The correct answer is (b) -3.41 V
The best explanation: Applying KCL to node A, \(\frac{V_A-10}{10} + \frac{V_A}{20} + \frac{V_A}{7}\) = 0
Or, VA (0.1 + 0.05 + 0.143) = 1
Or, VA = 3.41 V
The voltage across the 2 Ω resistor due to 10 V source is V2 = \(\frac{V_A}{7} × 2\) = 0.97 V
V2Ω due to 20 V source, \(\frac{V_A}{10} + \frac{V_A}{20} + \frac{V_A-20}{7}\) = 0
Or, 0.1 VA + 0.05VA + 0.143VA = 2.86
∴ VA = \(\frac{2.86}{0.293}\) = 9.76 V
V2Ω = \(\frac{V_A-20}{7}\) × 2 = -2.92 V
The current in 2 Ω resistor = 2 × \(\frac{5}{5+8.67}\)
= \(\frac{10}{13.67}\) = 0.73 A
The voltage across the 2 Ω resistor = 0.73 × 2 = 1.46 V
V2Ω = 0.97 – 2.92 -1.46 = -3.41 V.