Question

09. 06) U 06 (12 p.) The parameters in the following circuit àre: Ri= 0.06 Xi = 0.4 R2 = 0.05 X2=0.4 Xm= 300 V2 = 1200 V 12 = 500 with a pf of 0.866 Ry oor 1x 4 V lagging. www . Determine the input voltage, V1, the input current, I1, and the input power factor. Rc = 2000 50 R2.05 zw X2 4. + I v. 120 Eritjxm dimensionless! R2 + X₂ // Reti Xu) VitI,RitI , JX, + Is Ret 15xm to TL Retox Vit I, 6.06)+1(.4) +T (2000) +,(,4): Ret SXM) | Ret 1X2) 0=1200-500l64) - sool. OG) - Zwols

Answer 1

Given circuit diagram is WWW Given information is R = 0.06 x =0.4 R = 2000 2 R = 0.05 x = 0.4 Xom = 3002 = 1200 V Dy = 500 with cosø = 0.866 = 3686° If you write loop Equation ; we get - vx + I (R₂ + jx) + ₂ = 0 x = I (R+jx2) + 2 = (5.00 136.94°) ( 0.057 jo-4) + 1200 L0° = (500 (36-86") ( 0.403 8287) + 1200 Leº - (201.5 119.73) + 1200 L = - 99.956 + j 174 9 + 1200 110004 +j174.9 = 1113.86 19-03 Now; = ( 1 ) = 1/13 | 7-3 3 300 = 3.72 19.03 113.66 19-03 = 0.556 19.03 w = 19.03 1 Re) 2000 = Iw - jau = [0.556 19.03) - j( 3-712 19.03 ) = (0.566 1965 - 3+2 199.05 )

= 0.549 + 30-087 +0.582 - ) 3.665 1131 - 33.578 = (1.131 – 33-575) + 500/36 86 = (1.181 – 13.575) + (400-05 †j 299.93) T = 401181 +j 296.35 = 498.76 36.410 Simillary write loop Eq? (2) -V + , (Ri+jx) + x = 0 v = vx + I (Rit jx) = 13.86 19:03 + [6498.76 /36.410)(0.404 181.49) = 113.86 1903 + (201.49 117.8+) -94.1 + 317811 + 1100.05 tj 174.82 1005 95 tj 362.93 10 66 06 19.3 y = 1066. 06 1 19.3 Input :0 2 power factor coso, (...) angle between , &C cose = cos (19.3 - cosø, = 0.955 36.410) lag.
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