Question

A gold wire 5.77m long and of diameter 0.820mmcarries a current of 1.27A .

Part A

Find the resistance of this wire.

Answer in  ?

Part B

Find the potential difference between its ends.

Answer: in V

Answer 1

We have to find the reistance first. For finding the resistance let us use this relation

$R=\rho*l/A$
Where l=length in metre
A=Area in metre
$\rho$ =Resistivity of gold=2.2*10^-8 $\ohmm$ $\ohm$ ohm.m

so by substituting,
$R=\rho*l/A=2.2*10^-8*5.77/(3.14*(0.410*10^-^3)^2)$ =0..240ohm

b)From Ohms law, I=V/R

I=1.27A
V=?
R=0.240ohm
so V=I*R
=>V=1.27*0.240=0.304 V