PHYS
29_InstSolManual_PDF_Part6

# 29_InstSolManual_PDF_Part6 - 29-6 Chapter 29 29.26 Set Up...

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29.26. Set Up: Let the polarity of the power supply voltage as shown in the figure in the problem be positive. Let clockwise currents be positive. Solve: (a) The I versus V graph is shown in Figure 29.26a. For this orientation of the diode a positive voltage for the power supply corresponds to a positive bias for the diode. Figure 29.26 (b) The I versus V graph is shown in Figure 29.26b. Now a positive applied voltage corresponds to a negative bias for the diode. (c) For a resistor a certain magnitude of applied voltage produces the same magnitude of current, regardless of the polarity of the voltage. The graph of I versus V would be as sketched in Figure 29.26c. (d) (i) In one case a positive voltage is forward bias for the diode and produces a large positive current and in the other case a negative voltage is forward bias for the diode and produces a large negative current. In each case the cur-
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Unformatted text preview: rent is in the same direction as the polarity of the voltage. (ii) For a resistor there is no forward or reverse bias direction. 29.27. Set Up: Solve: (a) This is in the infrared. (b) This is in the infrared. 29.28. Set Up: For an ordinary metallic conductor the resistance increases as T increases. The resistance of an ordinary conductor is not zero at For a superconductor the resistance is zero below the critical temperature, has a nonzero value just above and increases as T increases. Solve: The graphs are sketched in Figures 29.28a and b. Figure 29.28 R T (K) Ordinary conductor R T (K) T c Superconductor ( a ) ( b ) T c T 5 0 K. l 5 1 1.9 m m 2 1 0.67 eV 1.14 eV 2 5 1.1 m m. l 5 hc D E 5 1 4.136 3 10 2 15 eV # s 21 3.00 3 10 8 m / s 2 0.67 eV 5 1.9 m m. D E 5 hc l I V I V I V ( a ) ( b ) ( c ) 29-6 Chapter 29...
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• Spring '07
• Shoberg
• Current, Power, Resistor, power supply, Electrical resistance

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