Question

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Answer 1

As op amp is ideal so: v_a = v_b as v_a = 0 so v_a = v_b = 0 v apply model analysis at node (1) v/12 + v - 2/2 + v - v b/4 = 0 put vb = 0 v/12 + v - 2/2 + v - 0/4 = 0 v/12 + v/2 - 2/2 + v/4 = 0 \r\n v[1/12 + 1/2 + 1/4] = 1 v = 1.2 v apply nodal analysis at node (b) vb - v/4 + vb - v_0/10 = 0 put vb = 0 0 - v/4 + 0 - v_0/10 = 0 - v/4 = v_0/10 v_0 = -10/4 v times 1.2 Therefore v = 1.2 v_0 = -3 as i_10 k Ohm = vb - v_0/10 m A = 0 - (- 3)/10 m A = 3 m A \r\n and i_2 k Ohm = v_0/2 m A i_2 k Ohm = -3/2 m A i_2 k Ohm = -1.5 m A applying K C L at node (c) i_0 + i_10 k Ohm = i_2 k Ohm