Boolean Algebra and Logic Gates Questions in English

(A) The given circuit consists of an $AND$ gate and an $OR$ gate.
Let the inputs be $A$ and $B$.
The output of the $AND$ gate is $A \cdot B$.
This output is one of the inputs to the $OR$ gate,while the other input to the $OR$ gate is $A$ directly.
Therefore,the final output $Y$ is given by the Boolean expression: $Y = A + (A \cdot B)$.
Using the Boolean algebra identity $A + (A \cdot B) = A(1 + B) = A \cdot 1 = A$,we find that $Y = A$.
Let's verify this with a truth table:
- If $A=0, B=0$: $Y = 0 + (0 \cdot 0) = 0 + 0 = 0$.
- If $A=0, B=1$: $Y = 0 + (0 \cdot 1) = 0 + 0 = 0$.
- If $A=1, B=0$: $Y = 1 + (1 \cdot 0) = 1 + 0 = 1$.
- If $A=1, B=1$: $Y = 1 + (1 \cdot 1) = 1 + 1 = 1$.
Comparing this with the given options,option $A$ matches this truth table.

(B) Let the inputs be $A$ and $B$.
$1$. The first $NAND$ gate produces output $C = \overline{A \cdot B}$.
$2$. The second $NAND$ gate (top) takes inputs $A$ and $C$, producing $D = \overline{A \cdot C} = \overline{A \cdot (\overline{A \cdot B})} = \overline{A \cdot (\overline{A} + \overline{B})} = \overline{A \cdot \overline{A} + A \cdot \overline{B}} = \overline{0 + A \cdot \overline{B}} = \overline{A \cdot \overline{B}} = \overline{A} + B$.
$3$. The third $NAND$ gate (bottom) takes inputs $B$ and $C$, producing $E = \overline{B \cdot C} = \overline{B \cdot (\overline{A \cdot B})} = \overline{B \cdot (\overline{A} + \overline{B})} = \overline{B \cdot \overline{A} + B \cdot \overline{B}} = \overline{B \cdot \overline{A} + 0} = \overline{B \cdot \overline{A}} = \overline{B} + A$.
$4$. The final $NAND$ gate takes inputs $D$ and $E$, producing $Y = \overline{D \cdot E} = \overline{(\overline{A} + B) \cdot (A + \overline{B})} = \overline{\overline{A} \cdot A + \overline{A} \cdot \overline{B} + B \cdot A + B \cdot \overline{B}} = \overline{0 + \overline{A} \cdot \overline{B} + A \cdot B + 0} = \overline{\overline{A} \cdot \overline{B} + A \cdot B} = A \oplus B$ ($XOR$ gate).
The correct table is $B$.

(B) The given circuit consists of two $NOT$ gates (formed by shorting the inputs of two $NOR$ gates) followed by a $NOR$ gate.
Let the inputs be $A$ and $B$.
The output of the first $NOR$ gate with inputs $A$ and $A$ is $Y_1 = \overline{A+A} = \overline{A}$.
The output of the second $NOR$ gate with inputs $B$ and $B$ is $Y_2 = \overline{B+B} = \overline{B}$.
These outputs $Y_1$ and $Y_2$ are fed into the final $NOR$ gate.
The final output $Y$ is given by $Y = \overline{Y_1 + Y_2} = \overline{\overline{A} + \overline{B}}$.
Using De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.
Thus,the final output is $Y = A \cdot B$,which represents an $AND$ gate.

(D) Let the output of the first $NAND$ gate be $P = \overline{A \cdot B}$.
The inputs to the top $NAND$ gate are $A$ and $P$,so its output is $Q = \overline{A \cdot P} = \overline{A \cdot \overline{A \cdot B}} = \overline{A} + (A \cdot B) = \overline{A} + B$.
The inputs to the bottom $NAND$ gate are $P$ and $B$,so its output is $R = \overline{P \cdot B} = \overline{\overline{A \cdot B} \cdot B} = (A \cdot B) + \overline{B} = A + \overline{B}$.
The final output $Y$ is the output of the last $NAND$ gate with inputs $Q$ and $R$:
$Y = \overline{Q \cdot R} = \overline{Q} + \overline{R} = \overline{\overline{A} + B} + \overline{A + \overline{B}} = (A \cdot \overline{B}) + (\overline{A} \cdot B)$.
This is the expression for an $XOR$ gate.
$1$. For $(A = 1, B = 0)$: $Y = (1 \cdot \overline{0}) + (\overline{1} \cdot 0) = (1 \cdot 1) + (0 \cdot 0) = 1 + 0 = 1$.
$2$. For $(A = 1, B = 1)$: $Y = (1 \cdot \overline{1}) + (\overline{1} \cdot 1) = (1 \cdot 0) + (0 \cdot 1) = 0 + 0 = 0$.
$3$. For $(A = 0, B = 0)$: $Y = (0 \cdot \overline{0}) + (\overline{0} \cdot 0) = (0 \cdot 1) + (1 \cdot 0) = 0 + 0 = 0$.
Thus,the outputs are $1, 0, 0$.

(A) The circuit consists of two $NAND$ gates. The top $NAND$ gate has inputs $A$ and $A$ (implied by the connection), but looking at the diagram, the top gate has input $A$ and the bottom gate has input $B$. Wait, let's re-examine the diagram: The top gate has input $A$ (connected to both terminals), acting as a $NOT$ gate. The bottom gate has input $B$ (connected to both terminals), also acting as a $NOT$ gate. The outputs of these two $NOT$ gates are fed into a third $NAND$ gate.
Let the output of the top $NAND$ gate be $Y_1 = \overline{A \cdot A} = \overline{A}$.
Let the output of the bottom $NAND$ gate be $Y_2 = \overline{B \cdot B} = \overline{B}$.
The final output $C$ is the $NAND$ of $Y_1$ and $Y_2$:
$C = \overline{Y_1 \cdot Y_2} = \overline{\overline{A} \cdot \overline{B}}$.
Using De Morgan's theorem, $\overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B$.
Thus, the circuit represents an $OR$ gate.

(D) The circuit consists of four $NOR$ gates $(G_1, G_2, G_3, G_4)$.
$G_1$ and $G_2$ act as $NOT$ gates because their inputs are tied together. Thus,the outputs are $\bar{A}$ and $\bar{B}$ respectively.
These are fed into $G_3$,which is a $NOR$ gate. Its output is $\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$ (by De Morgan's Law).
This output is fed into $G_4$,which acts as a $NOT$ gate. Its output is $\overline{A \cdot B}$.
Thus,the final output $Y = \overline{A \cdot B}$,which is the Boolean expression for a $NAND$ gate.

(D) The circuit consists of an $OR$ gate $(G_1)$ followed by a $NAND$ gate $(G_2)$.
Let the output of the $OR$ gate be $Y$. Then $Y = A + B$.
The final output $D$ is the output of the $NAND$ gate, which is $D = \overline{Y \cdot C} = \overline{(A + B) \cdot C}$.
Case $1$: $A=0, B=0, C=0$.
$Y = 0 + 0 = 0$.
$D = \overline{0 \cdot 0} = \overline{0} = 1$.
Case $2$: $A=1, B=1, C=0$.
$Y = 1 + 1 = 1$.
$D = \overline{1 \cdot 0} = \overline{0} = 1$.
Thus, the outputs are $1, 1$.

(B) By observing the given voltage waveforms,we can construct the truth table for the inputs $A$ and $B$ and the output $C$:

$A$	$B$	$C$
$1$	$1$	$1$
$1$	$0$	$0$
$0$	$1$	$0$
$0$	$0$	$0$

Comparing this truth table with the standard truth tables of logic gates,we find that the output $C$ is $1$ only when both inputs $A$ and $B$ are $1$. This is the characteristic behavior of an $AND$ gate. Therefore,the logic circuit gate is an $AND$ gate.

(C) The circuit consists of a $NOR$ gate followed by a $NOR$ gate acting as a $NOT$ gate (since both inputs are tied together).
Let the output of the first $NOR$ gate be $Y^{\prime} = \overline{A + B}$.
The second gate is a $NOR$ gate with both inputs as $Y^{\prime}$,so its output is $Y = \overline{Y^{\prime} + Y^{\prime}} = \overline{Y^{\prime}} = \overline{\overline{A + B}} = A + B$.
This is the Boolean expression for an $OR$ gate.
The truth table for $Y = A + B$ is:

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

Comparing this with the given options,option $C$ is correct.

(A) Let the inputs be $A$ and $B$.
$1$. The first gate is a $NOR$ gate,so its output is $X = \overline{A+B}$.
$2$. The second gate is a $NAND$ gate with both inputs tied together. $A$ $NAND$ gate with tied inputs acts as a $NOT$ gate. Thus,the output of the $NAND$ gate is $Z = \overline{X \cdot X} = \overline{X}$.
$3$. Substituting $X = \overline{A+B}$ into $Z$,we get $Z = \overline{\overline{A+B}} = A+B$.
$4$. The third gate is a $NOT$ gate,so its final output is $Y = \overline{Z} = \overline{A+B}$.
$5$. The expression $\overline{A+B}$ represents a $NOR$ gate.
Therefore,the circuit is equivalent to a $NOR$ gate.

(A) By observing the given logic gate symbols:
$(i)$ represents a $NAND$ gate.
(ii) represents a $NOT$ gate.
(iii) represents an $AND$ gate.
(iv) represents an $OR$ gate.
Therefore,the symbols for $OR$,$NOT$,and $NAND$ gates are $(iv)$,$(ii)$,and $(i)$ respectively.

(D) By observing the given voltage waveforms,we can construct the truth table for the logic gate:

$A$	$B$	$Y$
$1$	$1$	$0$
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$

From the truth table,it is clear that the output $Y$ is $0$ only when both inputs $A$ and $B$ are $1.$ In all other cases,the output is $1.$ This behavior corresponds to the $NAND$ gate.

(D) Based on the standard logic gate symbols:
$(i)$ represents an $OR$ gate.
(ii) represents an $AND$ gate.
(iii) represents a $NOT$ gate (specifically,a $NOT$ gate with two inputs is often represented as a $NOR$ or $NAND$ variant,but in this context,it is identified as a $NOT$ gate symbol).
(iv) represents a $NAND$ gate.
Therefore,the sequence for $AND, NAND$,and $NOT$ gates is $(ii), (iv)$,and $(iii)$ respectively.

(A) To identify the logic gate,we analyze the truth table from the given waveforms:

Time Interval	Input $A$	Input $B$	Output $C$
$0$ to $t_1$	$0$	$0$	$0$
$t_1$ to $t_2$	$1$	$0$	$1$
$t_2$ to $t_3$	$1$	$1$	$1$
$t_3$ to $t_4$	$0$	$1$	$1$
$t_4$ to $t_5$	$0$	$0$	$0$
$t_5$ to $t_6$	$1$	$0$	$1$

Comparing this with the truth tables of standard gates:
- For an $OR$ gate,the output is $1$ if at least one input is $1$.
- The observed output $C$ follows the condition: $C = 1$ if $A=1$ or $B=1$,and $C = 0$ if $A=0$ and $B=0$.
- This corresponds exactly to the truth table of an $OR$ gate.

(C) The first gate is a $NAND$ gate with inputs $A$ and $B$. Its output is $\overline{A.B}$.
This output is fed into a $NOT$ gate (a $NAND$ gate with both inputs connected together acts as a $NOT$ gate).
If the input to a $NOT$ gate is $Y$,the output is $\overline{Y}$.
Here,$Y = \overline{A.B}$,so the final output $X = \overline{\overline{A.B}}$.
Using the Boolean algebra property $\overline{\overline{Z}} = Z$,we get $X = A.B$.

(C) The given circuit consists of an $OR$ gate followed by an $AND$ gate.
Let the output of the $OR$ gate be $X$. Then $X = A + B$.
The final output $Y$ is the output of the $AND$ gate,which takes $X$ and $C$ as inputs.
Therefore,$Y = X \cdot C = (A + B) \cdot C$.
For the output $Y$ to be $1$,both inputs to the $AND$ gate must be $1$.
This means $X = 1$ and $C = 1$.
Since $X = A + B$,for $X$ to be $1$,at least one of $A$ or $B$ must be $1$.
Checking the options:
Option $A$: $A=1, B=0, C=0 \implies Y = (1+0) \cdot 0 = 0$.
Option $B$: $A=1, B=1, C=0 \implies Y = (1+1) \cdot 0 = 0$.
Option $C$: $A=1, B=0, C=1 \implies Y = (1+0) \cdot 1 = 1 \cdot 1 = 1$.
Option $D$: $A=0, B=1, C=0 \implies Y = (0+1) \cdot 0 = 0$.
Thus,the correct option is $C$.

(D) Let the inputs be $A$ and $B$. The outputs of the two $NOT$ gates are $\overline{A}$ and $\overline{B}$.
These are fed into a $NOR$ gate,so the output of the $NOR$ gate is $\overline{\overline{A} + \overline{B}}$.
By De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.
This output $(A \cdot B)$ and the input $B$ are fed into a $NAND$ gate.
The final output $Y = \overline{(A \cdot B) \cdot B} = \overline{A \cdot (B \cdot B)}$.
Since $B \cdot B = B$,we have $Y = \overline{A \cdot B}$.
This is the Boolean expression for a $NAND$ gate. However,looking at the provided circuit image,the final gate is a $NAND$ gate,and the expression simplifies to $Y = \overline{A \cdot B}$,which is a $NAND$ gate.

(D) The circuit consists of a $NOR$ gate followed by another $NOR$ gate acting as a $NOT$ gate (both inputs tied together),followed by a $NOT$ gate. Let the output of the first $NOR$ gate be $C = \overline{A+B}$.
This signal $C$ is fed into the second $NOR$ gate,which acts as a $NOT$ gate,producing $\overline{C} = \overline{\overline{A+B}} = A+B$.
Finally,this signal is passed through a $NOT$ gate,giving the final output $Y = \overline{\overline{C}} = \overline{A+B}$.
Thus,the circuit is equivalent to a $NOR$ gate.

(B) The given circuit consists of two $AND$ gates,two $NOT$ gates,and one $OR$ gate.
$1$. The upper $AND$ gate receives inputs $A$ and $\overline{B}$ (since $B$ passes through a $NOT$ gate). Its output is $A \cdot \overline{B}$.
$2$. The lower $AND$ gate receives inputs $\overline{A}$ (since $A$ passes through a $NOT$ gate) and $B$. Its output is $\overline{A} \cdot B$.
$3$. These two outputs are fed into an $OR$ gate. Therefore,the final output $Y$ is the sum of these two expressions: $Y = A \cdot \overline{B} + \overline{A} \cdot B$.
This is the Boolean expression for an $XOR$ (Exclusive-$OR$) gate.

(B) Using De Morgan's Law,$\overline {\bar A \cdot \bar B} = \overline{\bar A} + \overline{\bar B} = A + B$.
Substituting this into the expression: $(A + B) \cdot A$.
Expanding the expression: $A \cdot A + B \cdot A$.
Since $A \cdot A = A$,we get $A + AB$.
Factoring out $A$: $A(1 + B)$.
Since $(1 + B) = 1$,the expression simplifies to $A \cdot 1 = A$.

(A) In the given circuit,the inputs $A$ and $B$ are passed through $NOT$ gates before being fed into an $OR$ gate.
Let $A$ and $B$ be the input signals where $1$ represents high and $0$ represents low.
From the waveform,the sequence for $A$ is $1, 0, 1, 0$ and for $B$ is $1, 1, 0, 0$.
After passing through the $NOT$ gates,the inputs to the $OR$ gate become $A' = 0, 1, 0, 1$ and $B' = 0, 0, 1, 1$.
The $OR$ gate performs the operation $Y = A' + B'$.
Calculating the output sequence:
For the first interval: $0 + 0 = 0$
For the second interval: $1 + 0 = 1$
For the third interval: $0 + 1 = 1$
For the fourth interval: $1 + 1 = 1$
Thus,the resultant sequence for $Y$ is $0, 1, 1, 1$. This corresponds to option $A$.

(A) The given circuit consists of two diodes connected in parallel with a common resistor connected to the ground.
Let us analyze the truth table for this circuit:

$A$	$B$	$C$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

$1$. When both inputs $A$ and $B$ are at low potential $(0)$,both diodes are reverse-biased (or non-conducting),so the output $C$ is at ground potential $(0)$.
$2$. When either $A$ or $B$ is at high potential $(1)$,the corresponding diode becomes forward-biased and conducts,pulling the output $C$ to a high potential $(1)$.
$3$. This behavior corresponds to the Boolean expression $C = A + B$,which is the characteristic operation of an $OR$ gate.

(D) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate.
Let the inputs be $A$ and $B$. The outputs of the $NOT$ gates are $\bar{A}$ and $\bar{B}$.
These are fed into a $NOR$ gate,so the final output $Y$ is:
$Y = \overline{\bar{A} + \bar{B}}$
Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
Thus,the circuit acts as an $AND$ gate.
The truth table for an $AND$ gate is:

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

Comparing the input waveforms $A$ and $B$ with the $AND$ gate logic,the output $Y$ is high $(1)$ only when both $A$ and $B$ are high $(1)$. Looking at the provided waveforms,this corresponds to option $D$.

(B) The circuit consists of two $NAND$ gates acting as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.
Let the inputs be $A$ and $B$.
The output of the first $NAND$ gate is $\bar{A}$.
The output of the second $NAND$ gate is $\bar{B}$.
These two outputs are fed into the third $NAND$ gate.
The final output $X$ is given by the boolean expression:
$X = \overline{\bar{A} \cdot \bar{B}}$
Using De Morgan's theorem,$\overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.
Since the expression $X = A + B$ represents an $OR$ gate,the correct option is $B$.

(B) Let the output of the first $NAND$ gate be $Y_1 = \overline{A \cdot B}$.
This $Y_1$ is fed into the next two $NAND$ gates.
The output of the second $NAND$ gate is $Y_2 = \overline{A \cdot Y_1} = \overline{A \cdot (\overline{A \cdot B})} = \overline{A} + (A \cdot B) = \overline{A} + B$.
The output of the third $NAND$ gate is $Y_3 = \overline{B \cdot Y_1} = \overline{B \cdot (\overline{A \cdot B})} = \overline{B} + (A \cdot B) = \overline{B} + A$.
The final output $Y$ is the $NAND$ of $Y_2$ and $Y_3$:
$Y = \overline{Y_2 \cdot Y_3} = \overline{(\overline{A} + B) \cdot (\overline{B} + A)} = \overline{(\overline{A} \cdot \overline{B} + \overline{A} \cdot A + B \cdot \overline{B} + B \cdot A)} = \overline{(\overline{A} \cdot \overline{B} + 0 + 0 + A \cdot B)} = \overline{\overline{A} \cdot \overline{B}} + \overline{A \cdot B} = (A + B) \cdot (\overline{A} + \overline{B}) = A \cdot \overline{B} + B \cdot \overline{A}$.
This is the Boolean expression for an $XOR$ gate.
The truth table for an $XOR$ gate is:
If $A=0, B=0$, then $Y=0$.
If $A=0, B=1$, then $Y=1$.
If $A=1, B=0$, then $Y=1$.
If $A=1, B=1$, then $Y=0$.
This matches option $B$.

(A) By observing the time graph,we can analyze the relationship between inputs $a, b, c, d$ and output $x$.
In the given graph,the output $x$ becomes high $(1)$ as soon as any of the inputs $(a, b, c, d)$ becomes high $(1)$.
Specifically,at the beginning,all inputs are $0$ and the output $x$ is $0$.
As soon as the first pulse occurs in input $d$,the output $x$ transitions to $1$ and remains $1$ for the rest of the duration,regardless of the states of other inputs.
This behavior,where the output is $1$ if at least one input is $1$,is the characteristic truth table behavior of an $OR$ gate.
Therefore,the gate is an $OR$ gate.

(D) For the first circuit,the input $A$ and $B$ are passed through $NAND$ gates acting as $NOT$ gates,resulting in $\bar{A}$ and $\bar{B}$. These are then fed into a $NAND$ gate,giving $\overline{\bar{A} \cdot \bar{B}} = A + B$. This is then passed through another $NOT$ gate ($NAND$ as $NOT$),resulting in $C_1 = \overline{A + B} = \bar{A} \cdot \bar{B}$. Thus,$C_1$ is high $(1)$ only when both $A$ and $B$ are low $(0)$. Looking at the input graphs,$A=0$ and $B=0$ only in the interval $t = 4 \ s$ to $5 \ s$. Therefore,$C_1$ is high from $4 \ s$ to $5 \ s$.
For the second circuit,the inputs $A$ and $B$ are passed through $NOR$ gates acting as $NOT$ gates,resulting in $\bar{A}$ and $\bar{B}$. These are then fed into a $NOR$ gate,giving $C_2 = \overline{\bar{A} + \bar{B}} = A \cdot B$. Thus,$C_2$ is high $(1)$ only when both $A$ and $B$ are high $(1)$. Looking at the input graphs,$A=1$ and $B=1$ only in the interval $t = 1 \ s$ to $2 \ s$. Therefore,$C_2$ is high from $1 \ s$ to $2 \ s$.
Comparing this with the given options,the correct output signals match option $D$.

(C) The given circuit consists of two $NOT$ gates at the inputs,followed by a $NOR$ gate,and finally another $NOT$ gate at the output.
Let the inputs be $A$ and $B$.
The outputs of the first two $NOT$ gates are $\bar{A}$ and $\bar{B}$.
These are fed into a $NOR$ gate,so its output is $\overline{\bar{A} + \bar{B}}$.
By De Morgan's Law,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
This output is then passed through a final $NOT$ gate,so the final output $Y = \overline{A \cdot B}$.
This is the Boolean expression for a $NAND$ gate.
$A$ $NAND$ gate gives a low output $(0)$ only when both inputs are high $(1)$. In all other cases,the output is high $(1)$.
Comparing this logic with the given input waveforms,the output waveform corresponds to option $C$.

(D) To find the correct circuit, we analyze the given truth table conditions where $Y = 0$:
$1$. For $(A, B, C) = (0, 0, 0)$, $Y = 0$.
$2$. For $(A, B, C) = (0, 0, 1)$, $Y = 0$.
$3$. For $(A, B, C) = (0, 1, 0)$, $Y = 0$.
Let's evaluate the circuit in option $D$ (image $814-$d281), which consists of an $AND$ gate with inputs $B$ and $C$, followed by an $OR$ gate with inputs $A$ and the output of the $AND$ gate. The Boolean expression is $Y = A + (B \cdot C)$.
Testing the inputs:
- For $(0, 0, 0)$: $Y = 0 + (0 \cdot 0) = 0 + 0 = 0$.
- For $(0, 0, 1)$: $Y = 0 + (0 \cdot 1) = 0 + 0 = 0$.
- For $(0, 1, 0)$: $Y = 0 + (1 \cdot 0) = 0 + 0 = 0$.
- For $(0, 1, 1)$: $Y = 0 + (1 \cdot 1) = 0 + 1 = 1$.
- For $(1, 0, 0)$: $Y = 1 + (0 \cdot 0) = 1 + 0 = 1$.
- For $(1, 0, 1)$: $Y = 1 + (0 \cdot 1) = 1 + 0 = 1$.
- For $(1, 1, 0)$: $Y = 1 + (1 \cdot 0) = 1 + 0 = 1$.
- For $(1, 1, 1)$: $Y = 1 + (1 \cdot 1) = 1 + 1 = 1$.
This matches the required output conditions perfectly. Therefore, the correct circuit is $D$.

(A) The given circuit consists of an $OR$ gate and an $AND$ gate.
$1$. The inputs to the $OR$ gate are $A$ and $B$. Therefore,the output of the $OR$ gate is $(A + B)$.
$2$. The inputs to the $AND$ gate are $A$ and the output of the $OR$ gate,which is $(A + B)$.
$3$. The output $Y$ of the $AND$ gate is the product of its inputs: $Y = A \cdot (A + B)$.
$4$. Using the Boolean algebra law of absorption,$A \cdot (A + B) = A$.
$5$. Therefore,the final output is $Y = A$.

(B) Let the output of the first $NAND$ gate be $P = \overline{A \cdot B}$.
The inputs to the upper $NAND$ gate are $A$ and $P$, so its output is $Q = \overline{A \cdot P}$.
The inputs to the lower $NAND$ gate are $P$ and $B$, so its output is $R = \overline{P \cdot B}$.
The final output $Y$ is the output of a $NAND$ gate with inputs $Q$ and $R$, so $Y = \overline{Q \cdot R}$.
Using De Morgan's Law: $Y = \overline{Q} + \overline{R} = \overline{\overline{A \cdot P}} + \overline{\overline{P \cdot B}} = (A \cdot P) + (P \cdot B) = P \cdot (A + B)$.
Substituting $P = \overline{A \cdot B} = \overline{A} + \overline{B}$, we get $Y = (\overline{A} + \overline{B}) \cdot (A + B) = \overline{A} \cdot B + A \cdot \overline{B}$.
This is the expression for an $XOR$ gate.
For $A = 0, B = 0$: $Y = 0 \oplus 0 = 0$.
For $A = 1, B = 1$: $Y = 1 \oplus 1 = 0$.
For $A = 0, B = 1$: $Y = 0 \oplus 1 = 1$.
Thus, the outputs are $(0, 0, 1)$.

(D) Let the inputs to the two $NAND$ gates be $P$ and $Q$.
From the circuit diagram,the upper $NAND$ gate has inputs $\bar A$ and $B$. Its output is $P = \overline{\bar A \cdot B} = A + \bar B$.
The lower $NAND$ gate has inputs $A$ and $\bar B$. Its output is $Q = \overline{A \cdot \bar B} = \bar A + B$.
The final gate is a $NOR$ gate,so the output $y = \overline{P + Q}$.
Substituting the values of $P$ and $Q$:
$y = \overline{(A + \bar B) + (\bar A + B)}$
$y = \overline{(A + \bar A) + (B + \bar B)}$
Since $A + \bar A = 1$ and $B + \bar B = 1$:
$y = \overline{1 + 1} = \overline{1} = 0$.
Thus,the output $y = 0$.

(D) Let the inputs be $X$ and $Y$. The top branch consists of a $NOT$ gate followed by a $NOR$ gate. The input to the $NOR$ gate is $\bar{X}$ and $Y$. Thus,$P = \overline{\bar{X} + Y} = X \cdot \bar{Y}$.
The bottom branch consists of a $NAND$ gate with inputs $X$ and $\bar{Y}$. Thus,$Q = \overline{X \cdot \bar{Y}} = \bar{X} + Y$.
The final gate is a $NOR$ gate with inputs $P$ and $Q$. Thus,$R = \overline{P + Q} = \overline{(X \cdot \bar{Y}) + (\bar{X} + Y)}$.
Using De Morgan's laws,$R = \overline{(X \cdot \bar{Y})} \cdot \overline{(\bar{X} + Y)} = (\bar{X} + Y) \cdot (X \cdot \bar{Y})$.
Expanding this,$R = (\bar{X} \cdot X \cdot \bar{Y}) + (Y \cdot X \cdot \bar{Y}) = 0 + 0 = 0$.
Since the output $R$ is always $0$ regardless of the inputs $X$ and $Y$,it is not possible to obtain a high output $(1)$ at $R$.

(B) The output of the logic circuit is given by the Boolean expression: $Y = AB + \bar{A}\bar{B} + \bar{A}BC$.
Let us evaluate the output for each option:
$A$. For $A = 1, B = 0, C = 0$: $Y = (1)(0) + (0)(1) + (0)(0)(0) = 0 + 0 + 0 = 0$. (Incorrect)
$B$. For $A = 0, B = 1, C = 1$: $Y = (0)(1) + (1)(0) + (1)(1)(1) = 0 + 0 + 1 = 1$. (Correct)
$C$. For $A = 1, B = 1, C = 0$: $Y = (1)(1) + (0)(0) + (0)(1)(0) = 1 + 0 + 0 = 1$. (Incorrect)
$D$. For $A = 0, B = 1, C = 1$: $Y = (0)(1) + (1)(0) + (1)(1)(1) = 0 + 0 + 1 = 1$. (Incorrect)
Thus,the correct option is $B$.

(B) The circuit consists of two $NPN$ transistors connected in parallel. The collectors are connected to $+V_{cc}$,and the emitters are connected together to the output terminal,which is also grounded through a resistor.
$1$. If both inputs $A$ and $B$ are at low voltage $(0)$,both transistors are in the cut-off state. The output is pulled to ground through the resistor,so $Out = 0$.
$2$. If input $A$ is high $(1)$ and $B$ is low $(0)$,the top transistor conducts,and the output voltage becomes high $(1)$.
$3$. If input $A$ is low $(0)$ and $B$ is high $(1)$,the bottom transistor conducts,and the output voltage becomes high $(1)$.
$4$. If both inputs $A$ and $B$ are high $(1)$,both transistors conduct,and the output voltage becomes high $(1)$.
The truth table is:
$A=0, B=0 \implies Out=0$
$A=1, B=0 \implies Out=1$
$A=0, B=1 \implies Out=1$
$A=1, B=1 \implies Out=1$
This truth table corresponds to the $OR$ gate.

(A) Let the output of the $NAND$ gate be $C$. The boolean expression for $C$ is $C = \overline{A \cdot B}$.
Next, the top $AND$ gate takes inputs $A$ and $C$, giving output $D = A \cdot C = A \cdot (\overline{A \cdot B})$.
Similarly, the bottom $AND$ gate takes inputs $B$ and $C$, giving output $E = B \cdot C = B \cdot (\overline{A \cdot B})$.
The final output $X$ is the $OR$ of $D$ and $E$, so $X = D + E = A \cdot (\overline{A \cdot B}) + B \cdot (\overline{A \cdot B}) = (A + B) \cdot (\overline{A \cdot B})$.
Using the truth table:

$A$	$B$	$C = \overline{A \cdot B}$	$D = A \cdot C$	$E = B \cdot C$	$X = D + E$
$0$	$0$	$1$	$0$	$0$	$0$
$0$	$1$	$1$	$0$	$1$	$1$
$1$	$0$	$1$	$1$	$0$	$1$
$1$	$1$	$0$	$0$	$0$	$0$

This truth table corresponds to the $XOR$ gate.

(B) The problem states that the sensor output is $0$ when the water level is below the set value, and $1$ otherwise.
We want the pump to turn on $(P = 1)$ if at least one tank has a water level below the set value.
This means $P = 1$ if $S_1 = 0$ OR $S_2 = 0$.
Using Boolean algebra, this is equivalent to $P = \overline{S_1} + \overline{S_2}$.
By De Morgan's Law, $\overline{S_1} + \overline{S_2} = \overline{S_1 \cdot S_2}$.
This is the Boolean expression for a $NAND$ gate.
Looking at the options:
Option $A$ is an $OR$ gate fed by two $AND$ gates: $P = (S_1 \cdot S_2) + (S_1 \cdot S_2) = S_1 \cdot S_2$.
Option $B$ is an $OR$ gate fed by two $NAND$ gates: $P = \overline{S_1 \cdot S_2} + \overline{S_1 \cdot S_2} = \overline{S_1 \cdot S_2}$.
This matches our requirement $P = \overline{S_1 \cdot S_2}$.
Therefore, the correct circuit is $B$.

(C) In the given circuit,the output is high only when the circuit is closed.
Let $A=0$ when switch $A$ is at $3$,and $A=1$ when switch $A$ is at $4$.
Let $B=0$ when switch $B$ is at $1$,and $B=1$ when switch $B$ is at $2$.
The circuit is closed if the path is completed. This happens when $A$ and $B$ are in different states (i.e.,one is at $1/3$ and the other is at $2/4$).
Specifically,the circuit is closed if ($A=1$ and $B=0$) or ($A=0$ and $B=1$).
This corresponds to the Boolean expression $A \cdot \bar{B} + B \cdot \bar{A}$,which is the logic for an $XOR$ gate.

(D) The logic circuit consists of a $NOR$ gate followed by an $AND$ gate. The inputs to the $NOR$ gate are $A$ and $B$. The output of the $NOR$ gate is $\overline{A+B}$.
This output is then fed into an $AND$ gate along with input $B$. Thus,the final output $C$ is given by:
$C = (\overline{A+B}) \cdot B$
Let's analyze the truth table for this expression:
If $A=0, B=0$,then $C = (\overline{0+0}) \cdot 0 = 1 \cdot 0 = 0$.
If $A=1, B=0$,then $C = (\overline{1+0}) \cdot 0 = 0 \cdot 0 = 0$.
If $A=0, B=1$,then $C = (\overline{0+1}) \cdot 1 = 0 \cdot 1 = 0$.
If $A=1, B=1$,then $C = (\overline{1+1}) \cdot 1 = 0 \cdot 1 = 0$.
Wait,let's re-examine the circuit diagram. The input $B$ is connected to both the $NOR$ gate and the $AND$ gate. The output of the $NOR$ gate is $\overline{A+B}$. The $AND$ gate takes this and $B$ as inputs. So $C = (\overline{A+B}) \cdot B$. Since $(\overline{A+B}) \cdot B = (\overline{A} \cdot \overline{B}) \cdot B = \overline{A} \cdot (\overline{B} \cdot B) = \overline{A} \cdot 0 = 0$,the output is always $0$.

(D) The circuit consists of two $NAND$ gates whose outputs are fed into a $NOR$ gate. Let the inputs be $X, Y, Z$. The output of the first $NAND$ gate is $A = \overline{X \cdot Y}$ and the output of the second $NAND$ gate is $B = \overline{Y \cdot Z}$. These are the inputs to the $NOR$ gate,so the final output is $R = \overline{A + B} = \overline{\overline{X \cdot Y} + \overline{Y \cdot Z}}$. Using De Morgan's theorem,$R = \overline{\overline{X \cdot Y}} \cdot \overline{\overline{Y \cdot Z}} = (X \cdot Y) \cdot (Y \cdot Z) = X \cdot Y \cdot Z$. This is the Boolean expression for an $AND$ gate with three inputs. Thus,the system is equivalent to an $AND$ gate.

(D) The given circuit consists of two diodes with their cathodes connected together to a common output terminal $C$,which is connected to the ground through a resistor $R_2$.
If we apply inputs $A$ and $B$ (where $0$ represents low voltage and $1$ represents high voltage),the output $C$ is determined as follows:

$A$	$B$	$C$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

When both $A$ and $B$ are at $0$ (low),both diodes are reverse-biased or at the same potential,resulting in $C = 0$. If either $A$ or $B$ is at $1$ (high),the corresponding diode becomes forward-biased,pulling the output $C$ to a high state $(1)$. This truth table corresponds to an $OR$ gate.

(B) Using De Morgan's Law,$\overline {\bar A \cdot \bar B} = \overline{\bar A} + \overline{\bar B} = A + B$.
Substituting this into the expression:
$(A + B) \cdot A = A \cdot A + B \cdot A$.
Since $A \cdot A = A$,we have $A + AB$.
Factoring out $A$,we get $A(1 + B)$.
Since $(1 + B) = 1$,the expression simplifies to $A \cdot 1 = A$.

(C) The circuit consists of two $AND$ gates and one $OR$ gate. Let the inputs be $A$ and $B$.
$1$. The upper $AND$ gate receives inputs $A$ and $B$. Its output is $C = A \cdot B$.
$2$. The lower $AND$ gate receives inputs $\bar{A}$ (from the $NOT$ gate) and $B$. Its output is $D = \bar{A} \cdot B$.
$3$. The final output $E$ is the $OR$ combination of $C$ and $D$,so $E = C + D = (A \cdot B) + (\bar{A} \cdot B)$.
$4$. Factoring out $B$,we get $E = B(A + \bar{A})$. Since $A + \bar{A} = 1$,the expression simplifies to $E = B \cdot 1 = B$.
$5$. Constructing the truth table:
- If $A=0, B=0$,then $E=0$.
- If $A=0, B=1$,then $E=1$.
- If $A=1, B=0$,then $E=0$.
- If $A=1, B=1$,then $E=1$.
This matches the truth table where $E$ is identical to $B$.

(C) Let the inputs be $A$ and $B$. The first gate is an $OR$ gate with one input inverted,so its output is $Y' = \bar{A} + B$.
The second gate is an $AND$ gate with inputs $Y'$,$A$,and $B$. Its output is $Y'' = Y' \cdot A \cdot B = (\bar{A} + B) \cdot A \cdot B$.
Using Boolean algebra: $Y'' = (\bar{A} \cdot A \cdot B) + (B \cdot A \cdot B) = (0 \cdot B) + (A \cdot B) = A \cdot B$.
The final gate is a $NOT$ gate,so the final output is $Y = \overline{Y''} = \overline{A \cdot B}$.
This is the Boolean expression for a $NAND$ gate.
Truth table for the circuit:

$A$	$B$	$\bar{A}$	$Y' = \bar{A} + B$	$Y'' = Y' \cdot A \cdot B$	$Y = \overline{Y''}$
$0$	$0$	$1$	$1$	$0$	$1$
$0$	$1$	$1$	$1$	$0$	$1$
$1$	$0$	$0$	$0$	$0$	$1$
$1$	$1$	$0$	$1$	$1$	$0$

(D) The circuit consists of a $NAND$ gate and an $OR$ gate feeding into a final $OR$ gate.
Let the inputs be $A$ and $B$.
The output of the $NAND$ gate is $\overline{AB}$.
The output of the $OR$ gate is $A+B$.
The final output $Y$ is the $OR$ combination of these two: $Y = (A+B) + \overline{AB}$.
Using Boolean algebra: $Y = A + B + \bar{A} + \bar{B}$.
Since $A + \bar{A} = 1$ and $B + \bar{B} = 1$,the expression simplifies to $Y = 1 + 1 = 1$.
Thus,for all input combinations $(A, B)$,the output $Y$ is always $1$.
Therefore,the correct truth table is option $D$.

(A) The $NOR$ gate is a combination of an $OR$ gate followed by a $NOT$ gate.
The Boolean expression for an $OR$ gate is $Y = A + B$.
The Boolean expression for a $NOR$ gate is $Y = \overline{A + B}$.
According to this expression, if both inputs $A$ and $B$ are $0$ (low), then $Y = \overline{0 + 0} = \overline{0} = 1$ (high).
If any input is $1$ (high), the output becomes $0$ (low).
Therefore, the $NOR$ gate gives a high output only when both inputs are low.

(C) Let the inputs to the two $NOT$ gates be $A$ and $B$.
The outputs of the $NOT$ gates are $\bar{A}$ and $\bar{B}$ respectively.
These outputs are fed into an $EX-OR$ gate.
The output $Y$ of the $EX-OR$ gate is given by the Boolean expression:
$Y = \bar{A} \oplus \bar{B}$
Using the property of the $EX-OR$ gate,$X \oplus Y = X\bar{Y} + \bar{X}Y$,we get:
$Y = \bar{A}(\overline{\bar{B}}) + (\overline{\bar{A}})\bar{B}$
$Y = \bar{A}B + A\bar{B}$
This is the Boolean expression for an $EX-OR$ gate.
Therefore,the given combination is equivalent to an $EX-OR$ gate.

(A) Let the output of the circuit be $y$. The circuit consists of three $NAND$ gates $(G_1, G_2, G_3)$ and two $NOT$ gates.
$1$. The input to the first $NOT$ gate is $1$,so its output is $0$. This $0$ is fed to the second $NOT$ gate,making its output $1$. Thus,the input $B$ to gate $G_2$ is $1$.
$2$. The input to gate $G_3$ is $0$ (from the first $NOT$ gate) and $y$ (feedback from the output). The output of $G_3$ is $\overline{0 \cdot y} = \overline{0} = 1$.
$3$. The gate $G_1$ has inputs $0$ and $y$. Its output is $\overline{0 \cdot y} = \overline{0} = 1$. This output $1$ is the input $A$ to gate $G_2$.
$4$. Now,gate $G_2$ has inputs $A = 1$ and $B = 1$. Its output is $y = \overline{A \cdot B} = \overline{1 \cdot 1} = \overline{1} = 0$.
$5$. Since the output $y$ is $0$,the circuit is consistent. Therefore,the value of $y$ is $0$.

(C) $NAND$ gate performs the operation $Y = \overline{A \cdot B}$.
If the two inputs $A$ and $B$ are shorted, then $A = B = X$.
Substituting this into the $NAND$ expression, we get $Y = \overline{X \cdot X}$.
Since $X \cdot X = X$, the expression becomes $Y = \overline{X}$.
This is the Boolean expression for a $NOT$ gate.
Therefore, a $NAND$ gate with shorted inputs acts as a $NOT$ gate.

(A) The given logic circuit consists of two $NOT$ gates,two $AND$ gates,and one $OR$ gate. The output expression is derived as follows:
$1$. The upper branch gives the output $\overline{A} \cdot B$.
$2$. The lower branch gives the output $A \cdot \overline{B}$.
$3$. The final $OR$ gate combines these to give $Y = \overline{A}B + A\overline{B}$,which is the Boolean expression for an $XOR$ gate.
The truth table is:

$A$	$B$	$\overline{A}$	$\overline{B}$	$\overline{A} \cdot B$	$A \cdot \overline{B}$	$Y = \overline{A}B + A\overline{B}$
$0$	$0$	$1$	$1$	$0$	$0$	$0$
$0$	$1$	$1$	$0$	$1$	$0$	$1$
$1$	$0$	$0$	$1$	$0$	$1$	$1$
$1$	$1$	$0$	$0$	$0$	$0$	$0$