PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365215
In the truth table of the above circuit the value of \(X\) and \(Y\) are
1 0,0
2 0,1
3 1,0
4 1,1
Explanation:
\(y_{1}=A \cdot B, y_{3}=\bar{A}, y_{2}=\bar{B}, y_{4}=y_{2} \cdot y_{3}\) and \(E=\overline{y_{1}+y_{4}}\) For \(X: A=0, B=1\) and for \(Y: A=1, B=0\)
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365216
Given below are four logic gate symbols. Those for OR, NOR and NAND are respectively.
1 (1), (4) and (3)
2 (4), (1) and (2)
3 (1), (3) and (4)
4 (2), (3) and (4)
Explanation:
Gate (1) \(\rightarrow\) AND, Gate (2) \(\rightarrow\) OR,Gate (3) \(\rightarrow\) NOR and Gate (4) \(\rightarrow\) NAND
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365217
What is the output \(Y\) in the following circuit, when all the three inputs \(A\), \(B\), \(C\) are first 0 and then 1 ?
1 \(0,1\)
2 \(0,0\)
3 \(1,0\)
4 \(1,1\)
Explanation:
\(Y = \overline {(A \cdot B) \cdot C} \)
NEET - 2016
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365218
An AND gate is followed by a NOT gate in series. With two inputs \(A\) & \(B\), the Boolean expression for the output \(Y\) will be :
1 \(A.B\)
2 \(A + B\)
3 \({\rm{\bar A + \bar B}}\)
4 \({\rm{\bar A}} \cdot {\rm{\bar B}}\)
Explanation:
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365219
The output of a 'NAND' gate is shown in the truth-table (\(A\) and \(B\) are inputs, \(Y\) is output)
1 \(P\)
2 \(R\)
3 \(Q\)
4 \(S\)
Explanation:
Truth-table of ' \(N A N D\) ' gate is \(Y=\overline{A B}\) Note: For output of \((A N D)\) to be 1, all inputs to be 1 , \(\Rightarrow\) For output of (NAND) to be 0 all of inputs to be 1 . So, correct option is (4)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365215
In the truth table of the above circuit the value of \(X\) and \(Y\) are
1 0,0
2 0,1
3 1,0
4 1,1
Explanation:
\(y_{1}=A \cdot B, y_{3}=\bar{A}, y_{2}=\bar{B}, y_{4}=y_{2} \cdot y_{3}\) and \(E=\overline{y_{1}+y_{4}}\) For \(X: A=0, B=1\) and for \(Y: A=1, B=0\)
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365216
Given below are four logic gate symbols. Those for OR, NOR and NAND are respectively.
1 (1), (4) and (3)
2 (4), (1) and (2)
3 (1), (3) and (4)
4 (2), (3) and (4)
Explanation:
Gate (1) \(\rightarrow\) AND, Gate (2) \(\rightarrow\) OR,Gate (3) \(\rightarrow\) NOR and Gate (4) \(\rightarrow\) NAND
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365217
What is the output \(Y\) in the following circuit, when all the three inputs \(A\), \(B\), \(C\) are first 0 and then 1 ?
1 \(0,1\)
2 \(0,0\)
3 \(1,0\)
4 \(1,1\)
Explanation:
\(Y = \overline {(A \cdot B) \cdot C} \)
NEET - 2016
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365218
An AND gate is followed by a NOT gate in series. With two inputs \(A\) & \(B\), the Boolean expression for the output \(Y\) will be :
1 \(A.B\)
2 \(A + B\)
3 \({\rm{\bar A + \bar B}}\)
4 \({\rm{\bar A}} \cdot {\rm{\bar B}}\)
Explanation:
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365219
The output of a 'NAND' gate is shown in the truth-table (\(A\) and \(B\) are inputs, \(Y\) is output)
1 \(P\)
2 \(R\)
3 \(Q\)
4 \(S\)
Explanation:
Truth-table of ' \(N A N D\) ' gate is \(Y=\overline{A B}\) Note: For output of \((A N D)\) to be 1, all inputs to be 1 , \(\Rightarrow\) For output of (NAND) to be 0 all of inputs to be 1 . So, correct option is (4)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365215
In the truth table of the above circuit the value of \(X\) and \(Y\) are
1 0,0
2 0,1
3 1,0
4 1,1
Explanation:
\(y_{1}=A \cdot B, y_{3}=\bar{A}, y_{2}=\bar{B}, y_{4}=y_{2} \cdot y_{3}\) and \(E=\overline{y_{1}+y_{4}}\) For \(X: A=0, B=1\) and for \(Y: A=1, B=0\)
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365216
Given below are four logic gate symbols. Those for OR, NOR and NAND are respectively.
1 (1), (4) and (3)
2 (4), (1) and (2)
3 (1), (3) and (4)
4 (2), (3) and (4)
Explanation:
Gate (1) \(\rightarrow\) AND, Gate (2) \(\rightarrow\) OR,Gate (3) \(\rightarrow\) NOR and Gate (4) \(\rightarrow\) NAND
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365217
What is the output \(Y\) in the following circuit, when all the three inputs \(A\), \(B\), \(C\) are first 0 and then 1 ?
1 \(0,1\)
2 \(0,0\)
3 \(1,0\)
4 \(1,1\)
Explanation:
\(Y = \overline {(A \cdot B) \cdot C} \)
NEET - 2016
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365218
An AND gate is followed by a NOT gate in series. With two inputs \(A\) & \(B\), the Boolean expression for the output \(Y\) will be :
1 \(A.B\)
2 \(A + B\)
3 \({\rm{\bar A + \bar B}}\)
4 \({\rm{\bar A}} \cdot {\rm{\bar B}}\)
Explanation:
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365219
The output of a 'NAND' gate is shown in the truth-table (\(A\) and \(B\) are inputs, \(Y\) is output)
1 \(P\)
2 \(R\)
3 \(Q\)
4 \(S\)
Explanation:
Truth-table of ' \(N A N D\) ' gate is \(Y=\overline{A B}\) Note: For output of \((A N D)\) to be 1, all inputs to be 1 , \(\Rightarrow\) For output of (NAND) to be 0 all of inputs to be 1 . So, correct option is (4)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365215
In the truth table of the above circuit the value of \(X\) and \(Y\) are
1 0,0
2 0,1
3 1,0
4 1,1
Explanation:
\(y_{1}=A \cdot B, y_{3}=\bar{A}, y_{2}=\bar{B}, y_{4}=y_{2} \cdot y_{3}\) and \(E=\overline{y_{1}+y_{4}}\) For \(X: A=0, B=1\) and for \(Y: A=1, B=0\)
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365216
Given below are four logic gate symbols. Those for OR, NOR and NAND are respectively.
1 (1), (4) and (3)
2 (4), (1) and (2)
3 (1), (3) and (4)
4 (2), (3) and (4)
Explanation:
Gate (1) \(\rightarrow\) AND, Gate (2) \(\rightarrow\) OR,Gate (3) \(\rightarrow\) NOR and Gate (4) \(\rightarrow\) NAND
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365217
What is the output \(Y\) in the following circuit, when all the three inputs \(A\), \(B\), \(C\) are first 0 and then 1 ?
1 \(0,1\)
2 \(0,0\)
3 \(1,0\)
4 \(1,1\)
Explanation:
\(Y = \overline {(A \cdot B) \cdot C} \)
NEET - 2016
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365218
An AND gate is followed by a NOT gate in series. With two inputs \(A\) & \(B\), the Boolean expression for the output \(Y\) will be :
1 \(A.B\)
2 \(A + B\)
3 \({\rm{\bar A + \bar B}}\)
4 \({\rm{\bar A}} \cdot {\rm{\bar B}}\)
Explanation:
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365219
The output of a 'NAND' gate is shown in the truth-table (\(A\) and \(B\) are inputs, \(Y\) is output)
1 \(P\)
2 \(R\)
3 \(Q\)
4 \(S\)
Explanation:
Truth-table of ' \(N A N D\) ' gate is \(Y=\overline{A B}\) Note: For output of \((A N D)\) to be 1, all inputs to be 1 , \(\Rightarrow\) For output of (NAND) to be 0 all of inputs to be 1 . So, correct option is (4)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365215
In the truth table of the above circuit the value of \(X\) and \(Y\) are
1 0,0
2 0,1
3 1,0
4 1,1
Explanation:
\(y_{1}=A \cdot B, y_{3}=\bar{A}, y_{2}=\bar{B}, y_{4}=y_{2} \cdot y_{3}\) and \(E=\overline{y_{1}+y_{4}}\) For \(X: A=0, B=1\) and for \(Y: A=1, B=0\)
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365216
Given below are four logic gate symbols. Those for OR, NOR and NAND are respectively.
1 (1), (4) and (3)
2 (4), (1) and (2)
3 (1), (3) and (4)
4 (2), (3) and (4)
Explanation:
Gate (1) \(\rightarrow\) AND, Gate (2) \(\rightarrow\) OR,Gate (3) \(\rightarrow\) NOR and Gate (4) \(\rightarrow\) NAND
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365217
What is the output \(Y\) in the following circuit, when all the three inputs \(A\), \(B\), \(C\) are first 0 and then 1 ?
1 \(0,1\)
2 \(0,0\)
3 \(1,0\)
4 \(1,1\)
Explanation:
\(Y = \overline {(A \cdot B) \cdot C} \)
NEET - 2016
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365218
An AND gate is followed by a NOT gate in series. With two inputs \(A\) & \(B\), the Boolean expression for the output \(Y\) will be :
1 \(A.B\)
2 \(A + B\)
3 \({\rm{\bar A + \bar B}}\)
4 \({\rm{\bar A}} \cdot {\rm{\bar B}}\)
Explanation:
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365219
The output of a 'NAND' gate is shown in the truth-table (\(A\) and \(B\) are inputs, \(Y\) is output)
1 \(P\)
2 \(R\)
3 \(Q\)
4 \(S\)
Explanation:
Truth-table of ' \(N A N D\) ' gate is \(Y=\overline{A B}\) Note: For output of \((A N D)\) to be 1, all inputs to be 1 , \(\Rightarrow\) For output of (NAND) to be 0 all of inputs to be 1 . So, correct option is (4)
