Electric Circuits 8th Edition 64

# Electric Circuits - 40 Cncuit tements E C L t Applying hm'sL awa ndK irchhoff's awso F inda n U nknown urrent o a U se K irchhoff'sl awsa nd O hm l

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Unformatted text preview: 40 Cncuit tements E C L t Applying hm'sL awa ndK irchhoff's awso F inda n U nknown urrent o a) U se K irchhoff'sl awsa nd O hm\ l aw t o f ind i , i n the c ircuit s howni n F ig.2.18. 10() We obtain the secondequation from Kirchhoff's vollage law in combination with Ohm's law. Nolingl rom O hm c l aw l hrt , - i c I nt. a nd d ' i s a 504,w e s um t he v oltages roundt he c losedp ath - )tzov 5oa (t 120+10t,+50ir=0. a In w riting t his e quation,we a ssigned p ositive sign t o v oltage d rops i n t he c lockwise d irec tion. S olving t hese t wo e quations f or i , a nd it y ields f or ? .8. Figur€ .18A l hec ircuit E xanrpte 2 b) Tcst the solution for i, by verirying that the total power generatedequalsthe total power dissipated. t.: 3A and ir=3A. Solution a) W e b egh b y r edrawingt he c ircuit a nd a ssigning an u nknown c urrent t o t he 5 0 O r esistor a nd unknown v oltages a cross t he 1 0 O a nd 5 0 O resisto\$.F igure2 .19s howst he c ircuit.T he n odes arel abeled , h , a ndc l o r id t he d iscussion. a 1 0() , ,, in b) f te p o\$ef d irsipaled r h( 5 0 Q r e.i.lorr s p5oo= ( 3)'(50) : 4 s0 w . i The p ower d issipatedn t he l 0 O r esistori s 21oo=( 3),(10)=90W. . - ) rzo v The p ower d eliveredt o t he 1 20V s ourcei s = p121,y - 120i" = 1 20(-3) : 3 60 W jn 2 wt 2 c s Figure ,19A T he ircuithown F ig. .18, ith h€ i 1, r uiknowns 0 ., a nd r d €fined. Because, a lso i s t he c urrent i n t he 1 20V i we have two unknown currents and source, e lherefore m ust d eive t wo s imultaneous qua involving l, and 4. We obtain one of the tions equatioN by applying Kirchhoff's current law to eithernodeb o r c . S ummingt he c urrentsa tnode b and assigning a positive sign to the cunents leavingt he n ode g ives il io-6:0. The p ower d eliveredt o t he 6 A s ourcei s pra = u (6), b ut tr = 5 0t1: 1 50 V P6a = - 1s0(6) : - 900 w ' The 6 A s ource i s d elivering 9 00 W , a nd t he 120 V s ource i s a bsorbing 3 60 W . T he 1 otal power a bsorbed i s 3 60 + 4 50 + 9 0: 9 00 W Therefore,t he s olution v erifies t hat t he p ower deliverede qualst he p ower a bsorbed. ...
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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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