Question #adf01

Let us use Ohm's law and rated Voltage and Current of the load to calculate its resistance `R_L=120/6=20Omega`

Let `R_W` be resistance of requisite wire.
Total resistance`=R_L+R_W`
Current in circuit`=120/(20+R_W)A`
Voltage drop across wire`=R_Wxx120/(20+R_W)V`

Equating to the given value we get
`5=R_Wxx120/(20+R_W)`
`=>(24R_W)/(20+R_W)=1`
`=>(24R_W)=(20+R_W)`
`=>R_W=20/23Omega` .....(1)

The electrical resistance of a wire is dependent on its length `L`, its area of cross section `A`, and upon the material of wire accounted through `rho` resistivity of material. Resistance of a wire can be expressed as

`R=(rhoxxL)/A` ......(2)

It is also temperature dependent. Resistivity of copper at `20^@ "C"` is `1.724xx10^-8 Omegam`. Equating (1) and (2) and inserting given values in SI units we get
`20/23=(1.724xx10^-8xx(200xx0.3048))/A`
`=>A=(1.724xx10^-8xx(200xx0.3048)xx23)/20`
`=>A=1.2xx10^-6m^2`

From (1) we also have
`R_"Wire"=20/23xx1000/200=4.35Omega" per 1000 ft"`
This corresponds to U.S. wire gauge of `16AWG` which has a resistance of `4.016Omega" per 1000 ft at "20^@"C"`