Question

a 5.6m wire at 20C a 2.8m section has a diameter of .65 mm and the other portion has a 1.5mm diameter. the current is 5.5A if the resistivity is 1.7x10^8 ohm m at 20 C what is E in each section? what is the potential drop across the wire?

Answer 1

I get 2.9 X 10 ^(-4) vm^-1 for the .65mm section and 5.3x10^(-5) vm^-1 for the 1.5mm section I get 9.61x10^(-4) V for the potential drop.

Answer 2

I used resistivity = E/J and J = I/A which gives me \[\frac{ \rho i }{ A } = \left| E \right|\]

Answer 3

E = phi/A

:O what is that forumla? :O
use Ohms law!

Answer 4

how would you use ohms law V=IR to find the e field in the different sections of the wire?

Answer 5

\[V= IR \ or \\ J = \sigma E\] ;-)

Answer 6

so is it not ok to use \[\rho I /A \] which is really just \[J = \sigma E \]

Answer 7

I only stuck with those because I was given rho

Answer 8

ohhh.. it was that

i thought you wrote rho into I .. . i thought it was phi.. was wondering what phi was.. lol xD

Answer 9

ok so lemme check ur answers!

Answer 10

oh lol ok

Answer 11

thank you

Answer 12

wow.. I am getting some absurdly high answer \[v = iR = i \frac{\rho l}{A} = 5.5 \times \frac {1.7x10^8 \times 2.8}{\pi \times ( \frac {0.65 \times 10^-3}{2})^2}\]
that gives me a ridiculously high number!

Answer 13

is it 10 power 8 or 10 power -8 ??
your resisitivity?

Answer 14

10 ^ (-8)

Answer 15

sorry about that I missed the - earlier ><

Answer 16

ok.. so plug in.. those values.. i get 0.79 V

Answer 17

@sseebeck hello? :P

Answer 18

is that your result for voltage drop over the entire length of the wire?

Answer 19

no only the first section! 2.8m long!

Answer 20

Im trying to figure out why my numbers are so far off from yours

Answer 21

oh ok I have 2.9 x 10 ^ ( -4 ) v/m

Answer 22

that is still wrong.. maybe you plugged in wrong!

Answer 23

its 0.65 mm.. don't forget to convert that to meters!

Answer 24

hmm I have \[E = \rho i / A = ((1.7\times 10^{-8}) (5.5\times10^{-3}))/ (3.22\times10^{-7}\]

Answer 25

why 5.5 X 10^-3?
its 5.5 A.. not mili A xD

Answer 26

ahh there I was supposed to be 5.5 m A

Answer 27

* it

Answer 28

i'm apparently getting tired

Answer 29

you .. you missy.. you need spectacles :-/
then your answers are right
you just need to convert the field to potential difference thats it!

Answer 30

I just did Ea*La+Eb*Lb = Va-Vb = V

Answer 31

and you are right I do need glasses! I was worried about this half of the problem as the professor wrote (this is a difficult problem) at the end which the way I did it it wasn't difficult at all just a tiny bit of arithmetic.

Answer 32

exactly lol.. its so easy peezy lemon squeezy. you nailed it..!

Answer 33

oh excellent thanks so much for the help!

Answer 34

No problem!