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MadeEasy WorkBook: Digital Logic - Adder
charul
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Oct 27, 2017
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Shubhanshu
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Oct 27, 2017
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only A = 1 is sufficient.
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A_i_$_h
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Oct 27, 2017
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@subhanshu
output of first nand gate is 0 ( already given)
A=0 will never satisfy that because then output of first nand gate will be 1
so A=1 ...is that so ?
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Shubhanshu
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Yes, because if we keep A = 0, then it will contradict the OP which is given 0 but we will get 1, So it is contradiction.
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As they have fixed the o/p for 2 nand gates, it is not the correct implementation of the half adder. But given the diagram we can say the given combination satisfy the circuit requirement:
A-1 B-1 C-0 D-1
So option (D)
srivivek95
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