ch31-p060 - 60. (a) The power consumed by the light bulb is...

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60. (a) The power consumed by the light bulb is P = I 2 R /2. So we must let P max / P min = ( I / I min ) 2 = 5, or I I Z Z Z Z RL R m m min max / / . F H G I K J = F H G I K J = F H G I K J = + F H G G I K J J = 22 2 2 2 2 5 ε ω b g We solve for L max : L R / . . == = × 2 2 120 1000 26 0 0 764 10 2 2 VW Hz H. bg b g π (b) Yes, one could use a variable resistor. (c) Now we must let RR R , + F H G I K J = bulb bulb 2 5 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.

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