Correct Option: D

Let i₁ and i₂ be two loop currents.
∴ 0.5vx = 500i₁ + 500(i₁ + i₂)
⇒ v_x = – 500i₁
∴ 5i₁ – 2i₂ = 0 .......(i)
500(i₂ – i₁) + 900(i₂ + 0.3) + 600(i₂ – 0.6) = 0
⇒ – 5i₁ + 20i₂ = 0.9 .......(ii)
Solving equation (i) and (ii), we get
i₁ = 20 mA,
and v_x = – 500 × 20mA = – 10V