JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 7)
The truth table given in fig. represents :
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
AND - Gate
OR - Gate
NAND - Gate
NOR - Gate
Explanation
Here Y = 1 when,
(1) A = 0 and B = 1 $$\left( {\overline A B} \right)$$
(2) A = 1 and B = 0 $$\left( {A\,\overline B } \right)$$
(3) A = 1 and B = 1 $$\left( {A\,B} \right)$$
$$ \therefore $$ Y = $${\overline A }$$ B + A $${\overline B }$$ + AB
= $${\overline A }$$ B + A ($${\overline B }$$ + B)
= $${\overline A }$$ B + A [as $${\overline B }$$ + B = 1]
= (A + $${\overline A }$$) (A + B)
= A + B
So, this represented OR gate.
Note :
X + $${\overline X }$$ Y = (X + $${\overline X }$$) (X + Y)
(1) A = 0 and B = 1 $$\left( {\overline A B} \right)$$
(2) A = 1 and B = 0 $$\left( {A\,\overline B } \right)$$
(3) A = 1 and B = 1 $$\left( {A\,B} \right)$$
$$ \therefore $$ Y = $${\overline A }$$ B + A $${\overline B }$$ + AB
= $${\overline A }$$ B + A ($${\overline B }$$ + B)
= $${\overline A }$$ B + A [as $${\overline B }$$ + B = 1]
= (A + $${\overline A }$$) (A + B)
= A + B
So, this represented OR gate.
Note :
X + $${\overline X }$$ Y = (X + $${\overline X }$$) (X + Y)
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