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Choose the correct alternatives (More than one may be correct).

Two NAND gates having open collector outputs are tied together as shown in below figure.

The logic function $Y,$ implemented by the circuit is,

  1. $Y=ABC + DE$
  2. $Y=\overline{ABC + DE}$
  3. $Y=ABC.DE$
  4. $Y=\overline{ABC.DE}$
in Digital Logic by Boss (30.2k points) | 1.2k views

3 Answers

+3 votes
Best answer

There should be bubbled connection b/w two gates 

$Y=((ABC)' + (DE)' )'$

$\implies Y=(ABC).(DE)$

NOTE: open gate works as NOR gate.

by (167 points)
selected by
+3 votes

A Wired Connection can create both an AND gate and an OR gate , however it cannot implement a not gate .

but to know whether this intersection acts as AND gate or OR gate , we should know whether the intersection is at a HIGH or a LOW.

i think here they have assumed it to be at a HIGH , hence the answer will be Y=(ABC)' . (DE)' .

https://en.wikipedia.org/wiki/Wired_logic_connection

by Active (2.1k points)
0
The collector outputs have been connected. Therefore, the intersection is at HIGH.
0 votes
I think it is a or operation so A and b I think because  if we assume three wires are connected in y tri  joint now if any of the wire is having the current it will come to down wire also na sooo why and operation
by Active (1.8k points)
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