I want the step by step method to solve this question using Kirchoff voltage and current law then solving it simultaneously to get I1 and I2

Let's define two clockwise mesh currents:

• $i_1$: Current in the left loop (containing the top 6V source, 2Ω resistor, 3Ω resistor, and middle 6V source).
• $i_2$: Current in the right loop (containing the middle 6V source, 3Ω resistor, 1Ω resistor, and 6Ω resistor).

Applying KVL to the left loop (Mesh 1, clockwise):
Starting from the bottom-left corner and moving clockwise:

Applying KVL to the right loop (Mesh 2, clockwise):
Starting from the bottom-right corner and moving clockwise:

From equation $(*)$:

Substitute this expression for $i_1$ into equation $(**)$:

Multiply by 5 to clear the fraction:

Now, substitute $i_2$ back into the expression for $i_1$:

• The current $I_1$ in the diagram flows through the 1Ω resistor. This corresponds to our mesh current $i_2$. $$I_1 = i_2 = \frac{30}{41} \text{ A}$$
• The current $I_2$ in the diagram flows from B to A through the 3Ω resistor and middle 6V source. In our mesh analysis, $i_2$ flows from B to A, and $i_1$ flows from A to B. Therefore, the net current from B to A is $i_2 - i_1$. $$I_2 = i_2 - i_1 = \frac{30}{41} \text{ A} - \frac{18}{41} \text{ A} = \frac{12}{41} \text{ A}$$

We can find $V_{AB} = V_A - V_B$ by tracing a path from B to A through the middle branch (3Ω resistor and 6V source).
The current $I_2$ flows from B to A.
Going from B to A:

• Voltage drop across the 3Ω resistor: $I_2 \times 3\Omega$.
• Voltage rise across the 6V source (from negative to positive terminal): $+6\text{V}$.

The final answers are:

• i) The current $I_1$ and $I_2$: $$I_1 = \boxed{\frac{30}{41} \text{ A}}$$ $$I_2 = \boxed{\frac{12}{41} \text{ A}}$$
• ii) The p.d. across AB: $$V_{AB} = \boxed{\frac{282}{41} \text{ V}}$$