Question

An AND gate is followed by a NOT gate in series. With two inputs A and B, the Boolean expression for the output Y will be:
(A) $ A \cdot B $
(B) $ A + B $
(C) $ \overline {A + B} $
(D) $ \overline {A \cdot B} $

Answer 1

Hint
Operations in series are performed successively in order of which they occur. The AND gate combines input in a multiplicative way. And the NOT gate reverses the input. So from this we can find the output.

Complete step by step answer
An AND gate acts on all inputs in such a way that it is high only if all inputs are high, and low if at least one input is low. This is similar to a multiplication operation where if one function is Zero (analogous to Low in the AND operation) the result of the operation is zero.
In logic gates operations, when two or more inputs go through the gate, the results after the operator acts on them are called the output. This output can be considered an input to another gate if this other gate is in series to the first gate.
Now, for two inputs A and B going through an AND gate, the resulting output is denoted as $ A \cdot B $. This output is then the input to the NOT gate which is in series to the AND gate. Hence, the NOT gate acts on the output as whole as in: NOT $ A \cdot B $ denoted as $ \overline {A \cdot B} $
Hence, the Boolean expression for the output Y is $ \overline {A \cdot B} $.
Thus, the correct option is (D).

Note
When an AND gate is succeeded by a NOT gate, this is specifically called a NAND gate. As opposed to an AND gate, a NAND gate is high if at least one of the inputs is low, and low if all inputs are high. In essence, it is high when AND gate is low, and low when AND gate is high.