Calculate the current through a 10.0-m long 22 gauge (the radius is 0.321 mm) nichrome wire if it is connected to a 12.0-V battery. (The resistivity of nichrome is 100 × 10^-⁸ Ω∙m.)

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The resistance of the wire is:$$R = ho \frac{L}{A} = ho \frac{4L}{\pi d^2}$$$$= (100 	imes 10^{-8} \Omega \cdot m) \frac{4(10.0 \ m)}{\pi (0.321 	imes 10^{-3} \ m)^2}$$$$= 0.388 \Omega$$The current through the wire is:$$I = \frac{V}{R} = \frac{12.0 \ V}{0.388 \Omega} = 30.9 \ A$$

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