Find the power dissipated in the 8 W resistor.

is = I₁

KVL around mesh 1.

7 I₁ + 4 I₁ - 4 I₃ + 16 I₁ - 16 I₂ = 0

27 I₁ - 16 I₂ - 4 I₃ = 0 (Eq. 1)

KVL around mesh 2.

16 I₂ - 16 I₁ + 7 I₂ - 7 I₃ + 8 I₂ - 80 = 0

- 16 I₁ + 31 I₂ - 7 I₃ = 80 (Eq. 2)

KVL around mesh 3.

4 I₃ - 4 I₁ - 24 is + 20 I₃ + 7 I₃ - 7 I₂ = 0

4 I₃ - 4 I₁ - 24 i1 + 20 I₃ + 7 I₃ - 7 I₂ = 0

- 28 I₁ - 7 I₂ + 31 I₃ = 0 (Eq. 3)

Solving the three equations yields:

I₁ = 4.352 A

I₂ = 6.022 A

I₃ = 5.291A

Finally, the power dissipated in the 8 W resistor is

P = I²R = I₂²R = 6.022² (8) = 290 W

and yet again, the book's answer is incorrect. (This is getting ridiculous!)