Question

#14 I the Cercutahswin find allremaining bravc Curentand vsltaszer and demsnohrti conatrvátor of pauer ZA 4A 2V Fond Lo en the Circuitohsun 4K 15 Z20K 5 k Ams:g=4.07mA 20 V 12K 16 Fend VB in the Cucutphown 5Ks z2K 12K VAB 3K 15K Ams: Vas=7.2 vd

Answer 1

#(14 USing kVL n the auter da lov LooP as shocun 4A 2A 16 VI0 -V o Vi V GVolt IA JSing kvL rn lower Triang le loop 2(-6) 2 = o V26-2= U Sng kvL Tn UPPer Trangulcr loop, havr we 10V-V-V3 = lo-(-4)-V3 l0 4-V O V3 14 Volt Usiog kcL at Node D 2A+L 4 2A Using KcL at Node B 2A t 1A = I2 I2=3A USIag kcL at Node A 13+ 4 A I2 1A 3 A= I 4 A Conservahon of power Sum of Po wer along any Netaork must be zew 16x (-1) 4A* 14V) (2Vx 2A) + (1A (6)+4oVA 9A) (2A(-4V) 2P PIGV PAAPt P Pov t P2A 2 v 16 56t 4 -630 8 - 60 + 6 0 s following law of Canser vahon o power hich Net power alasorbe d b passive elemen EP O Net Power odelivered bY Souree

Simplitying the circuit 4K 201 5k Using last Network A 2ov to I 20 4t4+4 1OMA 6 K 12K fach branch Volfage acnss V (0MAX4K=4ov Lets sh0w the Voltage lebelled i (ast netwrk 40v and into Middle NetwK 4 k Req 5120= 5x2D 35 K clearly 120V 20 1 e 5t20 40V Cunent Howing thrugh 4K each branch 12116 = (26 (2+6 Req 6K 40V 4uV 12K 8 mA 5 K = 4K I20k tov 20K 2mA Middle NetwK 40V 6K I6K MA 3 4K I2K= 4K 2 K 20Y ongincl using kCL tn our 4 K Network at Node A last Netwok I12k 8 MA 3 14 MA 240 mA 3 4.67 m A

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