raymw Posted January 1, 2019 Share Posted January 1, 2019 No need to introduce AC theory, but we could open the RMS compartment of Pandora's box Do you want to talk about hifi amps, and their power ratings? Link to post Share on other sites More sharing options...
Dungrange Posted January 1, 2019 Share Posted January 1, 2019 A "40VA" supply could be 40V (1 Amp) or 400V (1/10 Amp). The current in a fixed resistance is determined ONLY by the voltage applied and WILL satisfy ohms law. [A heater will not actually have a constant resistance due to temperature, but I am ignoring that for the sake of clarity] This is obviously at the limit of my electrical knowledge (which thankfully I don't do for a living), but what I don't know is whether overloading a 40VA supply to run a 50W device results in a reduction in the voltage supplied (below the rated voltage of the supply), a reduction in the current (compared to what would flow with an adequate supply) or both. My solution to such a problem would just be to buy a bigger power supply and not worry about such issues. 2 Link to post Share on other sites More sharing options...
RMweb Premium Pannier Tank Posted January 1, 2019 RMweb Premium Share Posted January 1, 2019 Please read Nearholmer's post (no. 66) which explains that under certain conditions what I stated can be correct. There is nothing in that post (66) that states "Ohms law states that if the voltage goes down then the current must go up if the resistance stays the same." Link to post Share on other sites More sharing options...
raymw Posted January 1, 2019 Share Posted January 1, 2019 (edited) Hi David, it really depends on the type of supply (and on the type of load). Even if you consider something thought of as being simple, say a battery, it depends on the chemistry, and its age, how far down the discharge path, etc. There is no general answer to this type of question. If you have a specific 40VA supply in mind, then the manufacturer may have the answer. I guess your answer is the simplest. Edited January 1, 2019 by raymw Link to post Share on other sites More sharing options...
Crosland Posted January 1, 2019 Share Posted January 1, 2019 Do you want to talk about hifi amps, and their power ratings? Noooooo,pleeeease, Noooooo. then we would also have to talk about gold plated mains plugs and oxygen free copper. Link to post Share on other sites More sharing options...
Crosland Posted January 1, 2019 Share Posted January 1, 2019 This is obviously at the limit of my electrical knowledge (which thankfully I don't do for a living), but what I don't know is whether overloading a 40VA supply to run a 50W device results in a reduction in the voltage supplied (below the rated voltage of the supply), a reduction in the current (compared to what would flow with an adequate supply) or both. My solution to such a problem would just be to buy a bigger power supply and not worry about such issues. Now you are getting into design details of the supply. A 50W load is more than the supply is rated for. If the supply continues to operate and limits the current then ohms law tells us that the output voltage must also fall, since the load resistance is fixed, so the answer is both. Some supplies will cutout altogether or use "foldback" current limiting in which the current and voltage are limited to much lower values (the answer is still both ). I leave further details, for those that want to know, to Mr Google. Link to post Share on other sites More sharing options...
Martin Shaw Posted January 1, 2019 Share Posted January 1, 2019 I have read this thread with some interest, and after 47 years as a professional electrical engineer one of the few things I definitely know is Ohm's law, and it's application in any electrical circuit will always have the same impact, as many others have already made plain. However counter to this is the proffered argument by the Sussex flat earth society which goes like this. If you take a 100% efficient electrical device that can give a 1000w of output work, you have by virtue of Newton's laws to feed it 1000w of electrical energy that could well be in the form of 10v at 100A. If you halve the supply voltage to 5 and at that PD the device will still give 1000w ouptut then the current will indeed rise to 200A, this is of course nothing to do with Ohm's law, and everything to do with a change in how the device works which electric heaters can't do. On a serious note the issue isn't moulded plugs per se but the relationship between the plug pin and the wiping spring contacts in the socket, these can become corroded in hostile environments which causes a high resistance and localised heating. As the springs heat they lose temper which worsens the state of affairs increasing the heating which manifests itself in the plug. In any case both the plug and socket should be changed as both present a fire risk which may ignite before a 13A fuse blows, as well as personal hazard. Whilst I recognise that BS1361 permits a maximum rating for a socket outlet of 13A I personally wouldn't want to use a device at that current for a prolonged period. In all electrical matters assume it's dangerous and make your own safety the highest priority. Regards Martin 1 Link to post Share on other sites More sharing options...
AndyID Posted January 1, 2019 Share Posted January 1, 2019 Things can get complicated, he is no longer in the land of the living. My memory is returning, back when I had 2 gaugemaster controllers the trains used to slow down slightly when the 2kw heater kicked in Presumably other electricians are not quite dead yet? The voltage drop created by the heater could result from poor plug/socket contact or it could be the cable that's not equal to the task. My shed runs on an extension from the house and I can run an arc-welder there without any problems, but that's because I used the right size of wire when I buried the cable. Link to post Share on other sites More sharing options...
davetheroad Posted January 1, 2019 Share Posted January 1, 2019 Presumably other electricians are not quite dead yet? The voltage drop created by the heater could result from poor plug/socket contact or it could be the cable that's not equal to the task. My shed runs on an extension from the house and I can run an arc-welder there without any problems, but that's because I used the right size of wire when I buried the cable. The cable appears to be half inch diameter armoured so is it 1.5mm squared wire? Link to post Share on other sites More sharing options...
locomad Posted January 1, 2019 Share Posted January 1, 2019 Presumably other electricians are not quite dead yet? The voltage drop created by the heater could result from poor plug/socket contact or it could be the cable that's not equal to the task. My shed runs on an extension from the house and I can run an arc-welder there without any problems, but that's because I used the right size of wire when I buried the cable. Couple of years ago in local aldi customer brought back arc welder with 13 amp plug melted into the house socket complaining its broke, on box clearly stated don't use 13 amp plug, couldn't stop laughing Link to post Share on other sites More sharing options...
AndyID Posted January 1, 2019 Share Posted January 1, 2019 The cable appears to be half inch diameter armoured so is it 1.5mm squared wire? That's maybe a bit light for 20 meters but I think most of the voltage drop is due to poor contacts. An electrician should be able to quickly identify what's causing the drop. Link to post Share on other sites More sharing options...
AndyID Posted January 1, 2019 Share Posted January 1, 2019 I have to say that I was always under the (obviously wrong) impression that fuses were supposed to stop plugs melting, as I thought that the fuse was supposed to be the weak point in the circuit. It's a little alarming to discover that is not the case. The main reason for the plug fuse is to limit the current that can be delivered to an appliance and its cable. Without a fuse an appliance would be able draw over 30 amps from the ring. Link to post Share on other sites More sharing options...
RMweb Premium phil-b259 Posted January 1, 2019 RMweb Premium Share Posted January 1, 2019 There is nothing in that post (66) that states "Ohms law states that if the voltage goes down then the current must go up if the resistance stays the same." Ok - if it satisfy your macho streak then yes I admit Ohms law only talks about the Voltage, Resistance and Current relationship - and therefore should not have been cited when talking about Power. However according to Wikipedia (other references sources are available) ”In the case of resistive (Ohmic, or linear) loads, Joule's law can be combined with Ohm's law (V = I·R) to produce alternative expressions for the amount of power that is dissipated: P=IV” This equation quite clearly sates power is a function of voltage multiplied by current, and that if the power figure is to be maintained when voltage lowers then the current must increase. Link to post Share on other sites More sharing options...
RMweb Premium phil-b259 Posted January 1, 2019 RMweb Premium Share Posted January 1, 2019 I have read this thread with some interest, and after 47 years as a professional electrical engineer one of the few things I definitely know is Ohm's law, and it's application in any electrical circuit will always have the same impact, as many others have already made plain. However counter to this is the proffered argument by the Sussex flat earth society which goes like this. If you take a 100% efficient electrical device that can give a 1000w of output work, you have by virtue of Newton's laws to feed it 1000w of electrical energy that could well be in the form of 10v at 100A. If you halve the supply voltage to 5 and at that PD the device will still give 1000w ouptut then the current will indeed rise to 200A, this is of course nothing to do with Ohm's law, and everything to do with a change in how the device works which electric heaters can't do. On a serious note the issue isn't moulded plugs per se but the relationship between the plug pin and the wiping spring contacts in the socket, these can become corroded in hostile environments which causes a high resistance and localised heating. As the springs heat they lose temper which worsens the state of affairs increasing the heating which manifests itself in the plug. In any case both the plug and socket should be changed as both present a fire risk which may ignite before a 13A fuse blows, as well as personal hazard. Whilst I recognise that BS1361 permits a maximum rating for a socket outlet of 13A I personally wouldn't want to use a device at that current for a prolonged period. In all electrical matters assume it's dangerous and make your own safety the highest priority. Regards Martin 47 years in the industry may well taught you Ohms law - but it doesn’t seem to have taught you respect when dealing with mistaken folk or the ability to recognise that when someone may be 100% correct with the equations / relationships but have given them the wrong name. Link to post Share on other sites More sharing options...
APOLLO Posted January 1, 2019 Share Posted January 1, 2019 Electrickery ?- This guy explains all !!!!!! https://www.youtube.com/user/msadaghd/videos Brit15 Link to post Share on other sites More sharing options...
raymw Posted January 1, 2019 Share Posted January 1, 2019 Hi Phil, I'm guessing the angst is/was caused by a typo in your earlier post. Ohms law states that if the voltage goes down then the current must go up if the resistance stays the same.Similarly if the voltage goes down then the current must go up to maintain the same amount of power. the first line should be if the voltage goes down, the current goes down, if resistance is constant. The second line is fine. - (you've not mentioned ohm's law in that sentence). In general- Afaik, the moulded plugs have the wire ends crimped to the pins. There is no way of non-destructively testing the electrical quality* of a crimped connection, in particular once embedded in plastic. I expect there is little if any quality control on the bought in leads by the appliance manufacturer, and if purchased on price alone, the quality of the materials will not be the same as a more expensive item, since more or less the assembly/packaging will be similar for both. The plug top fuse is to protect the cable. The cable should be sized to suit the appliance. The appliance may/may not have its own protection. A 13 amp fuse, as mentioned before, does not blow at 13A. it is designed to carry 13A without blowing. * by that, I mean you can obviously test by passing say, 20A through the connection for a few days, but you can't determine its maximum current for failure and expect to use it afterwards. Link to post Share on other sites More sharing options...
Nearholmer Posted January 1, 2019 Share Posted January 1, 2019 (edited) Phil "...... if the power figure is to be maintained when voltage lowers then the current must increase." True. But, I think that the root of the confusion here is an assumption that in most loads the power figure will be maintained when the voltage varies, when in fact it won't. In order for a load to maintain constant power dissipation, it must incorporate the capabilities to measure voltage and current, then self-adjust its effective resistance, so that the product of the two remains constant ........ and most loads simply don't incorporate those capabilities, and even those that do can only maintain constant power across a stated range of conditions. For most loads, the rated "power" is not what it will dissipate under all (or even a range) of conditions, but one of two things: - the maximum power that it can dissipate without doing harm by overheating; or, - the power that it will dissipate when supplied with a stated voltage. Does that make sense? Kevin Edit: As a PS and correction, because it has just come back to my mind, induction motors can present themselves as constant power loads across a wide voltage range, without the need for a control scheme as I've outlined, because the 'control function' is inherent in the action of the motor. Edited January 1, 2019 by Nearholmer Link to post Share on other sites More sharing options...
sharris Posted January 1, 2019 Share Posted January 1, 2019 Ok - if it satisfy your macho streak then yes I admit Ohms law only talks about the Voltage, Resistance and Current relationship - and therefore should not have been cited when talking about Power. However according to Wikipedia (other references sources are available) ”In the case of resistive (Ohmic, or linear) loads, Joule's law can be combined with Ohm's law (V = I·R) to produce alternative expressions for the amount of power that is dissipated: P=IV” This equation quite clearly sates power is a function of voltage multiplied by current, and that if the power figure is to be maintained when voltage lowers then the current must increase. Yes, power P = VxI From ohm's law, V = IxR or I = V/R Thus P = VxV/R, so if you know the voltage and the resistance of the element you can work out the power without knowing the current (or conversely P = IxIxR if you know the current and don't need to know the voltage), so you can see that power is proportional to the square of the voltage (or proportional to the square of the current if that's what you're measuring) - therefore if you have a heater designed for 240V mains and you plugged it into a 120V supply it would only get as quarter as warm. Remember also that from ohm's law R = V / I If your hypothesis that the heater is a constant power device held, and the current went up in proportion to the amount the voltage went down, by ohm's law you would be stating that the resistance of the element varied with voltage (in fact with the square of the voltage). However resistance doesn't work like that - it depends on the physical properties of the material, the length and cross-sectional area of the element wire, and a bit on the temperature (different materials behave differently). Link to post Share on other sites More sharing options...
RMweb Gold chris p bacon Posted January 1, 2019 RMweb Gold Share Posted January 1, 2019 All I know is that Electrical cable is full of smoke, as everytime I go through one with a digger bucket that's what comes out... 2 Link to post Share on other sites More sharing options...
sharris Posted January 1, 2019 Share Posted January 1, 2019 All I know is that Electrical cable is full of smoke, as everytime I go through one with a digger bucket that's what comes out... All electrical devices run on smoke - they stop working when you let the smoke out. Link to post Share on other sites More sharing options...
RMweb Gold simon b Posted January 1, 2019 RMweb Gold Share Posted January 1, 2019 That's an interesting video but it does not show how to put it back together again! I think I will stick to traditional plugs where I can open them up to check all is ok and to correct if not. Best place for those molded plugs is in the skip, the older style which has a proper brass screw to clamp the cable is safest. Link to post Share on other sites More sharing options...
Nearholmer Posted January 1, 2019 Share Posted January 1, 2019 Ray "......There is no way of non-destructively testing the electrical quality* of a crimped connection....." I'm not sure I totally agree with that. Herewith a range of variously expensive instruments for measuring very low resistances, which are commonly used to test all sorts of connections, crimped and otherwise. https://www.tester.co.uk/high-voltage-utility-and-rail/earth-and-insulation-test-equipment/low-resistance-ohmmeters There are limits to what can be learned from such a test, in that things can change under the thermal stresses of loading, but a 'ductor test' is a very good indicator of the state of a connection, and is exactly what I would want to apply to the plug and socket that started this whole debate. Kevin Link to post Share on other sites More sharing options...
AndyID Posted January 1, 2019 Share Posted January 1, 2019 All electrical devices run on smoke - they stop working when you let the smoke out. And resistance is futile. 1 Link to post Share on other sites More sharing options...
exet1095 Posted January 1, 2019 Share Posted January 1, 2019 A practical example of what happens when you run fixed resistance devices at low voltages. In America, electric kettles are rare. So is decent tea. I always take a Tesco Value (£4.95) kettle and Yorkshire Tea when I go for work. Normally, with 110V, the kettle takes 4 times as long to boil as it does here. Using Ohm’s law, we can deduce that as a 2200W (10 Amps x 220 volts) device on 220V, it has a resistance of 22 ohms. This means that at 110V it will only draw 5 Amps. It therefore becomes a very slow 550W kettle. Despite this, my American friends are usually so impressed with a device that actually boils water, that they are delighted to accept it as a parting gift, giving me more room in my bag for all the stuff I have ended up buying. Paul 3 Link to post Share on other sites More sharing options...
RMweb Premium melmerby Posted January 1, 2019 RMweb Premium Share Posted January 1, 2019 A practical example of what happens when you run fixed resistance devices at low voltages. Using Ohm’s law, we can deduce that as a 2200W (10 Amps x 220 volts) device on 220V, it has a resistance of 22 ohms. This means that at 110V it will only draw 5 Amps. It therefore becomes a very slow 550W kettle. It's not always as simple as that as the hot resistance on 110v is different to 220v and therefore will have a different power. We used to have a small "Travel" kettle which was suitable for 100-220v but the quoted power at 110v wasn't 25% of the power at 220v (I can't remember what it actually quoted) BTW Amazon.com lists dozens of 110v electric kettles, so they can hardly be rare. Keith Link to post Share on other sites More sharing options...
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