Hence ttre Th6venin voltage for the circuit is 32 VThe trext step is to place a sholt circuit affoss the terminals and calcu-late the rcsulting short-cftcuit culrent. Figue 4.46 shows the circuit vriththe short in place. Note that the shot-circuit curent is in tle direction ofthe opetr-circuit voltage drop auoss the terminals a,b. If the shon-circuitcurent is in tlle direction of the open-circuit voltage rise aqoss the temi-oals. a miDus sign must be insened in Eq.4.5o.The short-circuit curent (isc) is foutd easily once ,2 is known.Therefore tle problem reduces to finding 1,2 with the short in place. AgaiqiJ we use the lower node as the reforence node. the equation for ,, becomes\4.59)Solving Eq.4.59 for ?]2 givesu2:16Y'HeDce, the shofi-circuit current is(4.60)4.10 Th6v€nin and Norton Equivatentr 121Step 1:Source transformationStep 3:Souce kansformationj seriesresistors combined, producingthe Thevenin equivalent circuitStep 1:Souce traDsfornation, producingthe Norton equivalent circuiiFigure 4.48 A Step-by-step
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