1-Visitor

Question

Cable length

Forum|Forum|10 years ago
December 10, 2015
25 replies
9162 views

Hi

How would I solve the following problem on cable length

Thanks

Two cables, both with cross sectional area 5 mm², are connected in parallel across a common DC voltage of E = 36 V.

The resistivity of Cable 1 is ρ = 17 x 10^-9 Ω.m at 20 °C, while the resistivity of Cable 2 is ρ = 28 x 10^-9 Ω.m at 20 °C.

Cable 1 is 1641 mm in length.

If 44.9 % of the current passes through Cable 1, determine the length of Cable 2.

Assume the ambient temperature is 20°C.

Other

F

Fred_Kohlhepp

23-Emerald I

You can use the current sharing ratio and Ohm's law to determine the relative resistance of the two cables. They're both the same area. Resistance is made up of resistivity cross-section area (the same) and length.

21-Topaz II

23-Emerald IV

I tend need to disagree.

Note that 44.9% and 50.1% do NOT add up to 100%...You're violating Kirchoff's law.

Also note that the resistivity is (correctly) given in nΩ.m, not in nΩ/m.

Luc

21-Topaz II

18-Opal

The 0.449 and 0.551 is the current split ratio.

All other numbers are self evident from your initial question.

You can spruce it up with a better labels and definitions.

A

AlvaroDíaz

12-Amethyst

There are also a some cheat which can be do with this set of equations.

Actually, the "hide" is only for show the equations, but it not necessary.

Best regards.

Alvaro.

cable length - solve.gif

1_cable-length---solve-xmcd.zip

P

ppal_255687

18-Opal

Cable length is 1641mm not 1640mm

A

AlvaroDíaz

12-Amethyst

I loose a milimeter attaching the wire to the battery 🙂

P

ppal_255687

18-Opal

There is also a physical impossibility problem with the way this question is drafted. The cables are NOT "connected in parallel across a common DC voltage of E = 36 V." They would have to be connected across a much lesser voltage. While not mentioned in the question the sketch shown in the replies show a Zl1 where most of the voltage would be dropped (which would be a real world connection). If the cables were connected across 36V they would have to carry approximately 2.87 million Amperes. Good luck with that.

A

AlvaroDíaz

12-Amethyst

Hi. That's what mean a shortciruit. But still have some applications. For example, as lamp, with power about 250 kW for each wire, or to use the wires to cut something by heating them.

Also, a good time to remember that Ohm law isn't a true "law", it is a rule, which sometimes hold, but not always. If the wire explode, them Ohm law don't apply.

Best regards.

Alvaro.