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15V 5a {su 13A 350 1) Calculate the power delivered by the 5A source using the following techniques: a. [10] Using nodal anal

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ISV kel at node ē 5. +13 + Vx-15 5- y = 13 tahun 3.. zvix + 10 YA+ V2 = -25 = -12.50 x = -12.5ul At node VA VA - Vx (4+1) VAb) hesh analysis: 7. 9 sono VI! .. From loop) , 9, -54 -> ☺ Super mesh of loop ② and 6 1545 (92-1,) = 5.53 ::.-51, + 512 - 5c Soure Transforriation (SV ༩ ༠ ཚེ ། (1)[3A. ་>5A ༨/ Sin - <- ཙ ན •S. s - 85 ཀ ཕ .་5f6 V z |cY ༢•S =Qs V + J་ + VBy Kilo Vs = 5x4 + 5X215 – 25 +1X5 20+ 12.5.-25+5 Vs = 12.50 Power delivered by sa sourcea Vs Is = 125X5 p = 62.5We) Thevenin equivalent: so 13A Rthz RAB 3 4+1+(5115); = 5+0.5 = 7152 15 V: Vx 2 13A 35 Band KCL at no cle va Va-152 V lo = 0 V2 = -10X5 Va = -as v. Vih= Va= -250 Aisa AMY + - Vab = 25- 7.585 - Vab = -12.5 Vab = 12:51 Now , Pravee delivered

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