# Calculating current loss in a cable

Discuss Calculating current loss in a cable in the Commercial Electrical Advice area at ElectriciansForums.net

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#### valleyboy

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Hi all, new here so appologies if this is posted in the wrong section. Also, please bear with me as I am a Mechanical Engineer, not a Electrician/Electrical Engineer. Though I do understand the very basic concepts of electrical circuits and will be able to perform reasonably complex calculations.

I have a H3 automotive headlamp bulb that is part of a flood lamp setup. The bulb is 24Vdc/70W. I need to specify some cable for wiring this, our customer has specified a connector to go on the end that has 22 gauge contacts so I am restricted to 22AWG cable (I'm uncomfortable with this small gauge). Cable length must be 3m long.

I am concerned that the cable is likely to restrict enough current to degrade the performance of the lamp enough for it to be innefective (too dim). Worse still, overheat the cable and cause a fire. So I would like to calculate the current loss of the 22AWG/0.32mmsq cable over its 3m length. Cable resistance is 60ohms/km.

Based on the above information, I have calculated that the lamp will draw 2.92A when in use using the formula A=V/W, is this correct? And do I now have enough info to calculate the current loss?

Could somebody please provide a formula?

Many thanks,

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#### TFGtv

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Hi all, new here so appologies if this is posted in the wrong section. Also, please bear with me as I am a Mechanical Engineer, not a Electrician/Electrical Engineer. Though I do understand the very basic concepts of electrical circuits and will be able to perform reasonably complex calculations.

I have a H3 automotive headlamp bulb that is part of a flood lamp setup. The bulb is 24Vdc/70W. I need to specify some cable for wiring this, our customer has specified a connector to go on the end that has 22 gauge contacts so I am restricted to 22AWG cable (I'm uncomfortable with this small gauge). Cable length must be 3m long.

I am concerned that the cable is likely to restrict enough current to degrade the performance of the lamp enough for it to be innefective (too dim). Worse still, overheat the cable and cause a fire. So I would like to calculate the current loss of the 22AWG/0.32mmsq cable over its 3m length. Cable resistance is 60ohms/km.

Based on the above information, I have calculated that the lamp will draw 2.92A when in use using the formula A=V/W, is this correct? And do I now have enough info to calculate the current loss?

Could somebody please provide a formula?

Many thanks,
You seem to have got the right answer using the wrong formula! 24/70 gives 0.34!

Think 'PVC'..... The 'proper' electrical engineers here may want to arrange a lynch mob after this BUT.....

POWER (Watts) = VOLTAGE (volts) times CURRENT PVC! Arrange that into a triangle (with 'P' at the top) and even an idiot like me can transpose it easily! Personally I go for Emma Peel in some kinky boots as an aide memoire' but then I'm quite old an' rather weird!

Proper electrical engineers will tell you P= V times I ' I' being current... SO translating into YOUR 'language' actually 'A' (Current) = W/V So your 70W bulb will draw 2.92A.... 22AWG is rated at 10A? SO I think everything will be cool and lovely.... NOT as lovely as 'Emma' in that catsuit but......

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#### valleyboy

Appologies, that is how I had arrived at 2.92A................ I=E/P.

You say that 22AWG cable is rated at 10A. What cable are you talking about? Surely all cables have a different current rating?

I know what the resistance of the cable is (the cable I intend to use), 60ohms/km. Can I now calculate the current loss over a 3 meter length?

Thanks for the help!

#### TFGtv

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Sorry.... Took the 10A off a chart which I misread... Here's a better one...

American Wire Gauge table and AWG Electrical Current Load Limits

The basic current carrying capacity of a cable is defined by it's Cross Sectional Area... Other factors such as it's sheathing and configuration might change it's safe rating for a particular use but basic 'hook up' wire for electronics or automotive use you're thinking about the cross sectional area.... Not being American AWG is alien to me... But that DOES seem like an awfully thin cable for what is presumably a worklight in a truck or something?

Remember, you can invalidate the insurance on a vehicle by fitting an accessory improperly. Wiring's a favourite with the loss adjusters!

It'll be the VOLTAGE drop across the cable you probably think you want to calculate. Think of voltage as analogous to hydraulic pressure whereas current is analogous to rate of flow ... Not strictly correct but a good jumping off point.

As for the drop across the cable.... Insignificant BUT... As you insist....

60/1000 will give you the resistance of 1M of wire .06 Ohms, 3M will have a resistance of 0.18

Ohms law will tell us the resistance of the lamp... I =V/R so R= V/I 24/2.92 8.22 Ohms. Effectively the lamp and the cable are in series so... You now have a circuit here with 8.4 Ohms in circuit 24/8.4 = 2.86A flowing... .05 of an Amp less!

Ohm's law allows you to calculate the voltage drop across a given resistor if you know the current flowing through the resistor. So in this case the cable is the 'resistor.... V = 2.86 x 0.18 = 0.51v

And the Bulb V=2.86X8.22 23.5V

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