1. Find I in 4 Ω resistor.





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  • By: guest on 02 Jun 2017 12.54 am
    Apply KVL to first to first loop 6i1 - 2i2 = 10 (4 + R)i2 - 2i1 - 2i3 = 0 4i3 - 2i2 = 0 but i3 = 0.5 A 2 - 2i2 = 0 ∴ i2 = 1 A, 6i1 - 2 = 10 ∴ i1 = 2 A, (4 + R) - 2 X 2 -1 = 0 ∴ R = 1 Ω Using Nodal analysis for loop 2 At node A, VA - 20 + VA + VA - VB = 0 3VA - VB - 20 = 0 3VA - VB = 20 ...(i) At node B, 2VB - 2VA + 2VB + VB = 0 5VB - 2VA = 0 ...(ii) Multiplying (i) by 2 and (ii) by 3 6VA - 2VB = 40 - 6VA - 15VB = 0 13VB = 40 VB ≅ 3 V ∴
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