Explanation

(a) The current in the circuit are assumed as shown in the fig.
Applying KVL along the loop $ABDA,$ we get
$- 6i_{1} - 3\ i_{2} + 15 = 0$ or $2i_{1} + i_{2} = 5$ ….(i)
Applying KVL along the loop $BCDB,$ we get
$- 3\left( i_{1} - i_{2} \right) - 30 + 3i_{2} = 0$ or $- i_{1} + 2i_{2} = 10$ ….(ii)
Solving equation (i) and (ii) for $i_{2},$ we get $i_{2} = 5\ A$