Electric Circuits 8th Edition 55

Electric Circuits 8th Edition 55 - .2 E tectricat 0hm's aw...

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Unformatted text preview: ? .2 E tectricat ( 0hm's aw) R esistance L 31 Sometimes r esistor's aluew lll b e e r?rcssed sa c onductanceather a v a r than as a resistance.Usirg tlle relationship betweer resistance and conductance iveni n E q.2.5,w e m ay a lso w rire E qs.2 .9 a nd 2 .10i n l erms o f g the c onductance,r o i, (2.11) l?.12) Equations2 .G2.12providea v adety o fmethodsf or c alculatinghe p ower t absorbedby a resistor. Each yields the sameanswer.In analyzing a circuit, look a t t he i nformationp rovideda nd c hoose he p ower e quationthatu ses t lbal i nlormalion irecll). d Example2 .3 i llustratest he a pplicationo f O hm,s l aw i n c oniunction with a n i deal s ourcea nd a r esistor. owerc alculations t t he t erminalso fa P a iesistor also aie illustrated. Cal.cutating V ottage, urrent, ndP ou/er a S impte esistive ircuit C a f or R C Ir e achc ircuiti n F ig.2.8,e ittrert he v alueo f z o r, t s The c urrenti , i n t he r esistorw ith a c onductance of 0 .2 S i n F ig. 2 .8(b) i s i r t he d irection o f r he voltaged rop a crosshe r esistor. hus t T I ) 1 A 1 , , ,8 o (a) 50v (b) 0.2 s rb= ( 50x0.2) 1 0A . The v oltager ). i n F ig.2.8(c)i s a d se i n r he d irection of the current in the rcsistor. Hence 0c= -(1)(20)= 20v. The c urent i d i n t he 2 5 O r esisrori n F ig.2 .8(d) is i n t he d irection o f t he v oltaged se a ffoss t he resistor. herefoie T f ) r a " .: o o (c) ( )50 v ( d) 2 5n '.1 .50 '25 b) The power dissipatedin eachof the four resistorsis tigure2 .8A T he ircuih E xampL€ c f or ? .3. a) C alculate he v alueso f 0 a nd i . t b) D eterminet h€ p ower d issipatedn e achr esistor. i r8l ( 1 ) 1 8= 8 w . ) r s o= ; = pors: ( s0),(0.2) 5 00 , w Solution a) T he v oltageo ai n F ig.2.8(a)s a d rcp i n t he d ireci tion of the curent in the resistor.Therefore, = I 7n\) =w r:on _ fr = t 1t(20)2 0 , = { 5 0 ) 2 ( 2 )12s) 1 00w. = PL'a i - ,o : ( 1X8) ...
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