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HW5 solution

# HW5 solution - Problem ‘26 Since V2 60 V the current...

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Unformatted text preview: Problem ‘26 Since V2 : 60 V, the current through 60 Q branch is 1 A. The resistors 90 Q and 180 Q are in parallel. Their equivalent resistance is 90 X 180 Reg 2 90+180 = 60 Now 60 Q and 60 Q are in series. Thus the voltage drop across the parallel combination of 40 S2 and 120 Q is 120 V. Thus current through 40 Q resistor is 120/4023 A. Hence, I5 2 3+1 = 4A Applying KVL around the loop containing the source and 40 Q resistor, Vs—180><4—~120 = O :> V; = 840 V Power delivered by the source is given by, P5 = V; X Is 2 840 x 4 3360 W Problem‘32(a) Using voltage division, the following equations can be written7 R1+R2+Rs R1+R2 R1 + R2 +- R8 R2 R1+R2+Rs 2400 0.75 0.25 Using (1) and (3), R2 = 600 S2. Using (2), R1 = 600 Q. 1200 82. Using (1), R5 W WQIfKWWW—WWM WW \. ”W72 wm 3&9} ______ m .V W “1“,,” MM Wm. ”Wt/ﬁwmwmmmw 3M.“ wwwﬁf, , L n L ’ J». WWWWQWMEX4,1::wV/WL,‘W QM“, M_MW:‘gﬁéégéﬁéﬁgw.ngmwwmwwwmWWW.MmW,WWW.WW_.{MWWW u. m _ g ' 7 /‘ : W/ , g? “:7‘2, m ”J ~ I W: MN MW WWWiZmeMQWQéiga 7 7 €505 m ”>3 V WWWWM £3, Wilmm, ,W-aggMWMWWwésﬂﬁwig4¢3yMKV, igmwWAWMWWWAMMWWMWM ...
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