Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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*39. For the network of Fig. 183: a. Calculate 57. b. Compare 57 to half the period of the applied signal. c. Sketch v- 12 V + 0.1 μF Si R 56 k2 Vi 2 V -12 V f = 1 kHz FIG. 183 Problem 39. 39. (a) 1= RC = (56 kQ)(0.1 µF) = 5.6 ms 5T = 28 ms T (b) 57 = 28 ms » 2 1 ms = 0.5 ms, 56:1 2 (c) Positive pulse of v;: Diode “on" and v, = -2 V + 0.7 V =-1.3 V Capacitor charges to 12 V + 2 V – 0.7 V = 13.3 V Negative pulse of v;: Diode "off" and v, = -12 V – 13.3 V = -25.3 V -1.3 V 24 V -25.3 V
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Transcribed Image Text:*39. For the network of Fig. 183: a. Calculate 57. b. Compare 57 to half the period of the applied signal. c. Sketch v- 12 V + 0.1 μF Si R 56 k2 Vi 2 V -12 V f = 1 kHz FIG. 183 Problem 39. 39. (a) 1= RC = (56 kQ)(0.1 µF) = 5.6 ms 5T = 28 ms T (b) 57 = 28 ms » 2 1 ms = 0.5 ms, 56:1 2 (c) Positive pulse of v;: Diode “on" and v, = -2 V + 0.7 V =-1.3 V Capacitor charges to 12 V + 2 V – 0.7 V = 13.3 V Negative pulse of v;: Diode "off" and v, = -12 V – 13.3 V = -25.3 V -1.3 V 24 V -25.3 V
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