## Answers

Apply virtual ground concept v (t) = v (-) = v_1 Apply K C L at node B v_1 - 0/24 k + v_1 - v_0/96 k = 0 v_1 (1/24 + 1/96) = v_0/96 v_1 (96/24 + 1) = v_0 v_1 = v_0/5 Apply K C L at node A v_1 - 5/16 k + v_1 - v_g/24 k = 0 v_1 (1/16 + 1/24) - v_g/24 = 5/16 put v_1 in equation (2) v_0/5(1/16 + 1/24) = 5/16 + v_g/24 v_0 (0.020833) = 5/16 + v_g/24 v_0 = 15 + 2 v_g v_0 = -10 v - 10 = 15 + 2 v_g v_g = -10 - 15/2 = -12.5 v v_0 = 10 v 10 = 15 + 2 v_g v_g = 10 - 15/2 = -2.5 v range of v_g - 12.5 v to - 2.5 v

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