Question
why i01=-1.5mA???
is that i01=1.5mA??? why???
Homework Answers
Answer #1
KVL in the circuit -9 + 3ki + 3 ki = 0 6ki = 9 I = 9/6k = 1.5 mA I and i_bc are in opposite direction I = -i_D_1 = 1.5 mA i_D_1 = -1.5 mA \r\n I_02 = v/3k . . . . . .(1) (V = IR) KCL at node (v) v/3k + v/3k + 3m = 0 2v/3k = -3m v = -9/2 = -4.5v from (1) I_02 = v/3k = -4.5/3k = -1.5 mA I_02 = -1.5 A kcl at node (1) V_2/R_1 + v_2 - v_0^2/R_2 = 0 \r\n v_2(1/R_2 + 1/R_2) = v_0^2/R_2 v_2(1/R_1 R_2/R_1 + R_2) = v_02/R_2 implies V_02 = (R_1 + R_2/R_1 R_2/2) R_2 v_01 v_02 = (1 + R_2/R_1) v_2 (10) using super position 4 mA current source acting alone i_2 = 4mA KCL in loop (i_1) 3k i_1 + 3ki_1 = 0 6 ki_1 = 0 implies i_1 = 0/6k = 0 i_1 = 0A 3 mA current
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why i01=-1.5mA???
is that i01=1.5mA??? why???
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