Question

A 22-gauge copper wire has a radius of 0.645 mm, is 5.00 m long, and has a resistivity ρ = 1.72 × 10-8 Ωm. The maximum current for which this wire is rated is 0.920A. What is the maximum electric potential difference that may be applied to this wire without overloading it?

Answer 1

@siddharth18

Answer 2

Let max pot diff be V

V = I * R

R = (resistivity)(length of wire(in m)) / (Area of crossection)

I = Current

Area of crossection = pi * (radius)^2

Substitute the given quantities and solve

Answer 3

Make sure you convert all quantities into SI units before substituting

Answer 4

ho can i change from mm to m
*10^3 or what ?

Answer 5

\[1 mm = 10^{-3}m\]

Answer 6

i am trying now to solve it can u give me the final answer to know if i solve correct or not

Answer 7

oK. Ans is
\[3.903 * 10^{-5}\]

Answer 8

Volts

Answer 9

are u sure i had to choose one of these answer
0.112 V
0.0605 V
0.561 V
0.242 V
0.342 V

Answer 10

sorry. It is \[6.05*10^{-2}\]

Answer 11

i.e option B. I forgot to square the radius :P

Answer 12

:D thanx very much :)

Answer 13

Welcome :)

Answer 14

i will post another one now can u try to solve it :D

Answer 15

A 1-m copper wire with a radius of 0.5 mm is stretched to a distance of 2 m. What is the fractional change in resistance ΔR/R as the wire is stretched?

Answer 16

We know that(WKT)
R = (resistivity)(length of wire(in m)) / (Area of crossection)
ΔR/R= ((change in length)*(initial Area)) / (intial length)(change in area)

Now volume of wire remains constant whle stretching.

WKT vol = area* length

(initial area)(initial length)=(final area)(final length)
(Pi)(0.5 * 10^(-3) )^2 (1) = (final area)(2)
final area = (pi) (0.25/2 * 10^(-6) ) = (pi) (0.125 * 10^(-6) )

Therefore
ΔR/R= ((change in length)*(initial Area)) / (intial length)(change in area)
= ( 1* (Pi)(0.25 * 10^(-6) ) / (1 * (pi) ( 0.5-0.125 ) * 10^ (-6)

Now simplify.
You will get
ΔR/R= 1.33

I hope I have not made silly mistakes now :P