A possible solution to the first question (out of six) in the Dec 2011 final examination. A question on MNA and Thevenin equivalent methods 1A/2A. Another possibility, more laborious, would have been to include among the unknowns, the "evil" current in the 10V source; which would have brought in two additional equations, namely: a KCL4 for the additional node (4), and an additional EVL equation: V4=10. This solution is also correct, and the markers are aware of it.
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