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5 July, 01:21

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.

What is the current in the wire? Express your answer using three significant figures.

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  1. 5 July, 01:27
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    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
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