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#### xebo

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Here's a concept I'm stuck on:

Some load has a resistance to it. The less resistance it has, the more current it allows to flow through it. More current means more power draw. So less resistance should mean more power is drawn by the load.

But...

The more resistance a load has, the more energy is lost as current passes through it (In the form of heat, light, noise, etc). More energy lost means more power is drawn from the system. So more resistance should equate to more power drawn from the load.

Which is the problem. Less R = More I = More P. But Less R = Less energy loss = less P. So which is it? Does increasing resistance result in more power drawn or less? What is the flaw in my thinking?

Here's a concept that I THINK is in the right direction of where i want to be, but i can't quite connect the dots:

A single resistor draws a certain amount of power from the system. Now combine multiple resistors in parallel - the resistance of the group (as a whole) will go down, but the amount of energy lost from that resistance (and subsequent power draw) will go UP.

#### backstay

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xebo said:
The more resistance a load has, the more energy is lost as current passes through it (In the form of heat, light, noise, etc). More energy lost means more power is drawn from the system. So more resistance should equate to more power drawn from the load.
Only if current was constant, you're talking about voltage drop.

#### Shockdoc

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I would need to take up smoking weed to give you an answer . That's way too deep for a Sunday night.

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#### Wirenuting

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I pretwist and then use wire nuts. Solder pots rule.
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E=IxR except when R changes, then it is inversely proportional to E

Indian warfare when you put arrows to it.

#### wildleg

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consider an ordinary resistor on a circuit with nothing else.

V=IR, and the current passing through the resistor will be inversely proportional to the resistance. less resistance = more current.

what happens with the power going through, or "used up" by the resistor
is irrelevant to the circuit. the power used up by the resistor is the voltage
drop across the resistor times the current going through it.
P=VI. That power could be dissipated in heat, infrared light, quantum flux, who cares what, but that is the power used up by the resistor and that
is also the circuit solution.

So now think of the resistor as a black box, and let's say it's not a resistor anymore (non linear load, capacitive, inductive, motor, whatever).
The power at any instant being used is still the voltage across it times the
current going through it, and the power could be light, heat, or whatever, but the instantaneous power used by the black box is still V*I at any instant.
The only difference being that over time the power changes for the nonlinear load and you have a different way of figuring the power used to solve the circuit (next lesson).

hope this helps

dronai

#### 99cents

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Now you have me confused .

Think of the word "draw". We talk about loads "drawing" a certain number of amps. A 1000W heater, for instance draws about 8 amps and a 2000W heater draws about 16 amps at 120V. The 2000W heater has higher resistance but also higher current.

Electrons are very determined. It takes more of them to make it through the higher resistance.

#### dmxtothemax

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99cents;1321604 The 2000W heater has higher resistance but also higher current.[/quote said:
Quite wrong !

Higher current means LESS resistance !

Where did you get this idea from ?

A 2000w heater has less resistance than a 1000w heater.

#### 99cents

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Quite wrong !

Higher current means LESS resistance !

Where did you get this idea from ?

A 2000w heater has less resistance than a 1000w heater.

So you're telling me that, if I connect two 1000W heaters in series I have less resistance?

You know what? I AM confused!

I think I'm going to learn something here because I think I just posted bull chit .

#### backstay

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99cents said:
So you're telling me that, if I connect two 1000W heaters in series I have less resistance? You know what? I AM confused!
Two 1000 watt heaters connected in series isn't going to give you more current. You'll have about 30 ohm, which will give you about 4 amps. 4 times 120 volts is 480 watts, not 2000.

#### 99cents

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Two 1000 watt heaters connected in series isn't going to give you more current. You'll have about 30 ohm, which will give you about 4 amps. 4 times 120 volts is 480 watts, not 2000.
Thanks. Just when you think you know it all, you realize you don't.

#### Tsmil

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And the more you do know, the more you will realize just how much you don't know.

#### Safari

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So you're telling me that, if I connect two 1000W heaters in series I have less resistance?

You know what? I AM confused!

I think I'm going to learn something here because I think I just posted bull chit .
I take it this way
P=Vsqured/R

letting V be a unit =1

this means power is inversely proportional to restitance.

so basically 1000w heater should have more resistance than 2000w heater(power inversely proportional to resistance)

NOW

current is inversely proportional to resistance that means the 2000w wich has less resistance will draw more current. than 1000w heater

in any case power is directly proportional to current

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#### xebo

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I did some more googling and apparently a common misconception is that - using light bulbs for an example - the more resistance in the bulb, the brighter it will glow.

The misconception is that more resistance = more friction = more heat/light. The truth is that the brightness of a bulb is controlled by the friction caused by each electron passing through the filament, as well as the rate which this occurs at.So resistance and current both dictate power output.

If you were to increase the resistance of a bulb, you decrease the current passing through it, so it seems logical that the brightness (or power output) would stay about the same, but as it turns out brightness decreases as resistance increases, which seems to imply that power depends more on current than it does on resistance.

This agrees with the formula P = I^2*R.

So now the question that's on my mind is...why? why does current matter more than friction when it comes to drawing power?

#### Tsmil

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I was ready to chime in here but I'm on vacation, my helmet is at home and that wall looks pretty thick.

#### dronai

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Two 1000 watt heaters connected in series isn't going to give you more current. You'll have about 30 ohm, which will give you about 4 amps. 4 times 120 volts is 480 watts, not 2000.
Your right, and the same two heaters connected in parallel will give you the 2000W. In a series circuit it seems odd when using logic though. I need to follow shock docs advise and get stoned :laughing:

#### triden

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You're mixing up power loss and power delivered to the load. Look at the following diagram for an example:

Vs is constant. It's your source voltage feeding the load. Let's sat it's 120 VAc.

Rs is constant. It's the internal impedance of the load an never changes. Let's assume for this sake it's made up of the power lines and conductors that feed the load. Let's make it 5 ohms.

RL is your load. You can put whatever you want here. Let's put a heater that is 100 ohms here and do the calculation.

Current = V/R
Current = 120 / (5+100) = 1.14 amps. So this circuit draws 1.14 amps.

How many watts is the load? P = (I^2)*R, so P = 1.14^2*100 = 130.6 watts.

How much power is lost in Rs and given off as heat? P = 1.14^2*5 = 6.53 watts.

Now lets put a heater with a HIGHER resistance in the circuit. Let's go for 200 ohms (double).

Current = 120 / (5+200) = 0.585 amps (half than before)

How many watts is the load? P = (I^2)*R, so P = 0.585^2*200 = 68.53 watts (exactly half)

How much power is lost in Rs and given off as heat? P = 0.585^2*5 = 1.71 watts. (Almost 4 times lower!)

I tried to simplify this to make it easy to follow. Current is a big deal when it comes to power loss because the power in a circuit is the square of the current. Double the current and quadruple the power (with all other things remaining constant)

#### JRaef

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I think that the key element to your misunderstanding here is that you are directly equivocating resistance and load. They are NOT the same thing. LOAD is the work that you want to take place, RESISTANCE is consequential to your device doing that work, not the goal.

So let's take your light bulb for example. The work that you want to perform with that light bulb is to create lumens. To create lumens with an incandescent bulb, the filament provides resistance to the flow of current through it, which transmutes the electrical energy into light and heat, but more white visible light than heat (incandescence). If I use an LED lamp however, I am achieving the same lumen output with much less energy input. Since there is less energy being consumed in the circuit for the same amount of work (lumens) being performed, at the same input voltage, it does equate to there being a much higher resistance to current flow at the equivalent voltage input point. But it is not specifically the resistance that is doing the work here, it is the power converter and LEDs that do the conversion, but without the heat, a wasted byproduct of the resistance. The resistance in the LED driver circuit is just something that you can back-calculate if you are interested, and is representative of the majority of the losses in the circuit, but is MUCH MUCH less than the losses in the incandescent version.

Take another direction on the incandescent bulb. If instead of a tungsten element that incandesces with a certain voltage applied, you use a NiChrome heating element of the exact same resistance value, will you get the same amount of light from it? No? But isn't that the same voltage and resistance, therefore the same current? The point is, the resistance is NOT the work being performed here, it is a component of the work being performed.

In this new case, let's say the desired work is calorific heating of the surfaces surrounding the NiChrome heating element. But let's say the surface was a cup of water, now let's put that water into a microwave. I can achieve the same calorific heating of that water using less total energy via the microwave than by putting it in a pot on a stove. So again, the resistance will technically be higher with the microwave when you calculate it backward, yet the same work is performed with less total energy input, because we have increased the efficiency of the transfer of energy from the 120V input to the water (by not also heating up the surrounding air).

#### 220/221

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Here's a concept I'm stuck on:

Some load has a resistance to it. The less resistance it has, the more current it allows to flow through it.
I'd suggest that you stop right there and think of it from a different angle.

The more resistance it has, the more current it will draw.

#### backstay

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220/221 said:
I'd suggest that you stop right there and think of it from a different angle. The more resistance it has, the more current it will draw.
Huh?

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