Answer 1

Applying KVL in loop 1, we get:-

\[i_1 + \left( i_1 - i_2 \right) - 1 + 2 = 0\]

\[2 i_1 - i_2 = - 1 .............(1)\]

Similarly, for loop 2:-

\[i_2 - \left( i_1 - i_2 \right) - 2 + 3 = 0\]

\[2 i_2 - i_1 = - 1 ............(2)\]

Solving (1) and (2), we get:-

\[i_1 = i_2 = 1 A\]

Potential difference between A and B:-

\[V_A - V_B = E_2 - \left( i_1 - i_2 \right) \times 1 = 2 - \left( 1 - 1 \right) = 2 V\]

Current through the top branch = i₂ = 1 A

Current through the middle branch = i₁ - i₂ = 0 A

Current through the bottom branch = i₁ = 1 A