Question

Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of higher gauge. Thus the thin 24-gauge wire included in many electronics kits has a wire diameter of 0.51 mm, while a thicker 12-gauge wire used for many home audio speakers has a wire diameter of 2.05 mm. Different gauge wires have different maximum current ratings, with thicker wires being capable of safely handling larger current.
A. Consider a meter of 24-gauge copper wire. Calculate the resistance of this wire.

B. Imagine that this 1-meter length of 24-gauge copper wire were laid across the terminals of an ideal 12-volt battery. Calculate the current that would run through the wire.

C. The maximum safe current rating for 24-gauge copper wire is 2.0 Amps. What is the shortest length of this wire that could be safely placed across the ideal 12-volt battery?

Answer 1

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B )  Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

= 71.42 m

Answer 2

Answer:

Explanation:

diameter of 24 gauge wire = 0.51 mm

diameter of 12 gauge wire = 2.05 mm

A.

length of wire, l = 1 m

diameter of wire, d = 0.51 mm

radius of wire, r = 0.51 /2 = 0.255 mm

resistivity of copper, ρ = 1.7 x 10^-8 ohm metre

The formula for the resistance of wire is given by

$R=\rho \frac{l}{A}$

where, l is the length of wire and A be the area of crossection of wire

A = πr² = 3.14 x 0.255 x 10^-3 x 0.255 x 10^-3 = 2.04 x 10^-7 m²

The resistance of wire is

$R=1.7\times10^{-8}\frac{1}{2.04\times 10^{-7}}$

R = 0.083 ohm

B.

R = 0.083 ohm

V = 12 V

Let the current is I.

Use ohm's law

V = I R

12 = I x 0.083

I = 144.6 A

C.

I = 2 A

V = 12 V

Area of wire, A = 2.04 x 10^-7 m²

Resistivity of wire, ρ = 1.7 x 10^-8 ohm metre

Let the length of the wire is l.

V = I R

12 = 2 x R

R = 6 ohm

$6=1.7\times10^{-8}\frac{l}{2.04\times 10^{-7}}$

l = 72 m