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For the circuit in Fig. 5.93, find { v }_{ 0 }.

Step-by-step

\begin{aligned}\mathrm{v}_{\mathrm{o}} &=\frac{-100}{25}(6)-\frac{100}{20}\left(-\frac{40}{20}\right)(4)-\frac{100}{10}(2) \\&=-24+40-20=\underline{-4 \mathrm{V}}\end{aligned}

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