PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365207
Consider a two-input AND gate of figure below. Out of the four entries for the Truth Table given here, the correct ones are
1 All
2 1 and 2 only
3 1, 2 and 3 only
4 1, 3 and 4 only
Explanation:
Check the truth table
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365208
Identify the logic operation performed by the given circuit.
1 NAND
2 NOR
3 OR
4 AND
Explanation:
\(Y=\overline{\bar{A} \cdot \bar{B}}=\overline{\bar{A}}+\overline{\bar{B}}=A+B\) Thus, this combination is OR gate.
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365209
For the given digital circuit, write the truth table and identify the logic gate it represents
1 NAND gate
2 OR gate
3 AND gate
4 NOR gate
Explanation:
The output \(Y\) of the circuit is \(Y = \overline {\overline A + \overline B } = \overline {\overline A } .\overline {\overline B } = A.B\) It is Boolean expression for AND gate Hence the given digital circuit represents the AND gate.
KCET - 2014
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365210
To get an output \(Y = 1\) from the circuit shown, the inputs \(A\),\(B\) and \(C\) must be respectively
1 \(1,0,1\)
2 \(1,1,0\)
3 \(0,1,0\)
4 \(1,0,0\)
Explanation:
For \(Y\) to be 1, both \(C\) and \(D\) should be 1. For \(D\) to be 1, at least one of its inputs (\(A\) or \(B\)) should be 1. Hence option (1), where \(A = 1,B = 0\) and \(C = 1\), is correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365207
Consider a two-input AND gate of figure below. Out of the four entries for the Truth Table given here, the correct ones are
1 All
2 1 and 2 only
3 1, 2 and 3 only
4 1, 3 and 4 only
Explanation:
Check the truth table
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365208
Identify the logic operation performed by the given circuit.
1 NAND
2 NOR
3 OR
4 AND
Explanation:
\(Y=\overline{\bar{A} \cdot \bar{B}}=\overline{\bar{A}}+\overline{\bar{B}}=A+B\) Thus, this combination is OR gate.
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365209
For the given digital circuit, write the truth table and identify the logic gate it represents
1 NAND gate
2 OR gate
3 AND gate
4 NOR gate
Explanation:
The output \(Y\) of the circuit is \(Y = \overline {\overline A + \overline B } = \overline {\overline A } .\overline {\overline B } = A.B\) It is Boolean expression for AND gate Hence the given digital circuit represents the AND gate.
KCET - 2014
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365210
To get an output \(Y = 1\) from the circuit shown, the inputs \(A\),\(B\) and \(C\) must be respectively
1 \(1,0,1\)
2 \(1,1,0\)
3 \(0,1,0\)
4 \(1,0,0\)
Explanation:
For \(Y\) to be 1, both \(C\) and \(D\) should be 1. For \(D\) to be 1, at least one of its inputs (\(A\) or \(B\)) should be 1. Hence option (1), where \(A = 1,B = 0\) and \(C = 1\), is correct.
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365207
Consider a two-input AND gate of figure below. Out of the four entries for the Truth Table given here, the correct ones are
1 All
2 1 and 2 only
3 1, 2 and 3 only
4 1, 3 and 4 only
Explanation:
Check the truth table
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365208
Identify the logic operation performed by the given circuit.
1 NAND
2 NOR
3 OR
4 AND
Explanation:
\(Y=\overline{\bar{A} \cdot \bar{B}}=\overline{\bar{A}}+\overline{\bar{B}}=A+B\) Thus, this combination is OR gate.
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365209
For the given digital circuit, write the truth table and identify the logic gate it represents
1 NAND gate
2 OR gate
3 AND gate
4 NOR gate
Explanation:
The output \(Y\) of the circuit is \(Y = \overline {\overline A + \overline B } = \overline {\overline A } .\overline {\overline B } = A.B\) It is Boolean expression for AND gate Hence the given digital circuit represents the AND gate.
KCET - 2014
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365210
To get an output \(Y = 1\) from the circuit shown, the inputs \(A\),\(B\) and \(C\) must be respectively
1 \(1,0,1\)
2 \(1,1,0\)
3 \(0,1,0\)
4 \(1,0,0\)
Explanation:
For \(Y\) to be 1, both \(C\) and \(D\) should be 1. For \(D\) to be 1, at least one of its inputs (\(A\) or \(B\)) should be 1. Hence option (1), where \(A = 1,B = 0\) and \(C = 1\), is correct.
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365207
Consider a two-input AND gate of figure below. Out of the four entries for the Truth Table given here, the correct ones are
1 All
2 1 and 2 only
3 1, 2 and 3 only
4 1, 3 and 4 only
Explanation:
Check the truth table
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365208
Identify the logic operation performed by the given circuit.
1 NAND
2 NOR
3 OR
4 AND
Explanation:
\(Y=\overline{\bar{A} \cdot \bar{B}}=\overline{\bar{A}}+\overline{\bar{B}}=A+B\) Thus, this combination is OR gate.
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365209
For the given digital circuit, write the truth table and identify the logic gate it represents
1 NAND gate
2 OR gate
3 AND gate
4 NOR gate
Explanation:
The output \(Y\) of the circuit is \(Y = \overline {\overline A + \overline B } = \overline {\overline A } .\overline {\overline B } = A.B\) It is Boolean expression for AND gate Hence the given digital circuit represents the AND gate.
KCET - 2014
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365210
To get an output \(Y = 1\) from the circuit shown, the inputs \(A\),\(B\) and \(C\) must be respectively
1 \(1,0,1\)
2 \(1,1,0\)
3 \(0,1,0\)
4 \(1,0,0\)
Explanation:
For \(Y\) to be 1, both \(C\) and \(D\) should be 1. For \(D\) to be 1, at least one of its inputs (\(A\) or \(B\)) should be 1. Hence option (1), where \(A = 1,B = 0\) and \(C = 1\), is correct.
