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Electric Circuits 8th Edition 85

# Electric Circuits 8th Edition 85 - ofthe cjrcuit shown jn...

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Applying Series-Parall.et simpLifi cation Find i,. rr . aod i2 in lhe ci-rcuil shownin Fig.3.s. Solution We begin by noting that the 3 f,) resistor is in senes wirh rbe6 Q resistor.we Lherelore replace lhis series combination with a 9 O resistor, reducing the circuit to the one shown in Fig. 3.10(a). We now can replace the parallel combhation of the 9 O and 18 O resis- tors with a single resistance of (18 x 9)/(18 + 9), or 6 O. Egure 3.i0(b) shows t]Iis turtler reduction of the circuit.The nodes x and y marked on all diagmms facilitate tracing through t]le reduction of the circuit. From Fig. 3.10(b) you can veiify that iJ equals 120/10, or 12 A. Figure3.11 shows the iesult at this point in the analysis. We added the voltage ?)1 to help cladfy the subsequent discussion. Using Ohm's law we compute the value of ,1: 3.2 Relisto6in Panttet + ttl:18o ',{ 6 0 v Figur€ 3.9 a Th€ crrcujt for E,dmph 3.1. 4f) l20v r,l\$rso r,,f y G) 4 0 x 120V +\ '" 6 0 v (b) Figue 3,10 a
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Unformatted text preview: ofthe cjrcuit shown jn Fjg.3.9. +\ 12A v figuru 3.r1 A lhe circuit of Fig. 3.10(b) showins the numedcaL 61 3() 4 0 t20 v 9f) 1\ = (12)(6) - '72v. (3.16) But t 1 is the voltage drop fmm node x to node y, so we can return to the circuit shown in Fig. 3.10(a) and again use Ohm's law to calculate tr and ir. Thus, (3.17) (3.18) 120 V 18 72- 1 8 : 4 A ' : 8 A . 4f) 1)1 :72 9 9 6() We have found t}le thrce specified currents by using series-parallel reductions in combination with Ohm's law' Belore leaving Example 3.1, we suggest that you take the tlme to show that the solution satisfies Kirchloffs curent law at every node and Kirchloff's voltage law around every closed path. (Note that there are thee closed paths that can be tested.) Showing t]Iat the power delivered by the voltage souce equals the total power dissipated in the rcsistors also is informative. fsee Prcblems 3.3 and 3.4.)...
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