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In the circuit, the logical value of \mathrm{A=1 \: or\: B=1} when potential at \mathrm{A \: or \: B} is 5 \mathrm{~V} and the logical value of \mathrm{A=0 \: or \: B=0} when potential at \mathrm{A\: or \: B \: is \: 0 \mathrm{~V}}.


The truth table of the given circuit will be :

Option: 1

\begin{array}{lll} \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{array}


Option: 2

\begin{array}{lll} \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{array}


Option: 3

\begin{array}{ccc} \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}


Option: 4

\begin{array}{ccc} \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{array}


Answers (1)

The given circuit is AND gate realisation.

\mathrm{Y=A.B}
\therefore The truth table of given Circuit will be as given in option (1)
Hence (1) is correct option.

Posted by

Ramraj Saini

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