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(A) In the given circuit,switches $A$ and $B$ are connected in parallel. The output $Y$ is the state of the $LED$. When both switches $A$ and $B$ are

Vedclass · Accepted Answer

$NOR$

Vedclass · Answer

(A) In the given circuit,switches $A$ and $B$ are connected in parallel. The output $Y$ is the state of the $LED$. When both switches $A$ and $B$ are at position $0$ (connected to ground),the $LED$ is connected across the $+6V$ supply through resistor $R$,so the $LED$ glows $(Y=1)$. If either switch $A$ or $B$ is moved to position $1$,the circuit is shorted to ground,and the potential difference across the $LED$ becomes zero,so the $LED$ does not glow $(Y=0)$. The truth table is: $A$$B$$Y$$0$$0$$1$$0$$1$$0$$1$$0$$0$$1$$1$$0$ This truth table corresponds to the $NOR$ gate.

The circuit diagram shown here corresponds to which logic gate?

Similar Questions

What is the output $D$ for the inputs $(A=0, B=0, C=0)$ and $(A=1, B=1, C=0)$ respectively?

The following diagram performs the logic function of:

Find the correct combination of $A, B, C$ and $D$ inputs which can cause the $LED$ to glow.

The output $Y$ of the following digital logic circuit will be '$1$' (one) for the inputs:

To obtain the truth-table shown, from the following logic circuit, the gate $G$ should be
$A$ $B$ $Y$
$0$ $0$ $1$
$0$ $1$ $0$
$1$ $0$ $1$
$1$ $1$ $1$

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