P28_078

# P28_078 - into the above ormula to get V = 0 . 293 V or 1 ....

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78. (a) The power delivered by the motor is P =(2 . 00 V)(0 . 500 m / s) = 1 . 00 W. From P = i 2 R motor and E = i ( r + R motor ) we then fnd i 2 r i E + P = 0 (which also ±ollows directly ±rom the conservation o± energy principle). We solve ±or i : i = E 2 4 rP 2 r = 2 . 00 V ± p (2 . 00 V) 2 4(0 . 500 Ω)(1 . 00 W) 2(0 . 500 Ω) . The answer is either 3 . 41 A or 0 . 586 A. (b) We use V = E− ir =2 . 00 V i (0 . 500 Ω). We substitute the two values o± i obtained in part (a)
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Unformatted text preview: into the above ormula to get V = 0 . 293 V or 1 . 71 V. (c) The power P delivered by the motor is the same or either solution. Since P = iV we may have a lower i and higher V or, alternatively, a lower V and higher i . One can check that the two sets o solutions or i and V above do yield the same power P = iV ....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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