Electric Circuits 8th Edition 89

# Electric Circuits 8th Edition 89 - thc voltage divider...

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3.4 Vottage Division and Crrent Division 0bjective 3.2 ^) c) d) 2*(now how to design simple vottage-divider and Find the no-load value of?J, in the Find n, when nr is 150 kO. How muchpower is dissipated in thc 25 kO resistor ifthe load terminals are accidentally shon-circuited? What is the maximum power dissipatedjn the ?5 kO resistor? cuff€nt-divid€r circuits 3,3 a) Find the value of R that will cause 4 A of current to flow through the 80Q resistor in the circuit shown. b) How much power will lhe resistor R from part (a) nced to dissipale? c) How much power will the current source genemte lor the value of R froln pafi (a)? 20A Answ€r: (a) 30O; (b) 7680 w; (c) 33,600 W. 60() 100 It 800 NOTE: (a) 150 v; (b) 133.33 V (c) 1.6W (d) 0.3w Also tty Chaptet Pnblems 3.13,3.15, antl3.21. 3.4 tfqlltaqe fiivisisn a\$d eurrent
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Unformatted text preview: thc voltage divider cir-cuit in Fig.3.12 and tlre currentdivider circuil in Fig.3.15.The generaliza-tions will yield two addiiional and very usctul circuit analysis lechniques known as voltage division and current division. Consider the circuil shown inFig.3.18. The box on tlre left can cortain a single voltage source or any other combination of basic circuit elements that results in thc vollag ?) shownin the tigure. To the right of the box are r resistors conncctcd in series. We are interested in finding the voltage drop ol across an arbitrary resistor Rl in terms of the voltage r. We start by using Ohm's law !o calculate i, fte current through all of the resistors in serics, in lerms of the current o and n l + R 2 + - + R , , R.q ' (3.2e) figure 3.18 t Cjrcuit usd to ittustEie vottage division....
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## This note was uploaded on 07/05/2010 for the course EE 100 taught by Professor Boser during the Spring '07 term at University of California, Berkeley.

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