Load resistance and current “draw”

[tranceical]

Hi all,

Sorry in advance if my questions are stupid and the answers are obvious.

I can see from P=VI why a high power appliance requires more current to function. However looking at ohms law and I=V/R it’s telling me that greater resistance will lead to less current. I've always figured that high power appliance or the load on something like a motor would have an inherently higher resistance/impedance, which is why it requires more power. In my head if a motor is working harder or has more physical resistance on its output shaft then it requires more power to function.

From trying to find an answer I read this on Wikipedia:
“Mains power outlets provide an easy example: they supply power at constant voltage, with electrical appliances connected to the power circuit collectively making up the load. When a high-power appliance switches on, it dramatically reduces the load impedance.”

Why would turning on a high power appliance reduce load impedance? If something needs more power, more work to be done then surely higher impedance exists.

Please help me marry up the relationship between the current a load draws with its inherent resistance and ohms law.

Many thanks

 

[Guineafowl]

Hi all,

Sorry in advance if my questions are stupid and the answers are obvious.

I can see from P=VI why a high power appliance requires more current to function. However looking at ohms law and I=V/R it’s telling me that greater resistance will lead to less current. I've always figured that high power appliance or the load on something like a motor would have an inherently higher resistance/impedance, which is why it requires more power. In my head if a motor is working harder or has more physical resistance on its output shaft then it requires more power to function.

From trying to find an answer I read this on Wikipedia:
“Mains power outlets provide an easy example: they supply power at constant voltage, with electrical appliances connected to the power circuit collectively making up the load. When a high-power appliance switches on, it dramatically reduces the load impedance.”

Why would turning on a high power appliance reduce load impedance? If something needs more power, more work to be done then surely higher impedance exists.

Please help me marry up the relationship between the current a load draws with its inherent resistance and ohms law.

Many thanks

Because P=VI, at a constant voltage P is proportional to I.

So you need more current for more power. Because I=V/R, at a constant voltage, the amount of current is proportional to 1/R.

So the device’s resistance needs to be low to draw a high current and deliver high power.

Why would a high power appliance reduce the load impedance? As we’ve seen, the appliance will have a low resistance/impedance in order to draw more current and output more power. By switching it into the circuit, you are adding an extra path for the current to flow.

“If something needs more power... then surely higher impedance exists” - delete this from your registry; It’s not true.

 

[Merlin3189]

What a nice question. My own intuitions used to lead me up the wrong path here. When I was young, I had toy boats (now I have big boats, but don't call them toys.) The first had a spring to drive it and when I lifted it out of the water, naturally I stopped the prop to prevent the spring from unwinding. When I got an electric one, I followed the same practice, imagining I suppose, that by stopping the prop I was saving my battery. (It was easier to stop the prop before I reached the switch and disconnected the battery.) Of course I know now, that the better course is to let the prop run free until I can switch off. A free running motor draws less current than a stalled motor (by a long way.)

Your issue hinges on this constant voltage supply and P=VI. If V is constant, then P∝I. So if no current flows, no power is used - the device is switched off. When it is switched on, more current = more power. A light bulb has perhaps a 3A fuse (generous), but a 3kW heater needs a 13A fuse.
So the light bulb must have a higher resistance and the heater a lower resistance to allow a suitable current to flow.

Another formula you get from $P=IV and R= \frac {V} {I} is P = \frac {V^2}{R}$
So power is inversely proportional to resistance for a given voltage.

Finally, there is the maximum power transfer theorem, which can be easily derived from the above, that says the maximum power you can draw from a supply occurs when the load resistance equals the supply resistance. If the load resistance is higher or lower, then you get less power out.
The extremes are infinite load resistance with no current flowing = no power out (obvious again) and zero load resistance or short circuit with large current flowing. Not quite so obvious that no power is dissipated in the load, because in practice we don't get absolute zero resistance shorts. But the supply (eg. a battery) gets very hot because that is where nearly all the power is being dissipated.
The mains power supply has to have a very, very low resistance, otherwise as soon as someone started drawing current, the supply voltage would be reduced and everyone else would get less power. So any load you connect is going to have a higher resistance - above the resistance of the supply - and reducing your load resistance brings it closer to that supply resistance, allowing it to draw more power.

So much for the math /physics argument, which maybe convincing, but may not be satisfying intuitively. Let's get on with the arm waving and hope the real scientists don't get too upset.

Say I want to push something along a flat surface. If it is very heavy, there will be lots of resistance (friction) and I may not be able to move it. I will push hard, work up a sweat, dissipate lots of heat inside me, but the thing doesn't move, I'm doing no work on it, it doesn't get hot from the firction of the ground. OTOH if it is very light and the surface is smooth, I'll move it easily. If I do as before and push as hard as I can, I end up running flat out, again getting hot, but still not doing much work on the object, which also isn't getting hot from the low friction. Somewhere in between are things I can push with moderate effort against moderate resistance. Both I and the object will get warm, me from the effort and it from the frictional heat derived from the work I'm doing on it.
We are rather complex power sources, so I can't say the rules are as simple as with electricity,but you can imagine there is an optimum load, where you can do the most amount of useful work. Deviate from that and the useful work done must decrease.

A dam full of water has lots of potential energy. If we just open the valve and let water pour out, it will rush out fast, lots of turbulence and that energy ends up as heat in the water as it slows down in the lower pool. If you put a waterwheel (for intuitive simplicity) in that flow, the water will be slowed down by it, the wheel will experience torque, rotate and useful work can be extracted. Less energy wasted in turbulent heating of the water. A light wheel (just so that we can ignore efficiency) with a light load will whiz round easily and extract a little useful work. The water will still flow fast and generate turbulent heat. Increase the load and we slow the water more, get more work out and less waste turbulence. You reach a load where the pressure of the water is turning the wheel at a good speed, generating lots of work, but if you increase the load any more, the pressure is insufficient to turn the wheel faster and it starts to slow down. Either the flow is restricted and now less energy is coming out of the dam, or the water can flow past, so more and more is flowing past without doing useful work and more energy is again ending up in turbulent waste. In the end, if the load is too great the wheel stops and no work is done. Again, making the resistance of the wheel closer to the optimum for that supply enables it to extract more power.

(Somewhere on the web was a beautiful video of a Pelton wheel being adjusted to optimum load. At low loads or at high loads water splashes off at high speed forwards or backwards, but at the optimum speed it almost dribbles away sideways, its energy largely spent. Unfortunately I can't find it at the moment.)

Hope that might be some help. I'll mull it over and try to think of better explanations if I can.

 

[jim hardy]

Why would turning on a high power appliance reduce load impedance? If something needs more power, more work to be done then surely higher impedance exists.

Please help me marry up the relationship between the current a load draws with its inherent resistance and ohms law.

Intuition can fool us. We must show our brain the way.

Surely as a kid you played with your Mom's vacuum cleaner ? I sure did.
It took me a long time to figure out why the motor sped up instead of slowing down when i put my hand over the end of the hose blocking it off.
Finally i figured out that when there was no flow of air the motor wasn't working very hard . I had to allow a large air flow to load down the motor.
My hand was a large resistance to airflow. By just spreading my fingers i could feel the airflow increase and hear the motor slow down.

Think of mechanical power as Pressure X Flow Rate . With same pressure, less resistance let's more air flow.
They're analogous to Volts X Amps which is electrical power ... With same voltage, fewer ohms let's more amps flow.

I hope that little experiment will encourage you to look up and study the definitions of Volt Amp Joule and Coulomb. Relating them to something you can feel and hear, like feeling the air from a vacuum cleaner exhaust push against your palm and listening as the motor loads and unloads , will train your brain to the right thinking steps.

Then do a units analysis. It'll get you ready for Physics class.

old jim

 

[tranceical]

Thanks a lot everyone for your replies. I really appreciate you taking the time out to enlighten me! The analogies have been a great read. A way I can explain how I have thought about it up to now is using the water pipe analogy for electricity. The higher the restriction or resistance in the pipe the more pressure and water flow required to achieve 100ml/second through per second than a pipe with less restriction.

So what it boils down to then is that high power devices have less impedance. In a simple case if I was designing a 1200w heater and a 2400w heater, during the design I would be ensuring adequate levels of resistance given what my supply voltage is and my desired power output? E.g. in the case of a 2400w heater if I have a 240v supply i would ensure the overall resistance in the heater element was 24ohms drawing 10amps? And in a 1200w heater 48ohms resistance drawing 5 amps?

If the above is true it will take further reading and time for me to get my little brain around the physics of motors! Haha.

 

[sophiecentaur]

So what it boils down to then is that high power devices have less impedance.

This is, indeed, true when you are using the normal sort of Electrical Power source. Batteries and most generators do their best to provide a Voltage that is more or less independent of the load. We refer to them as Constant Voltage supplies.
But not all supplies are like that and at some stage the OP may come across one of the following. There are supplies that are referred to as Constant Current Power sources. These will supply the rated current for a range of load resistances. When using one of these, we find that Increasing the Load Resistance will Increase the Power dissipated.
The formula for power can be written P = I²R, which, if I is constant, will be proportional to R.
I only quoted this out of a bit of devilment but it's a warning to read the small print about a device before making too many assumptions about what it will do in a circuit.

 

[Merlin3189]

In a simple case if I was designing a 1200w heater and a 2400w heater, during the design I would be ensuring adequate levels of resistance given what my supply voltage is and my desired power output? E.g. in the case of a 2400w heater if I have a 240v supply i would ensure the overall resistance in the heater element was 24ohms drawing 10amps? And in a 1200w heater 48ohms resistance drawing 5 amps?

That is a good example. You can do the calculations, as you say, but also see what's happening intuitively.

If you made a heater with switchable power levels of 1200W and 2400W, you might well use two filaments (or lengths of wire of wire.) so that you could switch the current to only or to both of them. So a single filament would need to be 48 Ω as you say, providing 1200W at 5A. The second filament would be just the same 48Ω providing another 1200W at 5A. So when both were switched on they'd draw 5+5=10A and provide 1200+1200=2400W. Because they are in parallel, the supply would see a lower resistance through the two filaments than when it had to flow through a single filament.
If you add more filaments in parallel, each will draw the same current and provide the same power. You get more power by providing more paths for the electricity to flow, which means less resistance and more current.

But that depends on the supply being able to maintain its voltage and supply the extra current for the extra filaments - the constant voltage supply.

If you chose to connect the two filaments in series, you upset this deal. They have to share the voltage between them and half the voltage means half the current. But half the current and half the voltage means a quarter of the power. You end up with only half the power of the single filament . By putting the filaments in series you have made it more difficult for the electricity to get through, ie. increased the resistance.

All that because you are using a near constant voltage source with very low resistance. If you could use a near constant current source as Sophie mentioned, it all reverses. If you put two filaments in series the current does not change, but the voltage must double to force it through. So you get your wanted double power. This time if you put them in parallel, the constant current has to be shared between them, half each, and half the current needs only half the voltage meaning a quarter the power in each and between them only half the power of the single filament.

For constant voltage sources, decreasing resistance loads extract more power, $P = \frac {V^2} {R}$ and for constant current sources, increasing resistance loads extract more power, $P = {I^2} {R}$

Of course both types of supply are ideal and real sources are somewhere in between, mainly tending towards low resistance constant voltage, though there is an increasing number of low power near-constant current supplies for LEDs and battery chargers.

 

[tranceical]

Thanks ever so much guys I’ve learned loads! Much appreciated :).

 

[sophiecentaur]

Thanks ever so much guys I’ve learned loads! Much appreciated :).

You have been given the relevant formulae here. I suggest you get your pencil and calculator and try bolting a few values into them to get used to the sorts of figures involved in practical resistive loads. e.g. A 100W mains light bulb, a 3kW mains water heater, a 6V body warmer giving 20W.