IRWIN 9e 5_32 - Irwin, Basic Engineering Circuit Analysis,...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 1 5.32 Fiml if; in the Fig. P532 115ng Thc’wnin‘s thenmn. + 4 H11 V“ 2 m Figure P532 SOLUTION: + Voc' 2 m V, " L- ( "' oc - 6) -* 4L + iZk-r 2k . + H ‘ 4. an». X» 12kg. _____—__—______._—_~—_—_—————-————— Chapter 5: Additional Analysis Techniques Problem 5.32 2 mm. Basic Engineering Circuit Analysis, 9/E 3m A 4, 2m ll —- A”... 3m =O'63mn I ' 4k+llk+2k>< ) Vac = 2k + (0' '1, = <_.4_L————— (1m) : 0.222mn 4L+l2k+2k ‘ V0: =12k(0'222m): 2.61v Voc, s ++i4+2.6‘1 a 20.61v Problem 5.32 Chapter 5: Additional Analysis Techniques Irwin, Basic Engineering Circuit Analysis, 9/E 3 ET” = [pm 4k)H12k] T 2k: 6km 4k (Masazw 4‘L+6k Chapter 5: Additional Analysis Techniques Problem 5.32 V ...
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