Cable resistance and voltage drop

[Fiona]

A cable carries a current of 65A with a 13v drop. What must be the resistance of a cable which when connected in parallel with the first cable will reduce the voltage drop to 5v?

 

[Sharpend]

So where do you start Fiona? 
I take it this is one of you college questions? 

 

[Fiona]

Yes can u help me?

 

[Fiona]

A radial circuit feeds a 12Kw load with a two core copper cable 70.25m long.the csa of each core is 10mm squared and the load current is 40A   ( pcopper=1.78×10 to the power of - 8 ohms m.   Calculate the total cable resistance.     The total voltage drop in the cable.   The supply voltage at the load.  The supply voltage.  Power loss in the cable total power supply?

 

[Sharpend]

Help we can, do it for you we can’t. 
 

so where do you think you need to start? 

 

[ProDave]

Hint. Try starting by working out the resistance of the existing cable.

 

[Evans Electric]

Always remembering the Regs  referring to  "Parallel feeders"  .    Wink wink 

 

[Bruspark]

Show us your solution or even the approach you intend to take   

 

[Doc Hudson]

I have merged your two questions together as they very much inter-relate with each other. As has been said the forum is not a free homework answering service. (possibly some members may do the work for you if you made a bank transfer to cover their time first). But generally you will need to show the question, what you think the answer is with any relevant workings-out, to get the best support and guidance from the forum.

Doc H.

 

[Sharpend]

More importantly Fiona, don’t be embarrassed or intimidated as we all had to start here ourselves at one point. 
better you learn than be spoon fed and have no understanding. 

 

[binky]

A cable carries a current of 65A with a 13v drop. What must be the resistance of a cable which when connected in parallel with the first cable will reduce the voltage drop to 5v?

so from the V I R  triangle (of whatever they call is these days) R = V/I 

 

[Bruspark]

so from the V I R  triangle (of whatever they call is these days) R = V/I 

You'll not pass if you stop there lad  . .  

Find Current cable resistance as above
Calculate required (smaller) total cable resistance for 5V  rather than 13V drop  ( pro-rata will do it)
Solve parallel resistors formula for a resistance needed in parallel with current resistance to give the required total

That's your lot !!

Although in the real world the current won't stay the same as the voltage changes but hey !!
 

 

[TonyMitchell]

Let me have a go at this one, although I’m a little confused by the 12kW load and specified with a load current of 40A, since as U0=230V, I would have thought Ib would be 52A, not 40A ?

Using Ib=40A
(Table 4D2B) 10mm2 Two core single phase AC voltage drop = 4.4mV/A/m
Voltage Drop = (4.4*40*70.25)/1000=12.4V

Power Loss = (12.4*40)=496W

(GN3 Table B1) 10mm2 R1 1.83mΩ/m
R1 = (1.83*70.25)/1000=0.13Ω 

Reduce voltage drop to 5A overall by using a parallel conductor:
13V/5V=2.6 ratio
(10*2.6=26)-10=16mm2 parallel conductor required

16mm2 conductor resistance 1.15mΩ/m

Parallel conductor resistance (1.15*70.25)/1000=0.08Ω

 

[binky]

You'll not pass if you stop there lad  . .  

Find Current cable resistance as above
Calculate required (smaller) total cable resistance for 5V  rather than 13V drop  ( pro-rata will do it)
Solve parallel resistors formula for a resistance needed in parallel with current resistance to give the required total

That's your lot !!

Although in the real world the current won't stay the same as the voltage changes but hey !!
 

it was a starter for 10!  :^O

 

[Bruspark]

it was a starter for 10!  :^O

Ok. 10 marks then