# Fa06Soln9 - Physics 7B Fall, 2006 Homework 9 Solutions 25.7...

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Physics 7B Fall, 2006 Homework 9 Solutions 25.7 The rate at which electrons leave the battery is the current: I = V / R = [(9.0 V)/(1.6 Ω )](60 s/min)/(1.60 × 10 –19 C/electron) = 2.1 × 10 21 electron/min . 25.10 We find the potential difference across the bird’s feet from V = IR = (2500 A)(2.5 × 10 –5 Ω /m)(4.0 × 10 –2 m) = 2.5 × 10 –3 V . 25.21 The dependence of the resistance on the dimensions is R = ρ L / A . When we form the ratio for the two conditions, we get R 2 / R 1 = ( L 2 / L 1 )( A 1 / A 2 ) = ( ! )( ! ) = ( , so R 2 = ( R 1 . 25.39 The required current to deliver the power is I = P / V , and the wasted power (thermal losses in the wires) is P loss = I 2 R . For the two conditions we have I 1 = (520 kW)/(12 kV) = 43.3 A; P loss1 = (43.3 A) 2 (3.0 Ω )(10 –3 kW/W) = 5.63 kW; I 2 = (520 kW)/(50 kV) = 10.4 A; P loss2 = (10.4 A) 2 (3.0 Ω )(10 –3 kW/W) = 0.324 kW. Thus the decrease in power loss is Δ P loss = P loss1 P loss2 = 5.63 kW – 0.324 kW = 5.3 kW . 25.64 The maximum current will produce the maximum rate of heating. We can find the resistance per meter from P / L = I 2 R / L ; 1.6 W/m = (30 A) 2 ( R / L ), which gives R / L = 1.78 × 10 –3 Ω /m. From the dependence of the resistance on the dimensions, R = L / A , we get R / L = / ( π D 2 = 4 / π D 2 ; 1.78 × 10 –3 Ω /m = 4(1.68 × 10 –8 Ω · m)/ π D 2 , which gives D = 3.5 × 10 –3 m = 3.5 mm .

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26.11 We can reduce the circuit to a single loop by successively
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## This note was uploaded on 09/09/2008 for the course PHYSICS 7B taught by Professor Packard during the Fall '08 term at University of California, Berkeley.

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Fa06Soln9 - Physics 7B Fall, 2006 Homework 9 Solutions 25.7...

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