Question

The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current in the wire?

Answer 1

Answer:

I = 11.26 mA

Explanation:

given,

V = 0.7 V           length = 80 m

diameter   = 0.2 mm = 0.02 cm

radius = 0.01 × 10⁻² m

$resistance = \dfarc{\rho l}{A}$

ρ for gold wire = 2.44 × 10⁻⁸ ohm-m at 20 °C

A = cross sectional area = π r² = π (0.01 × 10⁻² )²

    = 31.4× 10⁻⁹ m²

$resistance = \dfarc{2.44\times 10^{-8} \times 80}{31.4\times 10^{-9}}$

R = 62.165 Ω

$I = \dfrac{V}{R}$

$I = \dfrac{0.7}{62.165}$

I = 11.26 mA