PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365220
In the combination of the following gates the output \(Y\) can be written in terms of inputs \(A\) and \(B\) as
1 \(\overline {A \cdot B} + A \cdot B\)
2 \(\overline {A + B} \)
3 \(A \cdot \overline B + \overline A \cdot B\)
4 \(\overline {A \cdot B} \)
Explanation:
\(A \cdot \overline B + \overline A \cdot B = Y\)
NEET - 2018
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365221
Identify the logic gate given in the circuit.
1 NOR gate
2 NAND gate
3 OR gate
4 AND gate
Explanation:
\(Y = \overline {{Y_1} \cdot {Y_2}} = \overline {\overline A \cdot \overline B } \) \( = \mathop A\limits^{\,\,\,\,\, = } + \mathop B\limits^{\,\,\, = } = A + B\) This combination is OR gate.
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365222
Assertion : The output voltage and the input voltage of the NOT gate have \(180^{\circ}\) phase difference. Reason : The logic NOT can be built using diode.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Output \(Y=\bar{A}\) if input is \(A\). The output of a NOT gate is in the opposite phase compared to its input. The logic function NOT cannot be implemented using only a diode. It can be done using transistor. So correct option is (3).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365223
Which logic gate is represented by the following combination of logic gates?
1 OR
2 NAND
3 AND
4 NOR
Explanation:
Combination of logic gates gives the output, \(Y = \overline {\bar A + \bar B} \quad [\because {\text{ Gate is NOR gate}}]\) \(\therefore \quad Y = \overline {\overline A } \cdot \overline {\overline B } = A \cdot B\) This is boolean expression for AND gate. So correct option is (3)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365220
In the combination of the following gates the output \(Y\) can be written in terms of inputs \(A\) and \(B\) as
1 \(\overline {A \cdot B} + A \cdot B\)
2 \(\overline {A + B} \)
3 \(A \cdot \overline B + \overline A \cdot B\)
4 \(\overline {A \cdot B} \)
Explanation:
\(A \cdot \overline B + \overline A \cdot B = Y\)
NEET - 2018
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365221
Identify the logic gate given in the circuit.
1 NOR gate
2 NAND gate
3 OR gate
4 AND gate
Explanation:
\(Y = \overline {{Y_1} \cdot {Y_2}} = \overline {\overline A \cdot \overline B } \) \( = \mathop A\limits^{\,\,\,\,\, = } + \mathop B\limits^{\,\,\, = } = A + B\) This combination is OR gate.
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365222
Assertion : The output voltage and the input voltage of the NOT gate have \(180^{\circ}\) phase difference. Reason : The logic NOT can be built using diode.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Output \(Y=\bar{A}\) if input is \(A\). The output of a NOT gate is in the opposite phase compared to its input. The logic function NOT cannot be implemented using only a diode. It can be done using transistor. So correct option is (3).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365223
Which logic gate is represented by the following combination of logic gates?
1 OR
2 NAND
3 AND
4 NOR
Explanation:
Combination of logic gates gives the output, \(Y = \overline {\bar A + \bar B} \quad [\because {\text{ Gate is NOR gate}}]\) \(\therefore \quad Y = \overline {\overline A } \cdot \overline {\overline B } = A \cdot B\) This is boolean expression for AND gate. So correct option is (3)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365220
In the combination of the following gates the output \(Y\) can be written in terms of inputs \(A\) and \(B\) as
1 \(\overline {A \cdot B} + A \cdot B\)
2 \(\overline {A + B} \)
3 \(A \cdot \overline B + \overline A \cdot B\)
4 \(\overline {A \cdot B} \)
Explanation:
\(A \cdot \overline B + \overline A \cdot B = Y\)
NEET - 2018
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365221
Identify the logic gate given in the circuit.
1 NOR gate
2 NAND gate
3 OR gate
4 AND gate
Explanation:
\(Y = \overline {{Y_1} \cdot {Y_2}} = \overline {\overline A \cdot \overline B } \) \( = \mathop A\limits^{\,\,\,\,\, = } + \mathop B\limits^{\,\,\, = } = A + B\) This combination is OR gate.
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365222
Assertion : The output voltage and the input voltage of the NOT gate have \(180^{\circ}\) phase difference. Reason : The logic NOT can be built using diode.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Output \(Y=\bar{A}\) if input is \(A\). The output of a NOT gate is in the opposite phase compared to its input. The logic function NOT cannot be implemented using only a diode. It can be done using transistor. So correct option is (3).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365223
Which logic gate is represented by the following combination of logic gates?
1 OR
2 NAND
3 AND
4 NOR
Explanation:
Combination of logic gates gives the output, \(Y = \overline {\bar A + \bar B} \quad [\because {\text{ Gate is NOR gate}}]\) \(\therefore \quad Y = \overline {\overline A } \cdot \overline {\overline B } = A \cdot B\) This is boolean expression for AND gate. So correct option is (3)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365220
In the combination of the following gates the output \(Y\) can be written in terms of inputs \(A\) and \(B\) as
1 \(\overline {A \cdot B} + A \cdot B\)
2 \(\overline {A + B} \)
3 \(A \cdot \overline B + \overline A \cdot B\)
4 \(\overline {A \cdot B} \)
Explanation:
\(A \cdot \overline B + \overline A \cdot B = Y\)
NEET - 2018
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365221
Identify the logic gate given in the circuit.
1 NOR gate
2 NAND gate
3 OR gate
4 AND gate
Explanation:
\(Y = \overline {{Y_1} \cdot {Y_2}} = \overline {\overline A \cdot \overline B } \) \( = \mathop A\limits^{\,\,\,\,\, = } + \mathop B\limits^{\,\,\, = } = A + B\) This combination is OR gate.
JEE - 2024
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365222
Assertion : The output voltage and the input voltage of the NOT gate have \(180^{\circ}\) phase difference. Reason : The logic NOT can be built using diode.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Output \(Y=\bar{A}\) if input is \(A\). The output of a NOT gate is in the opposite phase compared to its input. The logic function NOT cannot be implemented using only a diode. It can be done using transistor. So correct option is (3).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365223
Which logic gate is represented by the following combination of logic gates?
1 OR
2 NAND
3 AND
4 NOR
Explanation:
Combination of logic gates gives the output, \(Y = \overline {\bar A + \bar B} \quad [\because {\text{ Gate is NOR gate}}]\) \(\therefore \quad Y = \overline {\overline A } \cdot \overline {\overline B } = A \cdot B\) This is boolean expression for AND gate. So correct option is (3)
