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E=60/20 [x2=122 from the circuit total impedarice ₂ will be Z= Jizel (9.1 j122) Tie, impedance of Parallel branch 1 [z=jan) impedance of parallel brancha (22=91-312-). How = (512) (9.1-512) 144+j1092 = 156+jom 2 +9.1-312 (31) 156.40 IZ = 15602 We have Admittance Y = £ 3 → Y F156 = 6:41x16 o How total custent (13) e GO 130 0.385 30 Ampa .385 38 AM 1560 => 1s=0.385 Les ampo. Current through Parallel branch (22) i.e. Iz using current dividing rule Zitz2 = 0.385 230x Ji2+f:1-512) -0.385 236 x 112 = -0.25130:44 11 I2 = 0.5/120 Ampel =
(d) Now Vc = Voltage Across Capacitor Ve= Jate -bosbeéx (-12) 23-0-5428, *cz- -5.2 +j3 volts L 122 Ve= 6 230 volts.
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I C 124-90 e) Total power delivered to the Netwolk :- PEVI 360 Z38 x 0.285/38 P = 11.55+320 : VA Average power P = S(1.55) 560 T Paverage. = 23 walts.
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