Question

4. Pod V, in the network shown in the given figure, where P.6240" V. Peste report your answer to the magnitude in partive and at angles are in the range of negative 180 days to postive 180 degrees ΤΩ 20 20 w V j40 140 -jla 2/0A w P . The value of V, 2

Answer 1

cohere p-4240 V, Find Vase in the network shown in the given figure (Page 1 سرور 22 2R ا + V 20 )2LEA j4R telee 34 p=4248v - jan Iz Solution - By using Source transformation method. Current source and reactance across capacitor converted into voltage Soupee and series Capacitive reactance I= 2LOA X = - JIR > V = IXc V= 2 Lox(-1) V=– 2j I V ---2/ 900 → V=(-1)× 2490 B-1 2 v=j2 x 290 > =(1290) ²%240

Page > = 198 x 198 x 2900 n - 1/180 x 290° 90 V- 22700 360 180€ (anticlock wise rotation V-21360°+270 -27 v=22-90 v ó, - 180 **-98 (clockwise Dotanica) that that means we can say 42700 = -90 SLR 2R 2R M + x - jes j4R I + 2.gov 3413 P-4 Lov Applying KvL in mesh lol) - 4240° +26, +344 -5 1 I ₂ = 0

3 عی) یا-راز-1 (214) KVL in mesh (02) Applying 0 - 11ز- را 4 + 1,1290 - راد 220 - - راور-از- را 2 + +[ - - - - - = 3 + را 2 + ] 1 - 290 - 1 (ز2 م - 1 - - 14 (213)) - (1) x 250 2 در 2 پر(ن) = 1 (213) + دز - 3 -1 و F = ( 1 ) x2 (243) 1 - 1 L188 x 2.90 رد (213) + ] در- وو-می) 2 - راد(21) + ] در- - ) هود - را (213) + ] در-

Page 4 from equation (i) and equation in (2+34), - js = 4240 Polar Form (2+)4) I j' I2 = 4 [costo tj sinto] (2+j4) I-1 [2 = 4 [0.766 +j 0.64 27] (2+ j4) I, - ĵ I2 = 4X0.766 +1 0.6427X4 (+34) , -j 12 = 3.064 +j 2.5708 (Rectangular from so (2+34) I - J I ₂ = (3.064 +) 2.5708) - (1) i I +(2+3)) { = 2; solving above two equation by Above two equation written as a matsix forms Cramer Rule AX-B 2+j4 - j I 3.064 +) 2.5708 I 2 -j (2+3;) 29 2+34 -; (243)

pages A= (2+4)) (2+3)) - (-))*(-1) A = (4 +6j+8j +3jxuj) - (+j?) A= (4+14)- 12) - (-1) = (14)-8) tt 145-8+1 s = - 7+14) → Rektron forms Rectangular form) (7) +(1032 -+00' ( 4 ) +180° - teo(2) +180° s = 494196 L- a = √245 L180 - test2 s = 15.6524 15.6524 188-63.43 D = 15.6524 116.57° (polar forms] 2434 3. Oby+) 2.5708 A2 = -j 21

Page 6 D₂ = (2+14) x 2j ti (3.064 +12.5708) 12 - 4j-8 + 3.0643 – 2.5708 -8-2.5708+4j+3.06yj A = -10.5708 +1 7.064Reetongular. ular form (10.5708)² + (7.064)? 5'(7.064 7.064 10.5708 +180° Dz ter (0.6682) +1809 Sza1011.7418 + 49.90) - 33.75 +180° I = 161.6418 A2- 12.713842 / 180-33.75° [polar form O2 = 12.713842/146.25° Da I ₂ = Iz - 12.713892/146125° 15.6524 / 116.570 I, = 0.81 22615 /146.23° -116.7° I = 0.8122615 (29.68° A Approximate

Page 7 Vse x = 2x Vx = 2 x(0.812261529068°) Approximate x = 1.624523 229.680 v Magnitude = 1.624523 angle =/29.68 v Always Remos.ember atibaſ 92462 tasi' ( bla) a-iba a 2462 tant (la) - atib=a²+62 % tot (3/4) +180 a-ib=a2762 / tasil (3/4) - 180 Final Answer