HW 4s - HW 4 Solutions 2.26 If i15= i0 = 2A, then v = 16x2...

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Unformatted text preview: HW 4 Solutions 2.26 If i15= i0 = 2A, then v = 16x2 = 32 V i8=§=4A, i4=§=8A, 1,: ix =z'2 +i4+i8 +i16 =16+8+4+2=30 A P=Zi2R =162x2+82x4+42x8+22x16 =96O W or P = ixv = 30x32 = 960 W 2.28 Design a problem, using Fig. 2.92, to help other students better understand series and parallel circuits. Problem Find v1, v2, and v3 in the circuit in Fig. 2.92. Solution We first combine the two resistors in parallel 15||10 = 6 Q We now apply voltage division, 14 = 40:23 V1 14+6( ) — v-v- 6 (40)*12V 2 3 14+6 Hence, v1 = 28 V, v; = 12 VI v5 = 12 V 2.29 Req = 1 + 1//(1 + 1//2) = 1 + 1//(1+ 2/3) =1+ 1//5/3 = 1.625 O 2.35 Combining the versions in parallel: 70||30= 70X30=21Q, 20115: 20X5=4Q 100 25 i= 50 =2A 21+4 vi=21i=42V,vo=4i=8V i1: fl: 0.6A=i2= 12: 0.4A 70 20 At node a: KCL must be. satisfied i1=i2 +Io—+ 0.6=0.4+Io —’Io=0_2A Hence v0 =fl and Io = 0.2A ...
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This note was uploaded on 02/03/2010 for the course ECE 201 taught by Professor C during the Spring '10 term at Miami University.

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HW 4s - HW 4 Solutions 2.26 If i15= i0 = 2A, then v = 16x2...

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