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The load will reduce the further down the cable you go.

I get it, as the first part of the cable is still carrying current for all loads, then when you go to the next stage, you dont need to include the first load.

The overall voltage drop remains a total of all loads and lengths.

Thanks Dill
 
cheers Dillb, appreciate your help mate..Looking at what you have worked out is the voltage drop in each section which is great! the question does say the overall Voltage drop so i then went with Wallyankers method. But thanks to you both for your help has eased my stress levels for now lol
 
here is my second attempt:

Total current = 10+7+5 = lb 22A

Total metres of cable = 50+30+40 = 120m

mV/A/m value = 4.4

overall voltage drop = 4.4x22x120/1000 = 11.61 volts.
 
here is my second attempt:

Total current = 10+7+5 = lb 22A

Total metres of cable = 50+30+40 = 120m

mV/A/m value = 4.4

overall voltage drop = 4.4x22x120/1000 = 11.61 volts.
Not quite right yet:

You've got 3 voltage drops to calculate and then add then together.
1) 1st length is 50m with 22A flowing
2) 2nd length is 30m with 12A flowing
3) 3rd length is 40 with 5A flowing
 
if i were to calculate voltage drop for each section would it look something like this:

first section (50m@10A) = 4.4x10x50/1000 = 2.2 volts

second section (80m@7A) = 4.4x7x80/1000 = 2.46 volts

third section (120m@5A) = 4.4x5x120/1000 = 2.64 volts

overall voltage = 2.2+2.46+2.64 = 7.3 volts.
 
Not quite, section one is 50 meters @ 22 amps. Section 2 30 meters @ 12 amps. Section 40 meters @ 5 amps.
 
Last edited by a moderator:
yellowvanman thanks for advice, here i go again:

first section (50m@22A) = 4.4x22x50/1000 = 4.84 volts

second section (30m@12A) = 4.4x12x30/1000 = 1.58 volts

third section (40m@5A) = 4.4x5x40/1000 = 0.88 volts

overall voltage = 4.84+1.58+0.88 = 7.3 volts
 
it is the sum of the current flowing down the second section of cable, you lose 10 amps at the first point. So 22amps-10 amps= 12 amps
 
so the current halves to the second section and then drops again by the third section and would drop again the longer the run of cable, am i right?
 
It should be pointed out that in the exam there will be points to be scored along the way in this type of question for the working out, sometimes even for writing out the formula. So show them all every step, you won't lose much time and it could make a difference
 
so the current halves to the second section and then drops again by the third section and would drop again the longer the run of cable, am i right?

Not halves no, you just lose the current used at each point along the circuit.
 
thanks dill you have totally cleared this up for me, I'm feeling a bit stupid after asking so many questions but thanks for taking the time out to help me.
 
Stuart, no one minds helping out when the OP comes on and has a go at getting the answer, well most don't. It's the ones who come on and just post the question that gets everyone's backs up.
 
No such thing as a silly question. As Trev said you attempted it before asking for help. Which is alot more than some posters and one in particular do.
 
cheers fellas, have really made helped me out with this question, i do prefer to have a go so i understand where i'm going wrong and knowing i can ask questions and get some very much appreciated help on here is a massive bonus, once again THANKS A MILLION Dill and Trev :)
 

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