Answer 1

Applying KVL to mesh a
−3𝐼_𝑎 − 2𝐼_𝑎 − 4(𝐼_𝑎 − 𝐼_𝑏 ) + 10 = 0
−3𝐼_𝑎 − 2𝐼_𝑎 − 4𝐼_𝑎 + 4𝐼_𝑏 + 10

Applying KVL to mesh b

-10 + 6𝐼_𝑎 + 8𝐼_𝑏 − 4(𝐼_𝑏 − 𝐼_𝑎) = 0
6𝐼_𝑏 + 8𝐼_𝑏 − 4𝐼_𝑏 + 4𝐼_𝑎 = 10
-4𝐼_𝑎 + 18𝐼_𝑏 = 10 ………..(2)
From (1) and (2) we get,
𝐼_𝑎 = 1.5068A and 𝐼_𝑏 = 0.890A
For voltage:-
Apply KVL to mesh
-V+3𝐼_𝑎 + 8𝐼_𝑏 = 0
V = 3𝐼_𝑎 + 8𝐼_𝑏
V = 3× 1.5068 + 0.89 × 8
V = 11.640V