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Problem 4.23 - Enhanced - with Hints and Feedback Use the node-voltage method to find the branch currents i1.12, and is in th

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HERE IS THE SOLUTION

il IK 5K TiOmA an D500 4K 7s|iz 5-102 40 let the voltage at Point A lis VA So, now applying KCL at Point A we will get, -1; +22.8125 iz = 500 = 0·0456A - 456mA Now Calculating iz here curren through 75 4x103- 4X103 SQ 13-10- + 228125 – 75 103 - 006

ATTACHING SIMULATION FOR ANSWE VERIFICATION

3.44 mA OV 40 V 40 V 3.44 mA OXS 45.6 mA 500 Q 22.8 v 10 mA 52.2 mA VYT 18.8 ma 4 ΚΩ 75 V 60.9 mA +++i 75 V

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