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Q1As oparp is ideal,\begin{array}{l}V_{A}=V_{B} \\I_{1}=I_{2}=0 \\V_{B}=3.9 \times 4 \\V_{B}=15.6 \mathrm{volt}\end{array}KCL at node A\begin{array}{l}I_{f}=6 \\V_{A}=V_{B}=15.6 \text { volt }\end{array}\begin{aligned}I_{f}=\frac{V_{A}-V_{0}}{1.5} & =6 \\I_{f}=15.6-V_{0} & =6 \times 1.5 \\V_{0} & =6.6 V_{0} / \mathrm{t} \\V_{0} & =4.3 I_{0} \\I_{0} & =\frac{6.6}{4.3} \\I_{0} & =1.53 \mathrm{~mA}\end{aligned}\mathrm{Ar}_{3} ...