Determine the power delivered by the 30 A source.

First, express the 30 A source in terms of mesh currents.

I₃ = 30

Now removing the 30 A source leaves only two meshes, 1 and 2.

KVL around mesh 1.

4 I₁ + 16 I₁ - 16 I₂ + 5.6 I₁ - 5.6 I₃ - 600 = 0

25.6 I₁ - 16 I₂ - 5.6 (30) = 600

25.6 I₁ - 16 I₂ = 768 (Eq. 1)

KVL around mesh 2.

3.2 I₂ + 424 + 0.8 I₂ - 0.8 I₃ + 16 I₂ - 16 I₁ = 0

- 16 I₁ + 20 I₂ - 0.8 (30) = - 424

- 16 I₁ + 20 I₂ = - 400 (Eq. 2)

Solving the two equations yields:

I₁ = 35 A

I₂ = 8 A

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Now, to find the power delivered by the 30 A source, find the voltage across it using KVL.

First find V₁ and V₁ with Ohm's Law.

V₁ = 5.6 (I₃ - I₁) = 5.6 (- 5) = - 28 V

V₂ = 0.8 (I₃ - I₂) = 0.8 (22) = 17.6 V

KVL yields:

V_S = V₁ + V₂ = - 10.4 V

So the power delivered to the circuit by the 30 A source is

P = 30 (- 10.4) = - 312 W

so the 30 A source is actually absorbing power.