PHYS 102 1nd Edition Lecture 12 Outline of Last Lecture I. Electrical Energy and Power Outline of Current Lecture II. Ohm’s Law Current LectureII. Ohm’s Law A. Ohm’s Law Applications 1. A nichrome wire, with a radius of 0.791 mm, that is to be used on winding a heat coil must carry a current of 9.25 A and a voltage of 120 V. Find: a. Required resistance: R: I = V/R 9.25 = 120/R 120/9.25 = R 13 = R b. Length of the wire to wind the coil: R = (L/A) 13 = (150 X 10-8)(L)/ r213 = (150 X 10-8)(L)/ (0.791/1000)2= 7 L = 17 m 2. There are two light bulbs, one that is 25 W and the other 100 W. Which one has higher R? P = IV I = V/R Thus, P = (V/R) V. 25 Watt: P = V2/ R R= (120)2/ 25 = 576 These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.100 Watt: P = V2/ R R = (120)2/ 100 = 144 So, the 25 Watt light bulb has the higher R. Which one has the higher current? P = IV 25 Watt: I = P/V = 25/250 = 0.1 100 Watt: I = P/V = 100/250 = 0.25 So, the 100 Watt light bulb has the higher
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