Electric Circuits 8th Edition 145

# Electric Circuits 8th Edition 145 - 4.10 T h6vnin N...

This preview shows page 1. Sign up to view the full content.

Hence ttre Th6venin voltage for the circuit is 32 V The 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 vrith the short in place. Note that the shot-circuit curent is in tle direction of the opetr-circuit voltage drop auoss the terminals a,b. If the shon-circuit curent 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. Agaiq iJ we use the lower node as the reforence node. the equation for ,, becomes \4.59) Solving Eq.4.59 for ?]2 gives u2:16Y' HeDce, the shofi-circuit current is (4.60) 4.10 Th6v€nin and Norton Equivatentr 121 Step 1: Source transformation Step 3: Souce kansformationj series resistors combined, producing the Thevenin equivalent circuit Step 1: Souce traDsfornation, producing the Norton equivalent circuii Figure 4.48 A Step-by-step
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online