JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 15)
The output $Y$ for the inputs $A$ and $B$ of circuit is given by
_30th_January_Evening_Shift_en_15_1.png)
Truth table of the shown circuit is:
_30th_January_Evening_Shift_en_15_2.png)
_30th_January_Evening_Shift_en_15_3.png)
_30th_January_Evening_Shift_en_15_4.png)
_30th_January_Evening_Shift_en_15_5.png)
Explanation
_30th_January_Evening_Shift_en_15_6.png)
$$Y = \overline {\left( {\overline A + AB} \right)\left( {\overline B + AB} \right)} = \left( {A + B} \right)\left( {\overline A + \overline B } \right)$$
$$ = A\overline B + B\overline A $$ (XOR gate)
_30th_January_Evening_Shift_en_15_7.png)
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