The correct choice is (d) 2 Ω
The explanation is: \(R_N = \frac{V_{OC}}{I_{SC}}\)
VN = VOC
Applying KCL at node A, \(\frac{2I-V_N}{1} + 2 = I + \frac{V_N}{2}\)
Or, I = \(\frac{V_N}{1}\)
Putting, 2VN – VN + 2 = VN + \(\frac{V_N}{2}\)
Or, VN = 4 V.
∴ RN = 4/2 = 2Ω.