21A 302 ww 120½ ✓ 18 V 2A (b) (i) From mesh 1; Applying KVL 21A-36-12(4-(2)-8cc-(3)=0 ⇒ 2IA-236, +126+863 =0 From 5A Current Source, connected between. nesh 3 and mesh 4; 13-14=5 (iii) Apply KVL in mesh 1; ww 802 ⇒>> 2IA = 231,-1262-863 100 3 5 A 502 IA = 23 c; -66-413 E2 IA a) Around which mesh currents can KVL be applied? b) Using the variables i1, 12, 13, and 14 for the mesh currents: i) Express lA in terms of the mesh currents ii) Write the equations for the mesh currents from the current sources iii) Apply KVL to the loop(s) identified in step i iv) Write the equation relating the mesh currents using the simplest superloop ✓ Screenshot 202... => Also from mesh 4; IA = (4 (b) (1) Equations from current Sources: from 2A current source in mesh 2: 2=-2A 21A-36-12CC-1)-8(c;-(3)=0 23 c; - 1262-8 13 = 21A (iv) Since there is a current source of 5A in common branch of mesh 3 and mesh4. KVL can not be applied to individual mesh 3 and mesti4. Forming superloop using mesh 3 and mesh 4 and applying KVL: -1063-8 (3-1)-18-514= =0 86-1863-564 = 18 A+ V+ A- V- -1.586 A 10 247.216 mA 12 2A 5A -5.247 A t = 7.635 s time step 5 us

21A 302 ww 120½ ✓ 18 V 2A (b) (i) From mesh 1; Applying KVL 21A-36-12(4-(2)-8cc-(3)=0 ⇒ 2IA-236, +126+863 =0 From 5A Current Source, connected between. nesh 3 and mesh 4; 13-14=5 (iii) Apply KVL in mesh 1; ww 802 ⇒>> 2IA = 231,-1262-863 100 3 5 A 502 IA = 23 c; -66-413 E2 IA a) Around which mesh currents can KVL be applied? b) Using the variables i1, 12, 13, and 14 for the mesh currents: i) Express lA in terms of the mesh currents ii) Write the equations for the mesh currents from the current sources iii) Apply KVL to the loop(s) identified in step i iv) Write the equation relating the mesh currents using the simplest superloop ✓ Screenshot 202... => Also from mesh 4; IA = (4 (b) (1) Equations from current Sources: from 2A current source in mesh 2: 2=-2A 21A-36-12CC-1)-8(c;-(3)=0 23 c; - 1262-8 13 = 21A (iv) Since there is a current source of 5A in common branch of mesh 3 and mesh4. KVL can not be applied to individual mesh 3 and mesti4. Forming superloop using mesh 3 and mesh 4 and applying KVL: -1063-8 (3-1)-18-514= =0 86-1863-564 = 18 A+ V+ A- V- -1.586 A 10 247.216 mA 12 2A 5A -5.247 A t = 7.635 s time step 5 us

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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