Correct answer is (c) −30 + 10IO
Easy explanation: Using Superposition Theorem the voltage VX is given as,
VX = (3-IO) (2||40) + 5VX \(\frac{2}{40+2}\)
Or, VX = \(\frac{80}{42}(3 – I_O) + \frac{10}{42}\)VX
Or, VX = \(\frac{\frac{80}{42}}{1-\frac{10}{42}}\)(3 – IO) = 2.5(3 – IO)
∴ VO = VX – 5VX = −30 + 10IO.