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Safety Issue - Moulded 13A Plugs


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If there is resistance (there’s always resistance, I mean appreciable resistance, unacceptable resistance, let’s not get too picky) between the pins of the plug and the springy fingers of the socket, a few insertions & withdrawals will sort it out, unless the spring fingers are actually knackered. You can also clean the pins of the plug. Don’t try to clean the socket, unless you’re sure it’s not live!

 

In the OP, it seems likely that one or more of the connections between the appliance flex and the pins within the overmoulded plug were unacceptably poor, and these would be the source of a high resistance element within the circuit. Probably a bad crimp or a poor riveted joint.

 

Doesn’t need to be much - as pointed out in an earlier post, the heating is I^2 x R, so a resistance of 0.1 ohm at 13A would give you some 17W. (The current & heating effect would actually be slightly lower, given the marginal extra resistance - 0.5%) Given that this heat source is enclosed within a chunk of plastic, not the greatest thermal conductor, it’s pretty obvious that it’s going to get pretty hot, pretty quickly. (If in doubt, hold a lit car brake light bulb, 21W, it’ll burn your fingers quite quickly)

 

Add the fact that the resistance increases as the temperature increases, and the clamp force of any rivet or crimp is probably going to reduce as the rivet expands due to the heat, and you’re into a vicious circle. It gets warm, then it gets hot, then it gets hotter, and it continues until it fails or it gets turned off.

 

And I don’t think I’m exaggerating to say that pretty much every electrical fire is pretty much due to poor contact, getting worse as it gets hotter. Some actually arc, due to a combination of vibration, and looseness. That sets fire to anything flammable in close proximity very quickly.

 

So if you’re doing a bit of wiring, make your connections well, and make sure you do the screws up tight!

 

Happy 2019

Simon

A good analogy the 21 Watt bulb, as obviously a properly designed lens holder for it, doesn't get hot enough to melt the plastic, due to the air gap. If you were to replace the bulb with a 100 Watt one, you'd probably burn a hole through it!

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A heating element (however well it responds to heat) is still, as far as I am aware at its core, nothing more than a carefully selected coil of wire.

 

 

It's a carefully selected coil of wire that has constant resistance at any temperature, but please keep going. If you can disprove Ohm's Law that would be a really big deal.

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It's a carefully selected coil of wire that has constant resistance at any temperature, but please keep going. If you can disprove Ohm's Law that would be a really big deal.

Well a coil of wire does vary in resistance, according to temperature, but so minute as to not be worth considering in a heater coil.

 

https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity#Temperature_dependence

 

Obviously it doesn't change Ohm's Law.

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Phil

 

Since you wanted to know where I, and many others, think you’re wrong, I’ll have one more go...

 

You said “A resistance heating element IS a constant power device - its is fundamentally just a coil of wire and unless you introduce other electrical components to regulate it (i.e. a thermostat which alters the overall resistance and regulates current flow) or change its conductive proprieties by altering its molecular makeup (e.g. heating / cooling it) that wire will always pass a certain unchanging electrical current.”

 

Taking the first and last assertions;

 

“A resistance heating element IS a constant power device”.

“that wire will always pass a certain unchanging electrical current“

 

Firstly, for the purposes of the discussion, the resistance heating element is a constant (at a given temperature) resistance.

Secondly, you didn’t say “the voltage is constant”. I did. The National Grid spends a considerable effort in maintaining it within strict limits, though, it may vary a bit through the day, and the available voltage may well be lower at the end of a long extension lead, due to the resistance of that lead.

Thirdly, the current that the wire passes depends only upon the resistance and the voltage. I = V/R

 

If the voltage is reduced, the current will be lower and the power, being the product of the two, will be lower. P = V x I.

 

Therefore, a resistor is not a “constant power device”.

QED

 

Herewith some numbers:

 

the declared voltage in the UK is now 230V AC +10% to -6%, so from 216 to 253V.

Assume the heater is actually putting out the rated 3kW at nominal 230V

 

The current will be 3000 / 230 = 13.04A

The resistance is 230 / 13.04 = 17.66 Ohms. This is fixed by the length, thickness and composition of the wire, and does not change for a given temperature.

 

If the voltage increases to 253V, the current will increase to 253 / 17.66 = 14.34 A (ie, 10% more) and the power will now be 3630W

 

If the voltage drops to 216 V, the current will be 12.25A and the power will be only 2646W.

 

Not even close to constant...

 

Happy New Year

 

Simon

Edited by Simond
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Phil

 

There’s nothing wrong with your equations, or the transpositions of them.

 

Where you’ve got hold of the wrong end of the stick is in assuming that a heater is a constant power device.

 

It almost certainly isn’t.

 

While a constant power heater, one incorporating sophisticated self-regulation, is theoretically conceivable, it is vanishingly unlikely that anyone would go to the expense of making one to heat a caravan.

 

The thing that is, for all practical purposes, fixed is the resistance of the heater, because it consists of wire that is selected to have a very low change of resistance with temperature, something like nichrome.

 

The source voltage is best represented as a constant voltage with a resistance to represent ‘source impedance’ in series with it, and a further resistance representing the cabling between source and our heater. On a caravan site with low voltage distribution, the source impedance and cable impedance are likely to be quite substantial.

 

I’ve mixed resistance and impedance in the above, because I’m using common terminology ........ if we need to get into power factor, we can, but I honestly think to do so would obscure the basic point that a heater is, almost certainly, a constant resistance for all practical purposes.

 

Kevin

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It's hard to be sure without looking at subjects, but I suspect oxidation has a lot to do with it. It might be best to simply have the outlets replaced and have the electrician install new plugs on the cables. I'd be surprised if that does not solve the problem (although I am surprised quite often).

 

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

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Phil, think of this as a series DC circuit.  You can ignore the reluctance and capacatance, and it being 1 ph AC, because this is still a purely resistive load.... (therefore the phase angle difference = 0...)

 

The physics of this in no way match your statement:

"Ohms law states that if the voltage goes down then the current must go up if the resistance stays the same."

 

That's backwards- if the voltage drops, then the current MUST drop, because the formula is V=I*R, where I= current in Amps, R= Resistance in Ohm's, V= Voltage.  If you decrease the voltage, then the current MUST drop if you have not changed the resistance. 

 

220V=9.09A*24.2 Ohms  (2000w)

 

Change 

200V= A*24.2 Ohms

 

200/24.2=A

A=8.26

 

Since the suggestion was that the resistance had _increased_, then the current must _decrease_ because the voltage is constant.  That's the whole relationship in a nutshell...and why more than I are getting upset about this.  

 

-------------------

To put a bunch more math into it

 

So, you have an "infinitely" strong source, that produces exactly 220V RMS AC.  This is then connected via wires (with resistance) to a socket (with resistance), through a plug (with resistance) and through more wire (with resistance) to the load, and back.

 

Lets put some values in here- again, these are fairly arbitrary.  Lets say .1 ohm, .1 ohm, .1 ohm, .1 ohm, and since we want 2 kw,  ((V*A=W,   2000w/220v=9.09 amps , so 220=9.09*R   R=24 ohms).  Now, our circuit looks like this:  .4+24+.4 ohms series resistance, or a total of 24.8 ohms.  At this point, the actual demand will be 220V/24.8 ohms= 8.87 A, (i^2*R=W, 8.87^2*24.8= 1951W total power)  and the heater will be putting out (I^2R=W)  (8.87^2*24)=1888w, and about (8.87^2*0.8= 63w used to heat the wires & plugs.

 

If you change the incoming supply voltage, then the current MUST change, because that's what Ohm's law says.  So, if we drop the voltage by 20v at the supply, we end up with 200/24.8=8.06A flowing, or a total of (200*8.06=1613w real power consumed in the circuit.  Nothing can change this, as long as Ohm's law is true.  The % of where the losses will be the same as you have only changed the I in the equations.

 

What David was saying happens, and what logic suggests is happening based on evidence, is that in fact, the voltage source remained constant, and the resistance at the plug:socket increased.  Lets put that to be .4 ohms, with nothing else changing-

 

.1+.4+.1+24+.1+.4+.1 ohms, means that 220/25.2=8.73 A will flow, allowing (8.73^2*25.2=1920W to flow).  HOWEVER, the heater is only going to put out (8.73^2*24=1829 W of heat), and the plug:socket is going to put out (8.73^2*0.8=60W) with the cables putting out 30 watts (8.73^2*.4=30.5w).  The plug now gets 2x the energy in as before, with the same surface area.  AKA it is going to get hot rather than warm...possibly even melty hot.

 

This is all fundamental circuit analysis, and should be something anyone should be able to do if they passed electrotech 100, or its equal.  I'm just a Marine Engineer Artificer, not a Marine Electrician, and I am very confident in what I am saying.

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Where did I mention a stable voltage source?.

 

Here?

If something has a resistance that doesn't change and it is supplied with an unchanging voltage and an unchanging current then it uses a fixed and constant number of Watts (Power).

 

Trying to explain it a different way without getting into detailed maths: we came into this with a suggestion that a heater might be connected to an undervoltage supply at a campsite.  Therefore the voltage has changed and the heater cannot be assumed to be constant power, just constant resistance.  Ohm's Law states V=IR, so if R is constant and V reduces then I must also reduce. 

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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.

 

 

Phil,

 

I'm no expert in electrical matters, but I think the point that you seem to be missing is that a heater that is rated at 2000W only puts out that power at the specified operating voltage.  That is, it is a nominal figure derived by multiplying the specified operating voltage by the current that that the heating coil (a simple resistor) will allow to flow at that specified operating voltage.  The heater will effectively have a fixed resistance and as others have pointed out I = V/R, which means that the current will drop if the voltage drops.  As you have highlighted P = V * A, which means that if both the voltage and the current drop then the power output of the heater will drop.  The important point being that it is the resistance that is fixed, not the stated power output of the heater.

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For anyone who has molded plugs that feel warm.

 

https://youtu.be/GVy6LcqjZLU?list=PLVsHvs2Suqmoq4luzXWYr_zS4NiiRnUx5

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.

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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 problem with traditional fuses is that they can end up running above their rated maximum capacity provided the increase is gradual enough to not overheat the fuse wire to breaking point (boiling frog syndrome). Put a sudden extra large flow of current across one on the other hand and it can pop (hence the use of slow to blow fuses in applications with high inrush currents). An appliance fitted with a 3 Amp fuse in the plug will likely still work if its pulling 4A - providing the initial inrush current when it gets turned on is low enough).

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Phil, think of this as a series DC circuit.  You can ignore the reluctance and capacatance, and it being 1 ph AC, because this is still a purely resistive load.... (therefore the phase angle difference = 0...)

 

The physics of this in no way match your statement:

"Ohms law states that if the voltage goes down then the current must go up if the resistance stays the same."

 

That's backwards- if the voltage drops, then the current MUST drop, because the formula is V=I*R, where I= current in Amps, R= Resistance in Ohm's, V= Voltage.  If you decrease the voltage, then the current MUST drop if you have not changed the resistance. 

 

220V=9.09A*24.2 Ohms  (2000w)

 

Change 

200V= A*24.2 Ohms

 

200/24.2=A

A=8.26

 

Since the suggestion was that the resistance had _increased_, then the current must _decrease_ because the voltage is constant.  That's the whole relationship in a nutshell...and why more than I are getting upset about this.  

 

 

 

 

If the voltage is reduced, the current will be lower and the power, being the product of the two, will be lower. P = V x I.

 

 

 

 

Then how do you explain this experiment.

 

Take a labatory power supply and set the current limit to say 2A and the voltage limit to 20V. This supply can only output a maximum of 40 Watts of power. Now connect a 50W heater and gradually turn it up to the maximum. Once that heater tries to pull more than 40Watts of power analysis of the electricity will show that the voltage falls towards zero while the current drawn by the heater increases.

 

The laws of physics have been respected - The maximum power drawn by the heater remains 40W (the maximum you are letting the power supply output), the resistance of the heater has not changed but the voltage and current have altered to keep all the other variables balanced.

 

This is why a typical short circuit track circuit failure will show a higher than usual current being drawn at the feed end and a low track voltage (reaching zero close to the point of failure) occurs. Too much current is being drawn and the power supply (or to be more accurate the isolating transformer in the feed side) has maxed out its VA rating.

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Erm you might like to rethink this 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.

 

This is why conductor rail electrification is so inefficient - the low voltage means high currents and extra meaty connections are needed.

 

If your RV heater is trying to consume XX Watts of power and the campsite voltage is low (120V vice 240V) , then more current will be pulled which could exceed the design limits of the plug - unless the RV features a power monitoring system that puts an absolute limit on the current drawn at any voltage.

 

 

Then how do you explain this experiment.

 

Take a labatory power supply and set the current limit to say 2A and the voltage limit to 20V. This supply can only output a maximum of 40 Watts of power. Now connect a 50W heater and gradually turn it up to the maximum. Once that heater tries to pull more than 40Watts of power analysis of the electricity will show that the voltage falls towards zero while the current drawn by the heater increases.

 

The laws of physics have been respected - The maximum power drawn by the heater remains 40W (the maximum you are letting the power supply output), the resistance of the heater has not changed but the voltage and current have altered to keep all the other variables balanced.

 

This is why a typical short circuit track circuit failure will show a higher than usual current being drawn at the feed end and a low track voltage (reaching zero close to the point of failure) occurs. Too much current is being drawn and the power supply (or to be more accurate the isolating transformer in the feed side) has maxed out its VA rating.

 

There is no mention of Voltage or Current Limit in your original post; introducing these conditions is a different ball game.

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There is no mention of Voltage or Current Limit in your original post; introducing these conditions is a different ball game.

 

No, but every power supply has a limit as to what it can supply - even the national grid!

 

On another thread we are told that the GWML electrification is being taken to Chippenham as Thingley Junction is the only place NR can tap into a sufficiently high power National Grid power line in the area.

 

A campsite is no different - it will have a finite limit as to how any amps and volts it can distribute (a function of the VA rating of the site transformer). If too much current tries to be drawn locally then the voltage will drop as a consequence.

 

Granted I may have incorrectly applied this principle to a RV heater, but a valid principle it remains when discussing electrical theory.

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Phil

 

Every source has a what is known as ‘source impedance’, so is usually represented as a firmly fixed voltage source, with a resistance/impedance in series.

 

A small source, like a lab supply transformer or a track-circuit transformer, or a small battery, has a very high source impedance, so that as current drawn increases, the voltage at the output falls steeply - a high proportion of the notionally fixed source voltage is dropped across the source impedance, rather than the load.

 

A large source, the output of a grid substation perhaps, has a small source impedance, so the output voltage falls far less steeply as current draw increases.

 

The effective source impedance is inherent to the design of the source and doesn’t involve ‘active’ components, but it can be used to effectively limit short-circuit current output from the source ....... conversely, failing to understand it properly can cause incorrect selection of protection (fuses etc), allowing short circuits to be supplied without interruption.

 

Kevin

Edited by Nearholmer
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Erm you might like to rethink this 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.

 

This is why conductor rail electrification is so inefficient - the low voltage means high currents and extra meaty connections are needed.

 

If your RV heater is trying to consume XX Watts of power and the campsite voltage is low (120V vice 240V) , then more current will be pulled which could exceed the design limits of the plug - unless the RV features a power monitoring system that puts an absolute limit on the current drawn at any voltage.

 

Using your Laboratory Power Supply set to 24V with a 12R Resistor across the Output what is the Current Flow?

 

Using the same Power Supply set to 12V with the Same 12R Resistor connected to the Output what is the Current Flow?

 

Do your answers confirm your theory that "Ohms law states that if the voltage goes down then the current must go up if the resistance stays the same."

 

EDITED to clarify whose theory it is for the above quote.

Edited by Pannier Tank
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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.

 

A plug top fuse is there to protect the lead between the socket and the appliance. Nothing more.

 

It can't protect the plug in the OPs case as the problem is a thermal one, not excessive current.

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Using your Laboratory Power Supply set to 24V with a 12R Resistor across the Output what is the Current Flow?

 

Using the same Power Supply set to 12V with the Same 12R Resistor connected to the Output what is the Current Flow?

 

Does your answers confirm "Ohms law states that if the voltage goes down then the current must go up if the resistance stays the same."

No.

Current drops from 2A to 1A.

 

If your theory was correct then a very low voltage would produce a very high current and a zero voltage woukd produce an infinite current.

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Then how do you explain this experiment.

 

Take a labatory power supply and set the current limit to say 2A and the voltage limit to 20V. This supply can only output a maximum of 40 Watts of power. Now connect a 50W heater and gradually turn it up to the maximum. Once that heater tries to pull more than 40Watts of power analysis of the electricity will show that the voltage falls towards zero while the current drawn by the heater increases.

 

The laws of physics have been respected - The maximum power drawn by the heater remains 40W (the maximum you are letting the power supply output), the resistance of the heater has not changed but the voltage and current have altered to keep all the other variables balanced.

 

 

This just proves what we are all saying, your 50W heater is not a constant power device. It never dissipates 50W in your example, only 40 or less. To be a meaningful experiment you need to specify the rated voltage of the heater at which it will dissipate 50W. It doesn't matter if you artificially limit the current (your example) or the voltage (the more realistic campsite example). 

 

The power output of a resistive heating element is determined ONLY by the voltage applied. I'll say it again, it obeys ohms law, reduce the applied voltage and the current, and hence the power, will reduce.

 

The original example of low voltage causing the plug to overheat is complete BS, unless the heater has much more complex control electronics to boost the internal voltage to it's rated value. Only in such a case would the input current increase. Think of a universal laptop power supply designed to work around the world on 120 - 240V. In the US the mains current will be double what it would be in Europe, for the same charging voltage and current. I put it to you that a simple electrical heater will not have anything so sophisticated, given the kW power levels involved.

 

Another thought experimen. Consider an incandescent light bulb.This is nothing more than a resistive heating element that is optimised to emit light at useful wavelength (almost 100 of the power is dissipated as heat). Now take a 240V 100W bulb and connect it to a variable transformer. Turn the voltage up slowly. The bulb is initially cold to the touch and dark, dissipating no power. When you reach 240V the bulb will be dissipating 100W and is working as the manufacturer intended. What happens when you continue to turn up the voltage? The bulb continues to dissipate more and more heat until it burns out.

 

No equations needed :)

 

Please, for everyones sake, just accept that the majority in this thread know what they are talking about :)

 

Happy New Year to all.

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Take a laboratory power supply and set the current limit to say 2A and the voltage limit to 20V. This supply can only output a maximum of 40 Watts of power. Now connect a 50W heater and gradually turn it up to the maximum. Once that heater tries to pull more than 40 Watts of power analysis of the electricity will show that the voltage falls towards zero while the current drawn by the heater increases.

 

The laws of physics have been respected - The maximum power drawn by the heater remains 40W (the maximum you are letting the power supply output), the resistance of the heater has not changed but the voltage and current have altered to keep all the other variables balanced.

 

Okay, I agree with this, but this has moved on from discussing the heater coil to discussing the power supply.  If the power supply is rated at 40 VA, which is the maximum power that can be supplied, then as the current drawn increases the voltage will fall.

 

This means that your 50 W heater will only be putting out 40 W of heat (constrained by the inadequate power supply).

 

The maximum output from your power supply is governed by P = V * A (as you have highlighted) but the current drawn by the heater is, I think, still based on the current required to provide the stated power at the stated voltage because the heater effectively has a fixed resistance.

 

Therefore a 50W heater at 240V would draw 0.204 Amp (I = P/V).  This would have a resistance of 1176 Ohms (R = V/I).

 

However, when you use your inadequate 40VA power supply, the resistance of the heater coil is still fixed at 1176 Ohms, but the voltage will drop below 240V because of the power supply.  What I'm not clear on is whether the current flowing from your 40VA power supply is still 0.204 Amp.

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No.

Current drops from 2A to 1A.

 

If your theory was correct then a very low voltage would produce a very high current and a zero voltage woukd produce an infinite current.

 

Just to clarify, it's not my theory but that of Phil-b259

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However, when you use your inadequate 40VA power supply, the resistance of the heater coil is still fixed at 1176 Ohms, but the voltage will drop below 240V because of the power supply.  What I'm not clear on is whether the current flowing from your 40VA power supply is still 0.204 Amp.

 

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]

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I’ve mixed resistance and impedance in the above, because I’m using common terminology ........ if we need to get into power factor, we can, but I honestly think to do so would obscure the basic point that a heater is, almost certainly, a constant resistance for all practical purposes

 

You're safe (unlike the OPs heater).

 

The same heater would work equally well on DC.

 

No need to introduce AC theory, but we could open the RMS compartment of Pandora's box :)

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Using your Laboratory Power Supply set to 24V with a 12R Resistor across the Output what is the Current Flow?

 

Using the same Power Supply set to 12V with the Same 12R Resistor connected to the Output what is the Current Flow?

 

Do your answers confirm your theory that "Ohms law states that if the voltage goes down then the current must go up if the resistance stays the same."

 

EDITED to clarify whose theory it is for the above quote.

 

 

Just to clarify, it's not my theory but that of Phil-b259

 

 

Please read Nearholmer's post (no. 66)  which explains that under certain conditions what I stated can be correct.

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