ch27-p018 - 18. (a) For each wire, Rwire = L/A where A =...

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18. (a) For each wire, R wire = ρ L/A where A = π r 2 . Consequently, we have R wire = (1.69 × 10 8 m Ω⋅ )(0.200 m)/ π (0.00100 m) 2 = 0.0011 . The total resistive load on the battery is therefore tot R = 2 R wire + R = 2( 0.0011 Ω) + 6.00 Ω = 6.0022 . Dividing this into the battery emf gives the current tot 12.0 V 1.9993 A 6.0022 i R ε == = . The voltage across the R = 6.00
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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