# 21 with this in mind we obtain from fig 32b v1 0 or i1

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Unformatted text preview: Fig. 2.1). With this in mind, we obtain from Fig. 3.2(b), v1 − 0 or i1 = G1 v1 R1 v1 − v2 i2 = or i2 = G2 (v1 − v2 ) R2 v2 − 0 i3 = or i3 = G3 v2 R3 (b) i1 = Figure 3.2 (3.4) Typical circuit for nodal analysis. Substituting Eq. (3.4) in Eqs. (3.1) and (3.2) results, respectively, in v1 v1 − v2 + (3.5) I1 = I2 + R1 R2 | v v I2 + | e-Text Main Menu v1 − v2 v2 = R2 R3 (3.6) | Textbook Table of Contents | Problem Solving Workbook Contents 78 PART 1 DC Circuits In terms of the conductances, Eqs. (3.5) and (3.6) become I1 = I2 + G1 v1 + G2 (v1 − v2 ) I2 + G2 (v1 − v2 ) = G3 v2 (3.7) (3.8) The third step in nodal analysis is to solve for the node voltages. If we apply KCL to n − 1 nonreference nodes, we obtain n − 1 simultaneous equations such as Eqs. (3.5) and (3.6) or (3.7) and (3.8). For the circuit of Fig. 3.2, we solve Eqs. (3.5) and (3.6) or (3.7) and (3.8) to obtain the node voltages v1 and v2 using any standard method, such as the substitution method, the elimination method, Cramer’s rule, or m...
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