HW 3s - Hw 3 Solutions 1.20 Since 2 p = 0...

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Unformatted text preview: Hw 3 Solutions 1.20 Since 2 p = 0 -30><6 + 6x12 + 3V0 + 28 + 28x2 - 3x10 = 0 72 + 84 + 3V0 = 210 or 3V0 = 54 W=l§¥ 2.7 La)6branches,4nodes , Cb) ’7 branC‘hesj s—r’odes 2.8 IZmA 8'“ fl 9 mA =i1+12, i1=—4mA 9=8+i2, iz=1mA 12+i3=9, i3=—3mA 2.12 Forloop 1= -20 -25 T10 +v1 = 0 —> v1 =35v Forloop 2= —10 +15 -V2 = 0 —> v» =5v For loop 3, 47; +172 +V3 = 0 —> v; = 30v 2.21 Applying KVL, -15 + (1+5+2)| + 2 V)( = 0 But V)( = SI, —15 +8| + 10| =0, I = 5/6 Vx = 5| = 25/6 = 4.167 V 2.22 At the no de: KCL requires that 33+10+2v0=0——> v0: 4.444v The current through the controlled source is i_= 2V0 = -8.888A and the voltage across it is v=(6+4) io (\vhereio =vox'4)= 10::— = —ll.lll Hence= mm: (-8.888)(—11.111) = 93.75 w ...
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HW 3s - Hw 3 Solutions 1.20 Since 2 p = 0...

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