Semiconductor Electronics Materials Devices and Simple Circuits - Result Question 100

102. The circuit

[2008]

is equivalent to

(a) AND gate

(b) NAND gate

(c) NOR gate

(d) OR gate

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Answer:

Correct Answer: 102. (c)

Solution:

  1. (c) Let $A$ and $B$ be inputs and $Y$ the output.

Then $Y_1=(\overline{A+B})=\overline{A} \cdot \overline{B}$

(By De-Morgan’s theorem)

$Y_2=\overline{Y_1}=(\overline{\overline{A} \cdot \overline{B}})=\overline{\overline{A}+B}=A+B$

$\therefore Y=\overline{Y_2}=(\overline{A+B})$

Hence, the given circuit is equivalent to a NOR gate.

NOR and NAND gates is a universal gates as they can be used to perform the basic logic functions. AND, OR and NOT gate.