Question

Q3. Solve for vo(t) in the circuit of Fig. 3 using the Superposition 62 2H 12 cos 31 V + 4 sin 21 A 10 V Solution:

Answer 1

62 27 000 W 120083t v 4sin(24) 12 10V Take Toy Sourla deactive other sources un 2.44 oon le 10 Here W=0 Under Do Xc: Jwe ->O-C XL = Jula J XOXLO-> Sic to Von apply kun We Now Take Sowie. cord deactivate all other get [vor 100 12 Cos (34) v sourees. Current Sowee voltage Source by open cruit in phason for (2 cos (3+) 12,00 V Here 3 redfree ey short crown't

then Xe= JWC Jx3x T24 4-5-6) Xc and XL Ejwls 6 jx3x2 = 16 J6 12L09 Voe- v 62 16 12/09 Vog Voz x 1220° from noiteye devison sule (J6211(-34) J611 (14) +6 -J6x.(-14) J6 -14 16X1-142 to T6-14 X1 24.0* 24 24 + 6(16-19) X12l6 : 2481226 24.+6X Rj

Noz 24x12/09 24 tj 12 2X12 Loo 24 j 510.733-2056° Voz 10.7334-260568" => Vo2 = 10.133 608 (37-26156) face row lastly take 4 sin 27 = 4/° Here W = 2 -6) then Xca IX 2X 1 Twe -61 kan man Xca ang XL JWL J X 2 X 2 = j 4 V03 60 14 Com 11 -j6 Nos 1.40 apply kit at nude 4100 -in til tis + voz V03 6 t NOB I ус je +15) 440 = Vos (Įsts 12x44053 Vos (? + 2j – 33). 3j) - 03(2-))

No3= -48 Loº 2-j = 21.466/26.56° Vort) = 21.466 608/ 27 +26.56') Voz It) = 21.466 Sin.(2++26.56°) than Total Vo (+) voltage Volt) = Voilt) + Vog (4) + Voilt) Volt = 10 + 10.733 cos (34-26.56") + 21.466 sin ( 24+26-560) Ans

Answer 2

A 2000 kW turbine –generator of 0.85 power factor operates at the rated load. An

additional load of 300kW at 0.8 power factor is added. Determine the required kVAR

of capacitor to operate the turbine generator but keep it from being overloaded.

Answer 3