Right answer is (d) 12 Ω
Best explanation: By KCL,
∴ \(\frac{V_P-40}{1} + \frac{V_P-100}{14} + \frac{V_P}{2}\) = 0
Or, 22 VP = 660
∴ VP = 30 V
Potential difference between node x and y = 60 V
∴ -I – 5 + \(\frac{40-30}{I}\) = 0
Or, I = 5 A
∴ R = \(\frac{60}{5}\)= 12 Ω.