Rear End Power Socket

[davewill]

Recently changed the fuse as the rear power socket wasn't working. After this it was fine, and i used it a few times with a small portable compressor to inflate the tyres on the kids bike.

 

Next time i went to use it, there was no life from it again?

 

Any thoughts as to why it has blown a fuse again

 

I have used the same compressor from the front fag lighter socket may times, with no problems at all

[RADIOTWO]

Recently changed the fuse as the rear power socket wasn't working. After this it was fine, and i used it a few times with a small portable compressor to inflate the tyres on the kids bike.

 

Next time i went to use it, there was no life from it again?

 

Any thoughts as to why it has blown a fuse again

 

I have used the same compressor from the front fag lighter socket may times, with no problems at all

 

Dave

 

There could be a number of reasons !

 

did you replace with the correct size fuse ? what size fuse does the hand book say, also what current does the compressor draw, as poss as the cable

length might mean it draws more at the back than the front.

 

Another thing there is some cheap fuse's going round and may blow sooner.

 

There could be a intermitiant fault on the compressor, if it does not blow with anything else

 

Radiotwo

[NikpV]

if the cable length to to the back is greater then the total resistance of the circuit would be bigger so the current drawn would be smaller :) so less likely to blow the fuse!!

 

Any chance that there is anything in the pocket above the sockets that may be shorting out the sockets at the back and so blowing the fuse - do you keep the compressor in there?

[RADIOTWO]

if the cable length to to the back is greater then the total resistance of the circuit would be bigger so the current drawn would be smaller :16: so less likely to blow the fuse!!

 

Any chance that there is anything in the pocket above the sockets that may be shorting out the sockets at the back and so blowing the fuse - do you keep the compressor in there?

 

If the cable length is longer (with the same size of cable) then there will be a greater volts drop, and the lower the voltage the higher the current, so more likly to blow the fuse !

 

Radiotwo

[JRWR]

If the cable length is longer (with the same size of cable) then there will be a greater volts drop, and the lower the voltage the higher the current, so more likly to blow the fuse !

Some good suggestions on things to check - have you checked that the fuse has actually blown again?

 

NikpV is correct that the longer the cable then the higher the resistance and so the lower the current. Ohm's law, V=IR, is alive and kicking. :16:

[davewill]

Hi Guys

 

Thanks for the replies

 

I did use the correct ampage fuse as stated in the Handbook. For the rear power socket it is a a 15A fuse Number 8 in the central fuse box. It is interesting that the Cigar lighter fuse in the car is a 25A fuse, but it does say in the handbook that this also controls the "Alternator and Controller".

 

The cable on the portable compressor from the cigar lighter adapter to the unit is about 2m (6ft)

 

So with that info do you think it is the cable length that could be blowing the fuse?

 

Cheers

[JRWR]

For the rear power socket it is a a 15A fuse ...

The compressor should be marked with its current requirements - quite possibly it's close to or over 15A. :16:

 

So with that info do you think it is the cable length that could be blowing the fuse?

No, the longer wires will (slightly) increase the total resitance of the circuit and reduce the current flowing in the fuse.

[insider]

So with that info do you think it is the cable length that could be blowing the fuse?

It's more likely to be a high current spike when the compressor is switched on or off. I've had fuses blow front and back while using portable compressors.

[NikpV]

Sorry that was an incomplete explanation :D I was being lazy (no comments from those wags who know my profession thank you :huh: )

 

This is the situation here .... as a current I flows through a wire with resistance r the potential difference is used up... lost volts so that the voltage appearing at the end of the cable is the original voltage - lost volts.

 

So assuming a perfect battery (no internal resistance) at 12V and a wire with a resistance of 1 Ohm(unrealistically high) an appliance drawing 2 amps the lost volts would be 2 x 1 = 2V so measuring the voltage at the end of the cable (socket) while appliance was working you would see 10V. To find the cuuent (in a simple way) you would divid the EMF (voltage at te battery with no current being drawn) by the total resistance so I=12/(internal r of battery + r of the cable +r of appliance).

 

Assuming the only thing changing is the length of the cable ... as the cable gets longer r increases and so 12/(r+r+r) gets smaller so the current i gets smaller so less likely to blow fuse.

 

As radiotwo says for the same power, using a lower voltage would need higher current P=IxV but this is more of a design constraint e.g. a 23W bulb on the mains would need about 0.1A but in a car at 12 it would need a current of 23/12 ... approximately 2A so in this case lower voltage would lead to higher current

[davewill]

Sorry that was an incomplete explanation :D I was being lazy (no comments from those wags who know my profession thank you :huh: )

 

This is the situation here .... as a current I flows through a wire with resistance r the potential difference is used up... lost volts so that the voltage appearing at the end of the cable is the original voltage - lost volts.

 

So assuming a perfect battery (no internal resistance) at 12V and a wire with a resistance of 1 Ohm(unrealistically high) an appliance drawing 2 amps the lost volts would be 2 x 1 = 2V so measuring the voltage at the end of the cable (socket) while appliance was working you would see 10V. To find the cuuent (in a simple way) you would divid the EMF (voltage at te battery with no current being drawn) by the total resistance so I=12/(internal r of battery + r of the cable +r of appliance).

 

Assuming the only thing changing is the length of the cable ... as the cable gets longer r increases and so 12/(r+r+r) gets smaller so the current i gets smaller so less likely to blow fuse.

 

As radiotwo says for the same power, using a lower voltage would need higher current P=IxV but this is more of a design constraint e.g. a 23W bulb on the mains would need about 0.1A but in a car at 12 it would need a current of 23/12 ... approximately 2A so in this case lower voltage would lead to higher current

 

Checked the compressor and its a 10A rating so to blow a 15A fuse i am assuming there is a spike?

I dont keep it in the car, and there is nothing else in there that would interfere with the rear end socket

I will change the fuse again at the weekend and see if that does the trick this time

Thanks all

Dave

[big_kev]

I have used the same compressor from the front fag lighter socket may times, with no problems at all

 

The front fag lighter has a much bigger fuse than the rear ones.

[davewill]

I have used the same compressor from the front fag lighter socket may times, with no problems at all

 

The front fag lighter has a much bigger fuse than the rear ones.

 

 

Yeah i know Kev

 

15A in the rear and 25A for the front cigar socket

[RADIOTWO]

Sorry that was an incomplete explanation :D I was being lazy (no comments from those wags who know my profession thank you :huh: )

 

This is the situation here .... as a current I flows through a wire with resistance r the potential difference is used up... lost volts so that the voltage appearing at the end of the cable is the original voltage - lost volts.

 

So assuming a perfect battery (no internal resistance) at 12V and a wire with a resistance of 1 Ohm(unrealistically high) an appliance drawing 2 amps the lost volts would be 2 x 1 = 2V so measuring the voltage at the end of the cable (socket) while appliance was working you would see 10V. To find the cuuent (in a simple way) you would divid the EMF (voltage at te battery with no current being drawn) by the total resistance so I=12/(internal r of battery + r of the cable +r of appliance).

 

Assuming the only thing changing is the length of the cable ... as the cable gets longer r increases and so 12/(r+r+r) gets smaller so the current i gets smaller so less likely to blow fuse.

 

As radiotwo says for the same power, using a lower voltage would need higher current P=IxV but this is more of a design constraint e.g. a 23W bulb on the mains would need about 0.1A but in a car at 12 it would need a current of 23/12 ... approximately 2A so in this case lower voltage would lead to higher current

 

Hi All

 

As radiotwo says for the same power, using a lower voltage would need higher current P=IxV but this is more of a design constraint e.g. a 23W bulb on the mains would need about 0.1A but in a car at 12 it would need a current of 23/12 ... approximately 2A so in this case lower voltage would lead to higher current

 

Just to further comment on this, the way round this would be to up the size of the cable, but I bet the cable size is the same as the front socket.

and to Dave, that why the fuse is lower at the back.

 

Dave a way round it, would be to run a new length of cable direct from the Battery to the rear of the car and either use a new socket or disconnect the original one and connect the new cable, and use something like 3 or 4mm Sq and fuse with a 30amp fuse.

 

Radiotwo

[RADIOTWO]

Sorry that was an incomplete explanation :D I was being lazy (no comments from those wags who know my profession thank you :huh: )

 

This is the situation here .... as a current I flows through a wire with resistance r the potential difference is used up... lost volts so that the voltage appearing at the end of the cable is the original voltage - lost volts.

 

So assuming a perfect battery (no internal resistance) at 12V and a wire with a resistance of 1 Ohm(unrealistically high) an appliance drawing 2 amps the lost volts would be 2 x 1 = 2V so measuring the voltage at the end of the cable (socket) while appliance was working you would see 10V. To find the cuuent (in a simple way) you would divid the EMF (voltage at te battery with no current being drawn) by the total resistance so I=12/(internal r of battery + r of the cable +r of appliance).

 

Assuming the only thing changing is the length of the cable ... as the cable gets longer r increases and so 12/(r+r+r) gets smaller so the current i gets smaller so less likely to blow fuse.

 

As radiotwo says for the same power, using a lower voltage would need higher current P=IxV but this is more of a design constraint e.g. a 23W bulb on the mains would need about 0.1A but in a car at 12 it would need a current of 23/12 ... approximately 2A so in this case lower voltage would lead to higher current

 

Hi All

 

As radiotwo says for the same power, using a lower voltage would need higher current P=IxV but this is more of a design constraint e.g. a 23W bulb on the mains would need about 0.1A but in a car at 12 it would need a current of 23/12 ... approximately 2A so in this case lower voltage would lead to higher current

 

Just to further comment on this, the way round this would be to up the size of the cable, but I bet the cable size is the same as the front socket.

and to Dave, that why the fuse is lower at the back.

 

Dave a way round it, would be to run a new length of cable direct from the Battery to the rear of the car and either use a new socket or disconnect the original one and connect the new cable, and use something like 3 or 4mm Sq and fuse with a 30amp fuse.

 

Radiotwo

[big_kev]

Dave a way round it, would be to run a new length of cable direct from the Battery to the rear of the car and either use a new socket or disconnect the original one and connect the new cable, and use something like 3 or 4mm Sq and fuse with a 30amp fuse.

 

Or alternatively you could move the bikes to the front of the car and use the power socket there.

 

Seems lot less work to me :huh:

[DaveJH]

Sorry that was an incomplete explanation :D I was being lazy (no comments from those wags who know my profession thank you :huh: )

 

This is the situation here .... as a current I flows through a wire with resistance r the potential difference is used up... lost volts so that the voltage appearing at the end of the cable is the original voltage - lost volts.

 

So assuming a perfect battery (no internal resistance) at 12V and a wire with a resistance of 1 Ohm(unrealistically high) an appliance drawing 2 amps the lost volts would be 2 x 1 = 2V so measuring the voltage at the end of the cable (socket) while appliance was working you would see 10V. To find the cuuent (in a simple way) you would divid the EMF (voltage at te battery with no current being drawn) by the total resistance so I=12/(internal r of battery + r of the cable +r of appliance).

 

Assuming the only thing changing is the length of the cable ... as the cable gets longer r increases and so 12/(r+r+r) gets smaller so the current i gets smaller so less likely to blow fuse.

 

As radiotwo says for the same power, using a lower voltage would need higher current P=IxV but this is more of a design constraint e.g. a 23W bulb on the mains would need about 0.1A but in a car at 12 it would need a current of 23/12 ... approximately 2A so in this case lower voltage would lead to higher current

 

Hi All

 

As radiotwo says for the same power, using a lower voltage would need higher current P=IxV but this is more of a design constraint e.g. a 23W bulb on the mains would need about 0.1A but in a car at 12 it would need a current of 23/12 ... approximately 2A so in this case lower voltage would lead to higher current

 

Just to further comment on this, the way round this would be to up the size of the cable, but I bet the cable size is the same as the front socket.

and to Dave, that why the fuse is lower at the back.

 

Dave a way round it, would be to run a new length of cable direct from the Battery to the rear of the car and either use a new socket or disconnect the original one and connect the new cable, and use something like 3 or 4mm Sq and fuse with a 30amp fuse.

 

Radiotwo

 

 

 

 

Radiotwo

 

Just for the record, the delivered power into a load of resistance R ohms is VxV/R, so if the votage at the front is 12V its 144/R watts, and if the voltage at the back is 10V its 100/R watts. Therfore connecting a resitive load R at the rear will draw less current ( V/R) and also result in less power in the load. The rated power of a device is the nominal power at 12v.

 

What is probably happening with the compressor is that the load is both inductive and resistive, with the inductive part offering less impedance when switched on hence the current spike.

 

DaveJH

[davewill]

[

 

 

Blimey

 

I didn't realise that a portable air compressor, and a blown fuse could cause so much response

 

Thanks for all the replies, i've certainly learned a bit about Ohms and voltages

 

Cheers