Question

(1 point) In the circuit below, i1 = -2 A, i2 = 8 A, and v1 = -2 V. Use nodal analysis to find vx and iy in the circuit below

i1 8Ω 15 Ω 36 Ω 2 iy y 14 Ω 20 Ω ν1 24 Ω 10 Ω i2 40 Ω 35 Ω 12 Ω 16 Ω 0.4Vx 2013 Paul Hummel CC BY NC SAΑ

Answer 1

sol: -2A 36n www 15n aiy 14n 20n -2V 011 Vo OV I2 8A 242 S4on 0 2n 35n 6n civeurt above Vo-VE ea(i VA+V 36+ 12) VA 2-VG 48 SO MU

Analysis At node Ve Nodh) -iy node VA At +yV-o) VA-VA 14 56 y + 4VA +Va -7VA 56 56 iy VA t + INA O - -0ig VA= VA = no de Ve A1 Vc- (Ve-0) +I, O t 15 20 Ne-Vo 0 Vc -2 20 15 3 Ve + 4Vc -4Vp -120 60

-120 7 V -4VD =0 4VD +120 7VC e = 4 VD +120 node Vo At (-VE -Ve) 12 sub eq0 35 5 VA Vo-V 35 15 7V-7Ve t 840 +3V I05 sub es) 7 VD-7V +3 V + 8 y0 =0 4 VD+120 3V +840 =O 7Vp 3Vt +840 =O 7VD 4 VD-120 3VD +3V+ 720 C 3 VD = -3V -720 VD = -3Vx -720 3 VD = -V-240 eg node Ve At (Ve-Vo (Ve-o) e-Y C + 35 40 Ve-V) Vo-Ve) sub eq + 35 40 -Va 35 VE-V VE + 40

28 Ve +7 Ve - Vf -8V 280 -8Vx + 35VE -7 V =O 35VE KA8+At 7Vf +8V Ve 3S A node V O.4Vt 40 16 5Vf - 5 VE 32 V 2 Vf + 80 sub eq6 32Vt7 Vf 5 VE =O 7Vf+8V 32Vd ナVF 35 32Vx 7Vf- 7f +8V 224V 49VF -7V 8 V 7 216 V 42 V =0 42 Vf -216V = -216 V 2 VF At node Va -+-0) O 4VMO 24

-241+V- 96 V 0 + 9.6 Vl e - fiome q sub eq Vo-VE SU eq E-240] +8Vx sub ey 35 -( -216 VX 42 8VM 35 V -240) C 36V+8V 35 ) -(-2A Vr -240 35 -V -240 + O.8V V V VA -O 8V = - 240 1-2 Vt = -240 V 240 2 200 V fom eg VA -2- 48 4siyVA-2- sub e [24y +96 sub -200 -2 4siyoly - iy ) - 2 4siy 0i-2-C24iy-1920]

48y70ig - 24y 264-iy Rot--Rish +A20 -2 22 21120 (i )0-26hig 1 $ XII -22 + 2)l 20 spsiy- -331y 21098 + 21098 36aiy 21098 21098 862 24. 4756 A VM =200V 24.4756A