Solução Capítulo 2 - Riedel, Nilsson. Circuitos Elétricos. Editora Pearson, 8ª edição., Exercícios de Eletrônica

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2 Circuit Elements Assessment Problems AP 21 is : + a t vg v(D) BA [a] Note that the current à, is in the same circuit branch as the 8 A current source; however, ip is defined in the opposite direction of the current source. Therefore, ip=-8A Next, note that the dependent current source and the independent current source are in parallel with the same polarity. Therefore, their voltages are equal, and wo -8 %g q= 4 2 fb| To find the power associated with the 8 À source, we need to find the voltage drop across the source, v;. Note that the two independent sources are in parallel, and that the voltages vz and v; have the same polarities, so these voltages arc equal: m=u=-2V Using the passive sign convention, ps — (8A)(vi) = (SA)(-2V) = —16W “Thus the current source generated 16 W of power. 24