Correct option is (b) 9.26 W
Easy explanation: Let us remove the 1 Ω resistor and short x-y.
At Node 1, assuming node potential to be V, \(\frac{V-10}{5}\) + ISC = 5
But ISC = \(\frac{V}{2}\)
∴ \(\frac{V-10}{5} + \frac{V}{2}\) = 5
Or, 0.7 V = 7
That is, V= 10 V
∴ ISC = \(\frac{V}{2}\) = 5 A
To find Rint, all constant sources are deactivated. Rint = \(\frac{(5+2)×2}{5+2+2} = \frac{14}{9}\) = 1.56 Ω
Rint = 1.56 Ω; ISC = IN = 5A
Here, I = IN \(\frac{R_{int}}{R_{int}+1} = 5 × \frac{1.56}{1.56+1}\) = 3.04 A
∴ Power loss = (3.04)^2 × 1 = 9.26 W.