The correct Boolean operation represented by the circuit diagram drawn is
The correct Boolean operation represented by the circuit diagram drawn is
The correct option is
C
NAND
As we know that current always chooses least resistance path, here if both of the switches goes high(0) then as the current will choose the least resistance path, it goes through both A & B switches. and therefore the LED will not glow in that case. In rest of the cases where one switches is high(1) or the other is low(0) and vise versa or both goes low (0) , the current goes through the LED, as it does not find any close circuit, other than the resistive path, and therefore illuminating the LED through the close path.
Truth table
A B Y 0 0 1 0 1 1 1 0 1 1 1 0
Now, The output (Y) looks like inverted output of AND gate, which is also known as NAND gate.
Truth table of AND and NAND gate
A B AND NAND 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0
∴ It is a NAND gate.
The correct option is
C
NAND
As we know that current always chooses least resistance path, here if both of the switches goes high(0) then as the current will choose the least resistance path, it goes through both A & B switches. and therefore the LED will not glow in that case. In rest of the cases where one switches is high(1) or the other is low(0) and vise versa or both goes low (0) , the current goes through the LED, as it does not find any close circuit, other than the resistive path, and therefore illuminating the LED through the close path.
Truth table
| A | B | Y |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Now, The output (Y) looks like inverted output of AND gate, which is also known as NAND gate.
Truth table of AND and NAND gate
| A | B | AND | NAND |
| 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 0 |
∴ It is a NAND gate.
