Example (cont.) KVL: Loop 1: - 10 + , + v3 = 0 Loop 2: - V3 + V2 + 12= 0 Loop 3 (outer loop): -10 + +V2 + 12=0 Note that Equation (3) is linearly...
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Employ Mathcad to solve symbolically the set of Equations 1,2, and 4. src="/qa/attachment/10454404/" alt="005.png" /> ATTACHMENT PREVIEW Download attachment 005.png Example (cont.) KVL: Loop 1: - 10 + , + v3 = 0 Loop 2: - V3 + V2 + 12= 0 Loop 3 (outer loop): -10 + +V2 + 12=0 Note that Equation (3) is linearly dependent on (can be derived from) Eqn. (1) and (2), so we ignore it. We have three unknowns and two equations. So, we need a third equation in the voltage variables. This can be obtained by applying KCL at node a: i1 - 12 - 13 = 0 - V1 V2 V3 R1 R2 R3 = 0 (4) Given numerical values for the resistors, Equations (1), (2) and (4) can be solved for the voltages v1, V2 and v3.

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