JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 22)
To get output 1 at R, for the given logic gate circuit the input values must be
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x = 0, y = 0
x = 1, y = 0
x = 0, y = 1
x = 1, y = 1
Explanation
P = $$\overline x + $$ y
Q = $$\overline {\overline y .x} $$ = y + $$\overline x $$
Output, R = $$\overline {P + Q} $$
To make R = 1
$$\overline {P + Q} $$ should be = 1
$$ \therefore $$ P + Q must be 0
$$ \therefore $$ ($$\overline x + $$ y) + (y + $$\overline x $$) = 0
Now by checking each option you can see
When y = 0 and x = 1 then
$$\overline x + $$ y = 0
$$ \therefore $$ ($$\overline x + $$ y) + (y + $$\overline x $$) = 0
Q = $$\overline {\overline y .x} $$ = y + $$\overline x $$
Output, R = $$\overline {P + Q} $$
To make R = 1
$$\overline {P + Q} $$ should be = 1
$$ \therefore $$ P + Q must be 0
$$ \therefore $$ ($$\overline x + $$ y) + (y + $$\overline x $$) = 0
Now by checking each option you can see
When y = 0 and x = 1 then
$$\overline x + $$ y = 0
$$ \therefore $$ ($$\overline x + $$ y) + (y + $$\overline x $$) = 0
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