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2009Chp31Soln

# 2009Chp31Soln - 8.8 mm A r π-× = = = × = 31.70 Solve The...

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31.68. Solve: (a) The charge delivered is ( 29 ( 29 3 6 50 10 A 50 10 s 2.5 C - × × = . (b) (Extra Credit) The current in the lightning rod and the potential drop across it are related by Equation 31.22. Using ρ for iron from Table 31.2, ( 29 ( 29 ( 29 8 3 4 2 9.7 10 m 5.0 m 50 10 A 2.43 10 m 100 V A LI I V A L V ρ ρ - - × × = = = = × This is the area required for a maximum voltage drop of 100 V. The corresponding diameter of the lightning rod is 4 2 3 2.43 10 m 8.8 10 m 8.8 mm A r
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Unformatted text preview: 8.8 mm A r π--× = = = × = 31.70. Solve: The total charge delivered by the battery is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 6 hours 6 hours 0 s 0 s 4 0.75 A 0.75 A 6 hours 0.75 A 6 3600 s 1.62 10 C t t Q dQ I dt e e ∞ ∞ ∞-- = = = =- = × = × ∫ ∫ ∫ The number of electrons transported is 4 23 19 1.62 10 C 1.01 10 1.6 10 C-× = × ×...
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