Calculate the current through a 15.0-m long 20 gauge (having radius 0.405 mm) nichrome wire if it is connected to a 12.0-V battery. The resistivity of nichrome is 100 × 10-8 Ω ∙ m.

Given Information:  

Radius of wire = r = 0.405 mm = 0.405×10⁻³ m

Length of wire = L = 15 m

Voltage = V = 12 V

Resistivity =  ρ = 100×10⁻⁸ Ωm

Required Information:  

Current = I = ?

Current = I = 0.412 A

Explanation:  

The current flowing through the wire can be found using Ohm's law that is

V = IR

I = V/R

Where V is the voltage across the wire and R is the resistance of the wire.

The resistance of the wire is given by

R = ρL/A

Where ρ is the resistivity of the wire, L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(0.405×10⁻³)²

A = 0.515×10⁻⁶ m²

So the resistivity of the wire is

R = ρL/A

R = (100×10⁻⁸×15)/0.515×10⁻⁶

R = 29.126 Ω

Finally, the current flowing through the wire is

I = V/R

I = 12/29.126

I = 0.412 A

Therefore, the current through a 15.0-m long 20 gauge nichrome wire is 0.412 A.

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1 a

2 b

3 b

4 d

5 d

6 d

explanation: