Question

Learning Goal: To use the node-voltage method to solve circuits with branches containing only a voltage source. The node-voltage method is a general technique for solving circuits. Fundamentally, it involves writing KCL equations at essential nodes. When the circuit contains a dependent source, you must write a constraint equation for each dependent source, in addition to the KCL equations. When the circuit contains one or more voltage sources that are the only components in branches connecting two essential nodes, the node-voltage method must be modified slightly. To begin this tutorial, you will use the node-voltage method to find the current through the 140 12 resistor in the circuit shown. In this circuit, vi = 8 V and v2 = 17 V. 30 Ω 25 22 140 Ω -i + Vil 1023 V2

Answer 1

As per HOMEWORKLIB POLICY I solved.if you have any doubts comment please.

at node u 301 M V=80 VX 140 o Ves 17 V a At nodevu, nodal Analysis Apply Vi Vz-V, Va & Vp -V2 + 28 ino Vy - 8 Vx-ly + Vu 10 140 25 + volás + tarot to ] = 25 17 Tyo 0.4uly Va (0.147 ) = V2 = 3.0027 N i = ₂-vu 17 -3.002) 0.09998 ~ OILA Ivo 140 Part A 4. shown in above Essential nodes are figuee parto 3.0 V Vm avo Node :8 V = Va Vi Node 2 = Vy= VD 3.0V = Vo Node 3 = V₂ 17 v - Ve Node u = ground voltage = refrence voltage

Part-c Зол Here Vc =V,= 17V va (що ara Vc V2-19v part- 17-8 - 0-3A 3o iso- igo Apply nodal at 30 Analysis at vo U6-17 ar Vb (५० Vo-8 25 100 ies Vio lino ve va 8 10 * 17 vols wat to) - ar is = 0.414 Vol 0,14 714) 17-3.0 -0.1A V6 ~ 3.ou - Vb ies 8 - 3.0 = OLA 25 25 Vb io 3 0.3 A kd at no de vc kal at node va izo +18= ܕܙܐ iso tilyo ier izr-iso 18 0.3 +0.! -0.2 -0.3 117 - Diya is -0.1 18.10.17

part-t at nodal ped Apply Vb Analysis nod Vor8 als vb-17 Tuo ar VD (0.14714) = 0.vuru Tu6= 3.0) part of 3ou 17-Vb i. 1 140 Tuo 8v Vb 17 2r- Vo already tindout = 3.00 10 17-3 is -O 100 Tuo To: 0.1A iolooma