What amp fuse for a fridge freezer

[Ihavenojob]

My brother is having trouble with electrics in his kitchen, and he's found the plug on the fridge freezer is burnt. I told him it's probably a loose neutral, so he's put another plug on, but with a 3amp fuse. I told him to put a 13amp one in due to current drawn when the motor/compressor kicks in. Am I correct or is 3amp ok?

 

[CBW]

Ohms law for wattage (power) https://www.electricalrebuilders.org/testing/ohms_law.php#:~:text=States%20the%20relationships%20between%20current,resistance%20(ohms)%20and%20voltage.&text=Watt's%20Law%3A,current%20(amps)%20and%20voltage.

 

[EFLImpudence]

My brother is having trouble with electrics in his kitchen, and he's found the plug on the fridge freezer is burnt. I told him it's probably a loose neutral,

Yes, likely a loose wire or loose fuse holder.

so he's put another plug on, but with a 3amp fuse. I told him to put a 13amp one in due to current drawn when the motor/compressor kicks in. Am I correct or is 3amp ok?

It doesn't really matter. 13A will be fine.

 

[JohnW2]

Ohms law for wattage (power) .... <link to Ohms Law and Watts Law calculator.>

For a start, to be a little pedantic, and as I'm sure you know, although 'wattage' can be calculated from any two of voltage, current and resistance, there is really no such thing as "Ohm's Law for wattage".

That aside, and more to the point, to calculate current from power and known voltage (which your calculator can do) requires one to know what figure for 'power' to use. As the OP clearly knows, for something with a significant motor, the 'instantaneous power' (instantaneous Voltage x Current) during 'start-up will be considerably higher than the 'running current' reported for the appliance, but will rarely, if ever be stated anywhere for a domestic appliance.

I don't know what the start-up current for a domestic fridge-freezer is likely to be, and it probably doesn't persist for long enough to blow a 3A fuse, but it could well be considerably greater than 3A, so the OP's question seems perfectly reasonable.

I would probably be inclined to use a 10A or 13A fuse, just to be sure' - after all, one doesn't want the freezer losing power whilst one is on a Spanish beach

Kind Regards, John

 

[ericmark]

I agree with @JohnW2 5 or 10 amp seems the norm for a non inverter drive refrigeration unit, but 3 amp is ample for inverter type.

So since we don't know if inverter drive or not, either answer could be right, but normally in the hand book.

 

[SUNRAY]

My non inverter F/F came with a 13A fuse but the power rating is not in the manual/paperwork (they only quote the annual consumption). At the time I replaced with a 5A as it is a thin flex.

 

[fixitflav]

For a start, to be a little pedantic, and as I'm sure you know, although 'wattage' can be calculated from any two of voltage, current and resistance, there is really no such thing as "Ohm's Law for wattage".

In any case, Ohm's law does not apply to an electric motor. Eg for a given power output, increased voltage gives smaller current, as can be checked in a motor catalogue. Opposite to what Ohm says.

I would probably be inclined to use a 10A or 13A fuse

I have a Bosch fridge freezer, handbook says use a 10 amp or higher fuse.

 

[EFLImpudence]

I have a Kunft, it just shares a 16A MCB.

 

[JohnW2]

My non inverter F/F came with a 13A fuse but the power rating is not in the manual/paperwork (they only quote the annual consumption). At the time I replaced with a 5A as it is a thin flex.

Fair enough. As I said, even if there was a 'power rating' (are you sure there isn't one on the machine's 'rating plate') it would relate to 'running power', and would therefore not help in telling you what the start-up current might be.

However, as I said, I doubt that the start-up current of a F/F, even if 'very high' would persist for long enough to blow a fuse of any credible size (e.g. 3A).

Kind Regards, John

 

[JohnW2]

In any case, Ohm's law does not apply to an electric motor.

Mr Ohm probably only knew (and thought) about what we would call resistive loads in DC circuits, but his Law will still work for a non-restive load in an AC circuit if one utilises the appropriate (instantenaeous, if necessary) impedance in place of 'resistance'.

Eg for a given power output, increased voltage gives smaller current, as can be checked in a motor catalogue. Opposite to what Ohm says.

I'm not quite sure what you are suggesting here. Ohm's Law certainly tells us that, for an unchanging load ('resistance'/impedance) increasing voltage will result in an increase in current (and hence an increase in power)..

However, you totally move the goalposts by adding "for a given power". In any circuit, even if only a simple resistor, if one constrains power to remain unchanged, when voltage increases current has to decrease (i.e. the resistance/impedance of the load must increase).

Kind Regards, John

 

[fixitflav]

Mr Ohm probably only knew (and thought) about what we would call resistive loads in DC circuits, but his Law will still work for a non-restive load in an AC circuit if one utilises the appropriate (instantenaeous, if necessary) impedance in place of 'resistance'.


I'm not quite sure what you are suggesting here. Ohm's Law certainly tells us that, for an unchanging load ('resistance'/impedance) increasing voltage will result in an increase in current (and hence an increase in power)..

However, you totally move the goalposts by adding "for a given power". In any circuit, even if only a simple resistor, if one constrains power to remain unchanged, when voltage increases current has to decrease (i.e. the resistance/impedance of the load must increase).

Kind Regards, John

I'm talking about an electric motor, 3-phase induction to be specific. If it's supplying a given load, and the voltage is varied over the allowable range, the current varies inversely. The load can be kept constant, any tendency to change can be corrected by in principle adjusting the coupling between motor and load. Change in PF is negligible over allowable voltage range.
I have an ABB catalogue showing this for voltages 380, 400 and 415. I believe motors are designed to work on nominal volts +/- 5%, but I haven't been into it for some time.

 

[JohnW2]

I'm talking about an electric motor, 3-phase induction to be specific. If it's supplying a given load, and the voltage is varied over the allowable range, the current varies inversely.

I'm still rather confused. I agree that, to achieve a given load (hence same electrical power, if efficiency remains unchanged), if one varies voltage, then" current would vary inversely" - that's what Watt's Law tells us, but you previously suggested that it was "the opposite of what Ohm says" (I presume you meant Watt).

What am I missing?

Kind Regards, John

 

[SUNRAY]

Fair enough. As I said, even if there was a 'power rating' (are you sure there isn't one on the machine's 'rating plate')

I'd be surprised if it is not on the plate but I'm not pulling the unit out to look

it would relate to 'running power', and would therefore not help in telling you what the start-up current might be.

However, as I said, I doubt that the start-up current of a F/F, even if 'very high' would persist for long enough to blow a fuse of any credible size (e.g. 3A).

Kind Regards, John

The original paper with the unit tells me 5 times the annual consumption is 291KWh. For BS88 fuses I usually set them 4-5 times the running current of a motor and will make the guess I did the same with the new fridge so ~230 - 280W would fit that, I'll make the guess it's probably <200W though.

 

[JohnW2]

I'd be surprised if it is not on the plate but I'm not pulling the unit out to look

Understandable!

The original paper with the unit tells me 5 times the annual consumption is 291KWh. For BS88 fuses I usually set them 4-5 times the running current of a motor and will make the guess I did the same with the new fridge so ~230 - 280W would fit that, I'll make the guess it's probably <200W though.

I must say that I usually struggle to reconcile these quoted "annual consumption" figures with the apparent reality!

Kind Regards, John

 

[SUNRAY]

I'm still rather confused. I agree that, to achieve a given load (hence same electrical power, if efficiency remains unchanged), if one varies voltage, then" current would vary inversely" - that's what Watt's Law tells us, but you previously suggested that it was "the opposite of what Ohm says" (I presume you meant Watt).

What am I missing?

Kind Regards, John

One has to look at the motors rating plate to see the running current at different voltages, they don't all follow the same power curves.