For the circuit shown in the figure below, the values of i₀ and i₁ respectively are

By applying voltage division in the input loop,

\({V_x} = 18\left( {\frac{6}{{12 + 6}}} \right) = 6\;V\)

i₀ will be zero, as there is no closed path.

By applying KCL in the right-side circuit,

\(\frac{{{V_x}}}{2} + {i_1} + {i_2} = 0\)

\(\Rightarrow {i_1} + {i_2} = - \frac{{{V_x}}}{2} = - 3\) ----(1)

As 10 Ω and 5 Ω are connected in parallel, voltage across them are equal.

⇒ 10 i₁ = 5 i₂ ⇒ i₂ = 2i₁ ----(2)

From equation (1) and (2)