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circuits_hw6SolutionsS

# circuits_hw6SolutionsS - 911A” The equivalent circuit is...

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Unformatted text preview: 911A” The equivalent circuit is: 100 x 103 4 W4+4 =33.33 With the switch open, we have: e, R V: 50mV= 4R 310%”ar (2) Dividing the respective Jr"sides of Equation (2) by those of Equation (1) we obtain: 0.5 = III 12;” +10“ Solving, we that that Rm =1MQ. ”1.16 For The amplifiers in The order A-B The equivalent circui‘l' is:R R, = “ =3kﬂ RD: 93 = 2m ’1‘“: —4OCJ40¢R Ry +320” 100(500 —__°6 = )106 +400 = 4933me For The amplifiers cascaded in the order S—A, we have: P11 .22* The two 15-V sources deliver power: e=(15v)><(1A)=15W g =(15V_)><(2A)=30W On the other hand, the 5-V source absorbs power: 1"; =(5V)x(—1A)=—5W Thus, The new power supplied to th ' ' - 2 am I : 3=ﬁ+€+g=4ow Plfler'ls H126" The equivalent circuit is: Wehuve: I! = mg, + g.) = 454.5,m ms t: = LR.- =9-090 "WW R, _ I9=As¢L “Ra-FRI. _.o_909Arms p; =g:,=4.545Vrms AI=IJL=EOOO 4=KM=500 \$4440" 3 = (HF/R: =4.131#W F: J.5:ﬁg"=12x2=24w a=g+g_g=19.37w q =:_oﬂoo%=17.2% (VLF/Rt =4-131W ll ...
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