EE110_W08_HW 1 Solution

EE110_W08_HW 1 Solution - Q2: For this circuit problem,...

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Unformatted text preview: Q2: For this circuit problem, nodal analysis will require 3 simultaneous nodal equations, then subtraction/ division steps to obtain the desired currents. Mesh analysis requires 1 mesh equation, 1 supermesh equation, 2 simple KCL equations and one subtraction step to determine the currents. If either technique has an edge in this situation, its probably mesh analysis. Thus, define four clockwise mesh equations: i a in the bottom left mesh, i b in the top left mesh, i c in the top right mesh, and i d in the bottom right mesh. At the a , b , c supermesh: -100 + 6 i a + 20 i b + 4 i c + 10 i c 10 i d = 0 [1] Mesh d: 100 + 10 id 10 i c + 24 i d = 0 [2] KCL: - i a + i b = 2 [3] and - i b + i c = 3 i 3 = 3 i a [4] Collecting terms & simplifying, 6 i a + 20 i b + 14 i c 10 i d = 100 [1] -10 i c + 34 i d = -100 [2] - i a + i b = 2 [3] -3 i a i b + i c = 0 [4] Solving, i a = 0.1206 A, i b = 2.121 A, i c = 2.482 A, and i d = -2.211 A. Thus, i 3 = i a = 120.6 mA and i 10 = i c...
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This note was uploaded on 03/06/2008 for the course EE 110 taught by Professor Gupta during the Winter '08 term at UCLA.

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EE110_W08_HW 1 Solution - Q2: For this circuit problem,...

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