Prior written consent of McGraw-Hill Education. Solution 2.19 Applying KVL around the loop, we obtain –(–8) – 12 + 10 + 3i = 0 i = –2A Power dissipated by the resistor: 2 p 3Ω = i R = 4(3) = 12W Power supplied by the sources: p 12V = 12 ((–2)) = –24W p 10V = 10 (–(–2)) = 20W p 8V = (–8)(–2) = 16W.

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