Correct option is DApplying Nodal analysis to node V₁-−2+V13+V1−102=0V13+V1−102=22V1+3V1−30=12V1=425I=V1−102=425−102I=42−5010I=−810I=−0.8 A-2 + \frac{V_1}{3} + \frac{V_1 - 10}{2} = 0\\\frac{V_1}{3} + \frac{V_1 - 10}{2} = 2\\2V_1 + 3V_1 - 30 = 12\\V_1 = \frac{42}{5}\\I = \frac{V_1 - 10}{2} = \frac{\frac{42}{5} - 10}{2}\\I = \frac{42 - 50}{10}\\I = -\frac{8}{10}\\I = -0.8 \text{ A}−2+3V1+2V1−10=03V1+2V1−10=22V1+3V1−30=12V1=542I=2V1−10=2542−10I=1042−50I=−108I=−0.8 A