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

I = 11.26 mA

Step-by-step 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* 10^(-8) * 80}{31.4* 10^(-9)}`

R = 62.165 Ω

`I = (V)/(R)`

`I = (0.7)/(62.165)`

I = 11.26 mA