(a) The larger iis 3.41 A. (b) We use V = ε– ir= 2.00 V – i(0.500 Ω). We substitute value of iobtained in part (a) into the above formula to get V= 0.293 V. (c) The smaller iis 0.586 A. (d) The corresponding Vis 1.71 V. 110. The power delivered by the motor is P= (2.00 V)(0.500 m/s) = 1.00 W. From P= i2Rmotor
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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.