beautiful circuit

Let us try KVL and hope that it will work.

From the first loop, we get:

\(\displaystyle 0.015 - 4000I_1 - \frac{v_o}{100} = 0\)

We also know that \(\displaystyle I_1 = -I_o\)

Then,

\(\displaystyle 0.015 + 4000I_o - \frac{v_o}{100} = 0\)
 
💀💀

\(\displaystyle 0.015 + 4000I_o - \frac{v_o}{100} = 0\)


\(\displaystyle 0.015 + 4000I_o - \frac{1000000I_o}{100} = 0\)


\(\displaystyle 0.015 + 4000I_o - 10000I_o = 0\)


\(\displaystyle 0.015 - 6000I_o = 0\)


\(\displaystyle 6000I_o = 0.015\)


\(\displaystyle I_o = \frac{0.015}{6000} = 2.5 \times 10^{-6} \ \text{A} = \textcolor{blue}{2.5 \ \mu \text{A}}\)
 
We continue like heroes.

\(\displaystyle v_o = 1000000I_o = 1000000(2.5)(10^{-6}) = \textcolor{blue}{2.5 \ \text{V}}\)
 
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