Find the currents I₁, I₂, I₃ using mesh analysis.

Correct option is C

Applying kVL on mesh (1), (2) and (3) respectively we have,

$$20 - 2I_1 - 2I_1 + 2I_2 = 0\\4I_1 - 2I_2 = 20 \qquad (1)\\(2 + 2 + 1 + 1)I_2 - 2I_1 - 2I_3 = 0\\6I_2 - 2I_1 - 2I_3 = 0 \qquad (2)\\(2 + 2 + 4)I_3 - 2I_2 = 0\\8I_3 = 2I_2 \qquad (3)\\I_3 = \frac{I_2}{4}$$​

$$\text{Substitute the value of } I_3 \text{ in equation (2)}\\6I_2 - 2I_1 - 2\left(\frac{I_2}{4}\right) = 0\\11I_2 - 4I_1 = 0 \qquad (4)\\\text{Solving equation (1) and (4) we get}\\I_1 = \frac{55}{9} \text{ A}\\\text{and} \quad I_2 = \frac{20}{9} \text{ A}\\\text{From equation (3) we get}\\I_3 = \frac{5}{9} \text{ A}\\$$​​

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