In the circuit shown here, the 20 V source has an internal resistance of 1 Ω. While the current source is ideal, the value of current I is _________.

Correct option is B

​$$\text{Applying KCL at node 'A':} \\[4pt]5 = \frac{V_A - 20}{8} + \frac{V_A}{2} \\[8pt]V_A = 12\,\text{V} \\[8pt]I = \frac{12 - 20}{8} = \frac{-8}{8} = -1\,\text{A}$$​​