Feb 6, 2017

The size of wire should have area of cross section $A \ge 1.2 \times {10}^{-} 6 {m}^{2}$

#### Explanation:

Let us use Ohm's law and rated Voltage and Current of the load to calculate its resistance ${R}_{L} = \frac{120}{6} = 20 \Omega$

Let ${R}_{W}$ be resistance of requisite wire.
Total resistance$= {R}_{L} + {R}_{W}$
Current in circuit$= \frac{120}{20 + {R}_{W}} A$
Voltage drop across wire$= {R}_{W} \times \frac{120}{20 + {R}_{W}} V$

Equating to the given value we get
$5 = {R}_{W} \times \frac{120}{20 + {R}_{W}}$
$\implies \frac{24 {R}_{W}}{20 + {R}_{W}} = 1$
$\implies \left(24 {R}_{W}\right) = \left(20 + {R}_{W}\right)$
$\implies {R}_{W} = \frac{20}{23} \Omega$ .....(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

$R = \frac{\rho \times L}{A}$ ......(2)

It is also temperature dependent. Resistivity of copper at ${20}^{\circ} \text{C}$ is $1.724 \times {10}^{-} 8 \Omega m$. Equating (1) and (2) and inserting given values in SI units we get
$\frac{20}{23} = \frac{1.724 \times {10}^{-} 8 \times \left(200 \times 0.3048\right)}{A}$
$\implies A = \frac{1.724 \times {10}^{-} 8 \times \left(200 \times 0.3048\right) \times 23}{20}$
$\implies A = 1.2 \times {10}^{-} 6 {m}^{2}$

From (1) we also have
${R}_{\text{Wire"=20/23xx1000/200=4.35Omega" per 1000 ft}}$
This corresponds to U.S. wire gauge of $16 A W G$ which has a resistance of $4.016 \Omega \text{ per 1000 ft at "20^@"C}$