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(2) Find Va and Vb using nodal analysis. This is an example of a problem where it is important to follow the directions usin

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@ Given circuit. SSA & 4 2A ht By nodal analysis. apply KCL at node @. 2 = va + Ya Vo io Va (4+1) - V6 = - Sva - Vo = 2. - Ov16 Va -5 V6 = 10 -46 + 5 V = 20 2. = 30 3 Va =15 V. eq 0 – a og 6Va -5 V6 = 10 6x15 -5% = 10 Vb = 16 v. .: Vo = 15 V, Z V = 162By nodal analysis, apply kcl at node@. Vans + Yes You Vary -0. - Ya u vet vo - G-za) 20 =0 > ts va +4 -15 =0. i va = 11 - Va33 V-va = 16 V =4V6 = 3V, – 446 = 16 > V6 = 16 = Vb =-16 v Va = 4 V = 4x-16 >> Va = -64 V4

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