Answer 1

Applying KCL at node 1

`-4+(V_1-V_2)/0.33+(V_1-V_2)/0.5=0`

0.83𝑉₁−0.5𝑉₂−0.33𝑉₃=0.66 …..(1)

Applying KCL at node 2

`V_2/0.2+V_2+5-V_3+(V_2-V_1)/0.33=0`

−0.2𝑉₁+0.596𝑉₂−0.066𝑉₃=−0.33 …….(2)
Applying KCL at node 3

`(V_3-V_1)/0.5+V_3+V_3-5-V_2=0`

−𝑉₁−𝑉₂+2.5𝑉₃=5 ………(3)
From (1), (2) and (3) we get,
𝑉₁=2.46𝑉       𝑉₂=0.63𝑉       𝑉₃=3.23𝑉
Applying KCL,

`I=(V_1-V_3)/0.5=(2.46-3.23)/0.5`

`I` = −1.54𝐴𝑚 𝐈(𝟎.𝟓Ω)= 𝟏.𝟓𝟒𝐀𝐦(←)