Question

Find V, in the network shown in the given figure, where P = 8/50° 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 2Ω + P+ j4Ω j4Ω -j1Ω 2/0° A The value of V, = V.

Find V_x in the network shown in the given figure, where P = 8<50° 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.

Answer 1

a 9112 ca 212 22 w FI +Vx 2Loa TI (4 8/50 34 34 عمان - P Given p=8250 Both currents I, and I leaving dot's, so, it is in magnetic aiding м м + Tri I st I I, IN Vz Xor и L.M IL L-M tom Is Keite) = + Vi M3 ts V₂ MP3) L = 34, 12734 Li-M = ju-j) = 93 L2- M - 34-3) = 93 33 I Iz 93 mm 2n L I t Iz + Vre Ут 8150* -31 220° I By KCL I = 2 -Iz I = 2 Loo - Iz (226=2) By KUL left most loop 8250 - 2I, - (93) I, - (;) (I,+I2) 20 to

=) 8250 - 25, -53 I, -5 I, -9 1 Iz co =) 8250 = (2+93) I + J I equation By KVL to middle loop -631) I +(81) (+12) + (93+2) I2 =0 g II + 9 (II t Iz) + (93+2) I2 = 0 - 31 (2 - I2) + 3 (I t Iz) +93 +2) Iz ca - ŷ I + (39+2) Iz +92 => - I ; I, 92 + (39 +2) I 2 + (3 st2) Iz 지 2 + (-377 -29 ) Iz I = 2 + (3-29) Iz T Substitute in equation SL500 = 3+33) C2+3-2012) +312 (6+3) + +53)(3-25) In + SI, = (6+56) + (15 + j;) Iz + JI 7 8250 = 6 (1+3) + (15+j4) Iz

3 5.14 + 6-12 6 (143) + (15+34) IL E SLsoºs514+612) -0.86 +0.123 = (15+34) I => -0.86 + 0.129 Iz 15.494 - ) Iz -0.05+ jo.021 Iz 0.056 L157.12° ㅋ from Circuit Va 2 I2 - 2x 2 x 0.056 157-12° = 2 L180° * 0.056/157 2154-12° = 0.112 | 337:12° Va . (360° - 337.12 ) = -22.789° - 337-12° = - Va= 0.112 L-22 22.78 9° Amps. TRIES YOUNEN