Sockets in my house MELTING my plugs!

sincere113

wildefalcon wrote: »

If the voltage drops, and the wattage remains the same, then the current (amps) increases.

Higher amps, higher temperatures.

robbie7730 wrote: »

Thats completely incorrect. When voltage reduces, the resistance of the item or appliance is what remains the same, so the amps reduce, and so the wattage is less.

.......................

You're wrong Robbie7730.....Wildefalcon is correct


Wildefalcon is correct because he states that the power (wattage) is the same..... In your example you are changing the power aren't you?

Example;-

Power (W) = Voltage X Current (I).

If W = 2000, Voltage is 200 (just for arguments sake for this example), then I = W/V = 2000/200 = 10A......

But like Wildefalcon states

"If the voltage drops, and the wattage remains the same, then the current (amps) increases"

So if now our Voltage is 100V;-

Again I = W/V = 2000/100 = 20A.

As you can see the current must go up to maintain 2000W power.....

M cebee

no that's not how it works

sincere113

M cebee wrote: »

no that's not how it works

In order to deliver the same amount of power to a load, if the input voltage reduces then the current must increase right?

M cebee

the power drops when voltage drops

rating of appliances is for a nominal voltage ie:230

sincere113

M cebee wrote: »

the power drops when voltage drops

rating of appliances is for a nominal voltage ie:230

Yes I agree that your statement is correct. What you are saying is correct that in practise if the supply voltage drops then output power drops proportionally.

However, what I am saying is that if the power consumed is constant and the voltage drops then the current drawn must increase. That is theoretically sound.

I work in an industry that provides a product which provides a power supply to a load. The power delivered must remain constant. e.g. 100W. The ESB (or whoever) electrical supply from the wall socket fluctuates greatly over the period of a day (210V-260V), so our product draws varying amounts of current depending on the supply voltage to maintain an output of 100W.

cast_iron

sincere113 wrote: »

However, what I am saying is that if the power consumed is constant and the voltage drops then the current drawn must increase. That is theoretically sound.

Perhaps, but makes no sense in the context of this thread, as there is no reason to think why the power would be constant. Which is probably why robbie assumed whitefalcon's input meant something other than what he was saying.

sincere113

cast_iron wrote: »

sincere113 wrote: »

However, what I am saying is that if the power consumed is constant and the voltage drops then the current drawn must increase. That is theoretically sound.

Perhaps, but makes no sense in the context of this thread, as there is no reason to think why the power would be constant. Which is probably why robbie assumed whitefalcon's input meant something other than what he was saying.

Yes we're maybe going abit OT but there are products which are designed to work off a residential supply which must deliver and maintain a specific power level. These products draw more current as the input voltage drops to a specific level and then they 'drop out' for safety purposes. It's all in PSU designs....

Therefore it is wrong to assume in all cases that as the nominal supply voltage drops that the power consumption drops and the particular product does not draw more current. Intentional or not, or in fact it is relevant to the OP query, wildefalcons statement is correct.

paddymick

sincere113 wrote: »

Yes we're maybe going abit OT.

That seems to be a common theme in these threads,The OP`s question will get answered quickly in the first few posts then the next few pages are like a competition to see who has the biggest gonads when it comes to electrical knowledge!!

Im not bashing just making an observation:)...

.G.

paddymick wrote: »

That seems to be a common theme in these threads,The OP`s question will get answered quickly in the first few posts then the next few pages are like a competition to see who has the biggest gonads when it comes to electrical knowledge!!

Im not bashing just making an observation:)...

Your wrong there Paddymick.

Utterly wrong:D

M cebee

2011 and robbie are the biggest troublemakers:D

M cebee

sincere113 wrote: »

Yes I agree that your statement is correct. What you are saying is correct that in practise if the supply voltage drops then output power drops proportionally.

However, what I am saying is that if the power consumed is constant and the voltage drops then the current drawn must increase. That is theoretically sound.

I work in an industry that provides a product which provides a power supply to a load. The power delivered must remain constant. e.g. 100W. The ESB (or whoever) electrical supply from the wall socket fluctuates greatly over the period of a day (210V-260V), so our product draws varying amounts of current depending on the supply voltage to maintain an output of 100W.

you're talking bout these gadgets
http://www.apc.com/resource/include/techspec_index.cfm?base_sku=LE1200I&tab=features

it's hardly relevant here anyhow

brightspark

sincere113 wrote: »

Yes we're maybe going abit OT but there are products which are designed to work off a residential supply which must deliver and maintain a specific power level. These products draw more current as the input voltage drops to a specific level and then they 'drop out' for safety purposes. It's all in PSU designs....

Therefore it is wrong to assume in all cases that as the nominal supply voltage drops that the power consumption drops and the particular product does not draw more current. Intentional or not, or in fact it is relevant to the OP query, wildefalcons statement is correct.

It's not relevant to the OP query at all, the OP query concerned melted plugs which are not caused by undervoltage/overcurrent.

AND the OP mentioned a hoover, are your power supplies ever installed in vacuum cleaners?
(I'm guessing they are 'Switched Mode Power Supplies')


We are now in excess of 40 posts, the OP stated back in post 20 that they are getting an electrician in and that their neighbour with a similar problem had cheap sockets replaced.

Can we not take it for granted that anyone who answers a post in this forum understands Ohm's law etc.?

Let's stop the guessing games here and wait for the OP to report back and let us know what the electrician discovers.

Bruthal

sincere113 wrote: »

.......................

You're wrong Robbie7730.....Wildefalcon is correct

Here is what wildefalcon said

wildefalcon wrote: »

If the voltage drops, and the wattage remains the same, then the current (amps) increases.

But the wattage cant remain the same. Your confusing this with electrical science questons to find currents in different situations.

For the wattage to remain constant as voltage reduces, the load resistance has to reduce as the voltage reduces. That doesnt happen.

As i said earlier, a motor is an example of an item which can draw higher current when it has its voltage reduced, when the motor is driving a load. But in general, wattage reduces as voltage reduces, and any decent electrical loads in houses will be heating element type loads.

For wattage to remain the same as voltage reduces, and so current increases, the appliance impedence has to reduce, which happens in motors as the rotor speed reduces. But motor loads in domestic situations are small.

Bruthal

sincere113 wrote: »

In order to deliver the same amount of power to a load, if the input voltage reduces then the current must increase right?

To deliver the same amount of power with a lower voltage, requires the item to be specifically designed for the voltage in question.

A 9kw 230v shower will not be 9kw if the voltage reduces to 220v. It is only 9kw at 230v.

Calculating 9kw at 115v and getting double the current of a 230v 9kw shower, is simply because the 115v one will have elements of 4 times lower resistance than the 230v version.

It does not mean if you connect the 230v 9kw shower to the 115v that you now get double the current. The current will be half, and so the output will be 2.25kw

Bruthal

cast_iron wrote: »

Perhaps, but makes no sense in the context of this thread, as there is no reason to think why the power would be constant. Which is probably why robbie assumed whitefalcon's input meant something other than what he was saying.

Whitefalcon said if the voltage drops, but the wattage remains the same, then the current increases. Thats not what happens with appliances in real life.

sincere113

robbie7730 wrote: »

Here is what wildefalcon said

But the wattage cant remain the same. Your confusing this with electrical science questons to find currents in different situations.

For the wattage to remain constant as voltage reduces, the load resistance has to reduce as the voltage reduces. That doesnt happen.

Yes it does in some products.

Some domestic products provide quite tight output power levels irrespective of input voltage. Thats my point exactly. If I need a product which provides a tightly specified output power level, but the nominal voltage varies over the course of a day your saying its not possible to deliver a specified output power because the device can not change its impedance. THAT IS WRONG..... Its also best to refer to impedance rather than resistance when talking about ac applications.

Your examples are correct because those examples are power rated at a nominal voltage as someone mentioned earlier in the thread. However some products have a rated power over a range of input voltage.

Brightspark:
Intentional or not, or in fact it is relevant to the OP query

Sorry, I meant to say or in fact if it is relevant, meaning its not very applicaple to the OP. However it is relevant as I corrected someone with incorrect information.

Bruthal

sincere113 wrote: »

Yes it does in some products.

Some domestic products provide quite tight output power levels irrespective of input voltage. Thats my point exactly. If I need a product which provides a tightly specified output power level, but the nominal voltage varies over the course of a day your saying its not possible to deliver a specified output power because the device can not change its impedance. THAT IS WRONG.....

Where did i say its not possible? You say if you need a product which has a regulated supply. What appliances in the usual domestic setup have regulated supplies? Now we are talking about appliances, not laptop transformers/power supplies.

Its also best to refer to impedance rather than resistance when talking about ac applications.

It passed you by that i refered to resistance in the examples of heating elements, and impedence in the motor references. An element has practicaly no reactance at 50hz, so resistance is good enough for elements. What is the PF of an immersion element? I`d say its 1. If its 1, then its impedence = its resistance.

Its astonishing you mention impedence, as if to use it as another method to argue your point, and yet you cant grasp the very basic theme of the thread. Telling us that wildefalcon saying w/v = amps is correct, and that means my statement, and other posters, that reducing voltage on a normal everyday domestic load does not result in a stable power output, but reduces power output, is incorrect accodring to you.

It is not incorrect to say an appliance power output reduces as its voltage reduces, simply because you state that voltage regulators can be got and used.

Your examples are correct because those examples are power rated at a nominal voltage as someone mentioned earlier in the thread. However some products have a rated power over a range of input voltage.

Brightspark:
Intentional or not, or in fact it is relevant to the OP query

Sorry, I meant to say or in fact if it is relevant, meaning its not very applicaple to the OP. However it is relevant as I corrected someone with incorrect information.

Which is it? Correct or incorrect? My examples are correct? Unbelievable this:D. You earlier said i was wrong in the thread, now my particular examples are correct. My examples are typical household items. And becuase you sell regulators, all of a sudden other posters are incorrect when they say a reducing voltage will mean a reducing power output.

sincere113

Once again, last time though...

robbie7730 wrote: »

Thats completely incorrect. When voltage reduces, the resistance of the item or appliance is what remains the same, so the amps reduce, and so the wattage is less.

That is wrong....There are items whos impedance changes to deliver a steady output power.

robbie7730 wrote: »

l will use an immersion as an example.

If its 3kw rated at 230v, then it will only be 3kw at the 230v.
3000w/230v = 13 amps

230v/13A = 17.7 ohms. This is fixed, and is what rates the immersion at 3kw only at 230v

Same immersion At 240v
240v/17.7 ohms = 13.6 amps
240v x 13.6A = 3.26 kw.

Now connect 115v to it.
115V/17.7 ohms = 6.2 amps

115 x 6.5 amps = 747 watts. Notice when the voltage is halved, the watts will be 1 quarter. If voltage is doubled to a fixed resistance, the wattage will be 4 times higher.

For an immersion to be 3kw at 115 volts.
3000w/115v = 26 amps
115v/26A = 4.42 ohms.

So the 3kw 115 volt immersion would be 1 quarter of the resistance of the 230v 3kw one.

So halving the voltage to a 230v 3kw appliance does not keep the appliance at 3kw and have a higher current to compensate. A 115v 3kw appliance will have a resistance 4 times lower than a 230v one, so it is designed to be 3kw only at the 115v.

Connect this 115v immersion to 230v and it would be 12kw, until it burnt out.

If the above 3kw 115v immersion was fed from a 230v/115v transformer, the secondary side would have 26 amps flowing into the immersion, and the primary side would have 13 amps, with both sides operating at 3kw. So the higher voltage side has a lower current flow. This is what can cause the confusion you are possibly having.

That example is correct since you refer to a device which does not provide a steady output power and only delivers that stated power at nominal voltage. I have not once implied that such equipment in your examples impedance changes, but my point is that not all devices/items behave in this manner. Therefore you can not state that Wildefalcons statement is completely nonsense, but in truth his statement holds validity.

When refering to the opposing force to electron flow in a conductor in an ac circuit it is correct to use the term impedance since impedance describes magnitude and phase.

You quoted "An element has practicaly no reactance at 50hz, so resistance is good enough for elements. What is the PF of an immersion element? I`d say its 1. If its 1, then its impedence = its resistance."

Not true, as resistance can not define the angular component.

My opinion now is that we should stop the debate here as its not really helping the OP. If you wish to continue maybe we should start a new thread in science or something?

Bruthal

sincere113 wrote: »

That is wrong....There are items whos impedance changes to deliver a steady output power.

Any examples of household appliances that do this? Or just keep repeating the sentence?

That example is correct since you refer to a device which does not provide a steady output power and only delivers that stated power at nominal voltage. I have not once implied that such equipment in your examples impedance changes, but my point is that not all devices/items behave in this manner. Therefore you can not state that Wildefalcons statement is completely nonsense, but in truth his statement holds validity.

I can, as all he told us was how to calculate power in a given circuit, this power alters in a circuit as the voltage alters, a factor you seem to disagree with, simply because a voltage regulator can be incorporated. But they are not in the supply line of general household appliances, are they?

The power in a circuit in domestic situations changes with voltage changes, so his reply to a post asking how a dropping voltage will damage sockets is not correct in context to the OP problem, but you said he was correct, simply because he states ohms law correctly. Now thanks to you he probably believes plugging a 230v 2kw kettle into 115v will still be 2kw but at double the current. Which is not correct.

So he seemed to be suggesting the current would increase for a voltage reduction, i said it wont for the same load, you said he was correct . Now you say my example was correct. I think you should read the entire thread again, more carefully.

When refering to the opposing force to electron flow in a conductor in an ac circuit it is correct to use the term impedance since impedance describes magnitude and phase.

Magnitude and phase? Id say you read all this recently, as thats like textbook terminology there.

Again, you brought this up, even though its of little relevance here. In a resistive circuit, current and voltage are in phase. So 230v AC or DC would give the same amps. PF = 1.

There is no reactance in a resistive circuit. So in a resistive circuit, impedence = resistance. You also dispute this, so you surely are reading this stuff, but dont understand it.

You quoted "An element has practicaly no reactance at 50hz, so resistance is good enough for elements. What is the PF of an immersion element? I`d say its 1. If its 1, then its impedence = its resistance."

Not true, as resistance can not define the angular component.

I still dont know why you brought it up. Im fully aware of what impedence is compared to resistance. Again, a purely resistive heating element such as an immersion will be completely in phase. Any reactance will be so negligible that it can be disregarded.

Here, read this, it might help you. Good examples of the impedence and resistance in a purely resistive load. The immersion element i used is a resistive load, and its very safe to treat it as a resistance load, and so using a heating element as an example, i used resistance. When taking about a motor, i used impedence. So your simply nit picking over terminology. I could of course have used impedence as the term, but you picking up on that is pedantic.

My opinion now is that we should stop the debate here as its not really helping the OP

You dragged it off by nit picking at irrevalent stuff.

Now, any chance of some examples of everyday appliances which will give a constant power output with different voltages in a house? Im guessing not.

What about the hoover mentioned by the OP.
Washing machine
Dishwasher
Cooker
Electric shower
Fridge
Tumble dryer

An induction motor can change its impedence due to changing slip (load or voltage changes). A universal one can change it due to forced slowing of the rotor due to more load, from the generated back EMF reducing as the rotor slows. More useless info for you to save you googling that too.

But are you telling us the above appliances will all output a steady power if the supply voltage drops from 230v to 220v, or vice versa?

If the ESB voltage went up to 250v from 230v for a household over a bill period, the bill would be higher for the exact same appliances used for the same times. Can you explain this for us? O wait, you said power changing with voltage is incorrect, so that cant happen of course:pac:

sincere113

I give up with you..........:o

2011

M cebee wrote: »

2011 and robbie are the biggest troublemakers:D

Can't argue there, but in this case I agree with Robbie 100%

If a 230V 50W GU10 lamp has 110V applied to it it will no longer consume 50W.

The power consumption is only constant if the voltage is.

That is Ohm's Law, not mine

Remember Ohm's Law can be expressed as:

P=(V^2)/R

Therfore if R is fixed (in the case of the GU10 it is) power is proportional to voltage.

Bruthal

2011 wrote: »

Can't argue there, but in this case I agree with Robbie 100%

If a 230V 50W GU10 lamp has 110V applied to it it will no longer consume 50W.

The power consumption is only constant if the voltage is.

That is Ohm's Law, not mine

Remember Ohm's Law can be expressed as:

P=(V^2)/R

Therfore if R is fixed (in the case of the GU10 it is) power is proportional to voltage.

What causes the confusion is, people see the P = V x I, and take from that that a 2kw load will still be 2kw even if plugged into a lower voltage, and so they assume the current increases. They are forgetting the lower voltage designed 2kw load has a lower resistance/impedence, to make it a 2kw load at the lower voltage.

Thats how it seemed to me with wildefalcons post. So i posted #30 to give my view on this, and try to clear that possible confusion.

Bruthal

sincere113 wrote: »

I give up with you..........:o

Why not simply give us examples of domestic appliances which have regulators in their power supplys then? Thats where you claim im wrong to say lower voltage will mean lower watts on appliances. If im wrong about that, give us the examples then. That seems simple enough. Maybe you can check my spelling too:)

sincere113

Most modern uP controlled appliances make use of constant power drives IC's like this one. http://www.maxim-ic.com/app-notes/index.mvp/id/4523 (I probably could have found better examples, but you can see that +Vcc is 8-24v and delivers a constant 20W output.


These type of circuits make use of an I (current) sense resistor. This I sense resitor/circuit is placed in series with the load. The voltage across the load measured and thus power output is calculated and controlled. The current sourced is controlled along with the voltage dropped thus controlling the power delivered.


5 examples;-

1) modern uP controlled microwave oven.
2) modern uP controlled washing machine.
3) modern uP controlled tuble dryer.
4) Modern uP controlled power washers.
5) Domestic/Industrial Application - Modern electronic lighting ballast. (Not the old magnetic/mechanical wire wound ones). These deliver x output power to the lamp. Changes in the supply voltage do not effect the required power to the lamp. If it did the lamp would not deliver the correct light output and the light level would be up and down.

I think that you're thinking more about electro/mechanical devices rather than modern electronic equipment.

PS making comment like "you're talking rubbish" isn't very constructive and not really in the spirit of the forum IMHO. You challenged me to give you examples, well there you go.....

Bruthal

sincere113 wrote: »

I think that you're thinking more about electro/mechanical devices rather than modern electronic equipment.

No, im thinking of the appliances that are in use in houses now.

The current flowing out of a socket wont increase when the voltage to the house decreases. Thats what i said. You said i was wrong when i posted a post in reference to that.

Thats fine. Im wrong then. The current does increase with a reduction in voltage so, because P = V x I says it does. Have a good day, as im rushing for a night shift now:mad:

2011

Sincere113, lets keep this simple.
The OP has an issue with socket outlets melting.

Lets assume that some devices plugged in are drawing a higher current than they should as you are suggesting.

The plugs used by the OP will be fused at a maximum of 13A. This BS1362 fuse limits the level of current that can pass through the plug/socket arrangement.

If devices were drawing a current that was in excess of the maximum design current of the sockets or plugs then the fuse would fail regardless of the reason for the increased current.

The reason for the sockets melting is simple and was stated very early in the tread:
There is a high resistance connection at the socket and/or plug.

They are not melting because of "uP controlled appliances". They are melting because of heat (I^2) R

Back to Ohm's law again!

sincere113

2011 wrote: »

Lets assume that some devices plugged in are drawing a higher current than they should as you are suggesting.

The plugs used by the OP will be fused at a maximum of 13A. This BS1362 fuse limits the level of current that can pass through the plug/socket arrangement.

If devices were drawing a current that was in excess of the maximum design current of the sockets or plugs then the fuse would fail regardless of the reason for the increased current.

The reason for the sockets melting is simple and was stated very early in the tread:
There is a high resistance connection at the socket and/or plug.

They are not melting because of "uP controlled appliances". They are melting because of heat (I^2) R

Back to Ohm's law again!

I agree with all you quoted apart from the bit in bold, I never said something was drawing a larger current than it should. That implies that something draws more current than it was designed for. I never said that. Plus I remember saying earlier in the thread that if the voltage drops enough the unit drops out to avoid overheating in the particular device/product (more current more heat) leading to thermal run away and destruction. Like you said if the current drawn exceeds the rating of the fuse then the fuse will fail, thats fairly simple.....

I also remember saying that this has very little to do with the OP....

All I am saying is that robbie7730 was wrong to say that Wildefalcons statement is completely wrong;-

Originally Posted by wildefalcon
If the voltage drops, and the wattage remains the same, then the current (amps) increases.

The debate between my and Robbie7730 has nothing to do with the OP but rather the above statement. Thats why I suggested stopping further debate here and moving it somewhere else......

It may be because I'm tired but it does seem as though your looking to add stuff to the thread for the sake of getting a response from me and turning it into another argument.

2011

Originally Posted by sincere113
I never said something was drawing a larger current than it should.

Fair enough.
In that case do you agree with wildefalcon's statement below?

wildefalcon wrote: »

If the voltage drops, and the wattage remains the same, then the current (amps) increases.

To me that statement suggests that if the voltage were to decrease then the current would increase. In Ireland most domestic appliances around the home are designed to use a 230V supply (RMS at 50Hz). When operating at this voltage they would draw a given current. Any current above this 230V current is a current in excess of the design value. This was what I meant by "drawing a larger current than it should". My impression was that you were agreeing with wildefalcon's statement above. Please confirm if

Generally a reduction in voltage will result in a corresponding reduction in current. Robbie's point reflected this and he pointed this out so that the OP would be put on the correct path to fixing his melting sockets issue.

There are some obscure (for a domestic installation) contradictions to this such as switched mode power supplies but they are generally units that do not consume very much power in the first place and therefore would not be lightly to be drawing a current high enough to melt a socket.

sincere113

My belief is this;-

1) Some products that have a power rating at a particular voltage draw a particular current. This is usually stated on the product eg. 2KW heater @ 230Vac. So for your 230V you consume 2KW.

2) If the device is not designed to maintain a certain power level to a component or certain part of the product then the current reduces proportionally as the supply voltage reduces as correctly stated by Robbie. e.g. heater....

3) A contradiction to this is certain modern household appliances that make use of 'power sensing' circuits to maintain delivery of a power level to a component or certain part of the product.

Do you agree thus far?

Now, if we consider P = IV then if P stays the same and V drops then I increases for some products. Right? This is what Wildefalcon stated. if the voltage drops and the wattage stays the same then the current must increase.

It is not correct to apply the thought that when the voltage drops then the current drops for all products.

I know for a fact that 3 certain types of common house hold appliances, microwave ovens, electric power washers and power showers make use of the approach explained in 3 above.

In a round about way I think that we agree and it boils down to the interpretation of Wildefalcons post. Here's why.

Wildefalcon states that "If the voltage drops, and the wattage remains the same, then the current (amps) increases."

Now I interpret the "and the wattage remains the same" to be the P in P=IV. I think that you and robbie assume the wattage he refers to is the power rating of a product. Since this could be the case, thats why I refer to the equation P=IV and not P (rating) = IV. The stated power rating becomes meaningless once you go beyond or below nominal voltage for certain products but not for all! For some products as Robbie correctly pointed out the impedance of the product does not change, and so the P is proportional to the voltage supply. But when the input impedance to the device changes in order to be able to deliver a steady output power, P remains almost a constant. So we are now back to P=IV.

I assume with Robbie earlier that he didn't think that a house hold product could change its input impedance and thats why he pushed for examples.

2011

sincere113 wrote: »

My belief is this;-

1) Some products that have a power rating at a particular voltage draw a particular current. This is usually stated on the product eg. 2KW heater @ 230Vac. So for your 230V you consume 2KW.

2) If the device is not designed to maintain a certain power level to a component or certain part of the product then the current reduces proportionally as the supply voltage reduces as correctly stated by Robbie. e.g. heater....

3) A contradiction to this is certain modern household appliances that make use of 'power sensing' circuits to maintain delivery of a power level to a component or certain part of the product.

Do you agree thus far?

Yes, although the contradiction is not as common and is generally for low power consumption units such as laptops (around 60W).

Now, if we consider P = IV then if P stays the same and V drops then I increases for some products. Right?

Yes in some cases very low currents may increase but will remain very low.

This is what Wildefalcon stated. if the voltage drops and the wattage stays the same then the current must increase.

I dont agree. What he stated suggests that the current will increase if the voltage decreases so that the power consumption will remain the same as though this was the norm, he made no mention of voltage sensing:

If the voltage drops, and the wattage remains the same, then the current (amps) increases.

Originally Posted by sincere113 It is not correct to apply the thought that when the voltage drops then the current drops for all products.

I agree. But in the context of multiple melting sockets in a domestic installation it is quite a safe bet that the problem is not caused by an increase in current caused by under voltage. Would you not agree?

I know for a fact that 3 certain types of common house hold appliances, microwave ovens, electric power washers and power showers make use of the approach explained in 3 above.

I won't argue, but I assume that you will agree that if they drew a current above the max design current of the socket the fuse should blow rather than the socket melting?

Wildefalcon states that "If the voltage drops, and the wattage remains the same, then the current (amps) increases."

Now I interpret the "and the wattage remains the same" to be the P in P=IV.

I don't. Because of other statements made that I disagree with like this:

It is highly likely that the wiring is dangerous.

Come on, the sockets were melting! Would you agree with this statement?

I think that you and robbie assume the wattage he refers to is the power rating of a product.

I take the rating to be what is written on the data sheet or the nameplate of the unit.

Since this could be the case, thats why I refer to the equation P=IV and not P (rating) = IV.

That formula is incorrect for AC voltage unless the power factor is =1
S = VI



The power should be taken as the manufacture states it is if you are carrying out a calculation.

Power= (V^2)/R
or
Power=V x I x Cosθ
or
Power= (I^2)R

The stated power rating becomes meaningless once you go beyond or below nominal voltage for certain products but not for all!

See above

For some products as Robbie correctly pointed out the impedance of the product does not change, and so the P is proportional to the voltage supply.

For every device I can think of in a typical domestic installation that draws a current large enough to melt a socket with a bad connection the resistance is a fixed value. The impedance is irrelevant as it is the resistive component of impedance that causes the heating effect i.e. heat generated = (I^2)R and not (I^2)Z

That is why domestic installations are charged for kW hours and not kVA hours.
There are financial penalties for bad power factor on larger installations, but thats another story...

But when the input impedance to the device changes in order to be able to deliver a steady output power, P remains almost a constant. So we are now back to P=IV.

See above
Except that power is only VI in DC or when the power factor is equal to 1

I assume with Robbie earlier that he didn't think that a house hold product could change its input impedance and thats why he pushed for examples.

I would think that both Robbie and I are of the view that Wildefalcon felt that some limited cases of low powered devices were the norm and that somehow under voltage could cause sockets to melt. This and other statements made were misleading in my opinion and take from the correct answer which is that the most lightly cause of the problem is a bad connection at the socket (and/or plug) and it could be dangerous. Call an electrician OP, should be an easy fix.