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The output (X) of the logic circuit shown in figure will be:
A
$X = \overline{A + B}$
B
$X = \overline{\overline{A\cdot B}}$
C
$X = \overline{A\cdot B}$
D
$X = A\cdot B$
Detailed Solution
The first NAND gate gives $\overline{A\cdot B}$; the second NAND gate with joined inputs acts as a NOT gate.
$X = \overline{\overline{A\cdot B}} = A\cdot B$
$X = \overline{\overline{A\cdot B}} = A\cdot B$