Question

Please write/explain every step in detail. Be careful of decimals/rounding. Please help asap.

Find Vx in the network shown in the given figure, where P = 4250° V. Please report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 180 degrees. j1Ω 2 Ω 2 Ω + V. x P(+ j4 Ω j4 Ω -j1Ω 2/0° A The value of Vx = Z * V.

Answer 1

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Please find the answer attached as under. Please give a Thumbs UP if you find the answer useful. Have a nice day Ahead:)

tve " Igini entering solenoid It is leaving solenoid sough dot Il using 2 VX 2 mor KUL in loop I:) 4250°= 2 I, + j4 I, j Iz 44250 a) 4480 = (2+84) I1 -j Iz jų ull 34 d. 220 Ij = 42.50 +y Iz + Vyd 2154 2 -j 4250 2 14 94 In 22-90 usung KVL in loop Din using current to voltage source (2-j+94). I2 +22-90-jIj = 0 transformation A 220 ; 220xd = 2.2590 Ba Bo 2) (2 +3%) Iz + 22-90-3 >> (2+31) 11 + Tz 3/4250 ) = 0 4250+; 12 24 j4 -22-90 + j4250 2+ j4 27 14 z) 2.1 + 2.8j) I2 0.2 + 2.86j I 0.6916 + 0.4444 0.822 L32.72 Vx =2I2 = 2x0.822 232.72 2 1.644 4.32.728

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