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[Circuits] Finding the Thevenin and Norton Equivalents #3
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[QUOTE="gneill, post: 4650036, member: 293536"] No doubt it has to do with the direction of the current that you're "injecting" into the circuit versus the polarity of the resulting voltage across the terminals. The resistance of a component (or circuit for that matter) is defined as the current [I]into[/I] the port divided by the resulting potential across that port. The polarity of the voltage "measurement" is such that the terminal where the current is injected is taken to be positive: [ATTACH=full]167116[/ATTACH] [/QUOTE]
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[Circuits] Finding the Thevenin and Norton Equivalents #3
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