HW1_sol - Applied Electronics HW1 - problems and solutions...

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Applied Electronics HW1 - problems and solutions
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(a) Node voltage method Apply KCL: Node a: i_ga-(6-v_f)/7-2-(6-v_c)/12=0 Node c: (6-v_c)/12+2-i_cd=0 Node d: i_cd+(v_f-v_c-9)/8-(v_c+4)/13=0 Node e: (6-v_f)/7-(v_f-5)/9-(v_f-v_c-9)/8=0 Node g: -i_ga+(v_f-5)/9+(v_c+4)/13=0 4 unknowns (v_c, v_f, i_cd, i_ga) and 5 equations -> 1 equation is redundant Solve with equations about nodes a, c, d and e: v_c=7.8039, v_f=9.2704, i_cd=1.8497, i_ga=1.3825 Additional calculations v_b=v_c+2*11-3=v_c+19 therefore v_b=26.8039 Therefore, the results are: (voltage: V, current: A) v_a v_b v_c v_d v_e v_f i_ac (12 Ohm) i_eg (9 Ohm) 6 26.8039 7.8039 11.8039 4.2704 9.2704 0.1503 0.4745
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(b) Mesh current method Apply KVL: mesh 1: 6-7*(i_1-i_3)-5-9*(i_1-i_2)=0 mesh 2: -9*(i_2-i_1)-8*(i_2-i_3)-13*i_2=0 Supermesh: meshes 3 and 4: -12*i_4+4-8*(i_3-i_2)+5-7*(i_3-i_1)=0 Or, instead of using supermesh, assume a current source as a voltage source mesh 3: -10*(i_3-i_4)+V+3-11*(i_3-i_4)+4-8*(i_3-i_2)+5-7*(i_3-i_1)=0 mesh 4: -12*i_4-11*(i_4-i_3)-3-V-10*(i_4-i_3)=0
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HW1_sol - Applied Electronics HW1 - problems and solutions...

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