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"#
