an infinite resistance should dissipate 100% of the power flowing through it.

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.

>But that is NOT what is taught. It is taught that as resistance increases, more power is dissipated.

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.

>Yet, for some reason, at infinite resistance...no power is dissipated at all (or so it is taught).

Because there is no such thing as infinite voltage.

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

 

On Thursday, October 30, 2014 12:33:58 PM UTC-7, Robert Roland wrote:

On Tue, 28 Oct 2014 15:02:20 -0700 (PDT), rb.rb642@gmail.com wrote:

Is measured current dissipated or undissipated current? In other words, is measured power "true power" or "imaginary power" (at least in AC circuits)?

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.

Yes, true. But if it is measurable after the resistor then it is like the water 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.

Forget about reactive power and imaginary numbers for now.

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

For example, a 60 watt incandescent lightbulb has more resistance than a 100 watt incandescent bulb. More current flows through a 100 watt bulb than through a 60 watt bulb. That is why the 100 watt bulb has a higher power rating.

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

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

So, since a 100 watt lightbulb has less resistance than a 60 watt bulb, is measured current the current dissipated by the resistance or is it the undissipated current?

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

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

Here's another (related) paradox: Resistance dissipates power. The higher the resistance the greater the dissipated power.

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.

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?

But what is the power dissipated by an infinite resistance? The (measurable) current flowing through an infinite resistance is zero amps.

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.

So, it is said that an infinite resistance acts like an open switch. But if an increase of resistance results in an increase of dissipated power then an infinite resistance should dissipate 100% of the power flowing through it.

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.

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

But that is NOT what is taught. It is taught that as resistance increases, more power is dissipated.

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.

My understanding is that the increased heat isn't due to the increased current alone but rather the property of the material used for the heating element which responds to the 60Hz frequency by emitting higher harmonics all the way up to the infra-red range. Or perhaps the resistance increases (exponentially?) as it heats up so that it starts out with low resistance but increases? 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? What happens when we add a resistance? Power is suddenly dissipated (as "true power"), right? Doesn't it make sense to conclude that as resistance increases (while using the same power supply with the same voltage) that the amount of power dissipated as "true power" should also increase?

Yet, for some reason, at infinite resistance...no power is dissipated at all (or so it is taught).

Because there is no such thing as infinite voltage.

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

 

[Bob Penoyer]

On Mon, 10 Nov 2014 01:29:06 -0800 (PST), rb.rb642@gmail.com wrote:

On Thursday, October 30, 2014 12:33:58 PM UTC-7, Robert Roland wrote:
On Tue, 28 Oct 2014 15:02:20 -0700 (PDT), rb.rb642@gmail.com wrote:

Is measured current dissipated or undissipated current? In other words, is measured power "true power" or "imaginary power" (at least in AC circuits)?

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.


Yes, true. But if it is measurable after the resistor then it is like the water 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.

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.

Forget about reactive power and imaginary numbers for now.


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

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

<Snip>

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.


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

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.