Question

The terminals of a 0.70 V watch battery are connected by a 110-m-long gold wire with a diameter of 0.200 mm.

Answer 1

Answer:

Current = 0.0082 A

Explanation:

We are given;

Voltage; V = 0.7 V

Diameter; d = 0.2mm radius ; r = 0.1mm = 0.0001

Length; L = 110m

This should be considered as a short circuit because of the absence of any legitimate impedance. However, in the circuit in the question, the length of the wire is very long and thus it's impedance should not be neglected

We know that, I= V/ Z

Where;

I = current

V = Voltage

Z= Impedance

However, we are considering only resistance in impedance for a long wire.

Thus,

R = ρL/A

Where;

ρ is resistivity

L is length

A is area.

Now, resistivity of gold wire at has a value of 2.44* 10^(-8) Ω⋅m

Area is πr² = π(0.0001²)

Thus resistance is

R = (2.44 x 10^(-8) x 110)/ (π(0.0001²))

R = 85.43 ohm

Now, formula for current is given as;

I= V/R

Thus, I = 0.7/85.43 = 0.0082 A