60. (a) The power consumed by the light bulb is P = I2R/2. So we must let Pmax/Pmin= (I/Imin)2= 5, or IIZZZZRLRmmminmax//.FHGIKJ=FHGIKJ=FHGIKJ=+FHGGIKJJ=2222225εωbgWe solve for Lmax: LR/..===×−22 12010002600764 1022VWHzH.bgbgπ(b) Yes, one could use a variable resistor. (c) Now we must let RRR,+FHGIKJ=bulbbulb25 or ..
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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.