Semiconductor Electronics Material Devices and Simple Circuits
151347
If \(A=1\) and \(B=0\), then in terms of Boolean algebra, \(\mathbf{A}+\overline{\mathbf{B}}=\)
1 \(\mathrm{B}\)
2 \(\overline{\mathrm{B}}\)
3 \(\mathrm{A}\)
4 \(\overline{\mathrm{A}}\)
Explanation:
C Given that, \(\mathrm{A}=1 \text { and } \mathrm{B}=0\) Then the value of Boolean expression, \(\mathrm{y}=\mathrm{A}+\overline{\mathrm{B}}\) \(\mathrm{y}=1+\overline{0}\) \(\mathrm{y}=1+1\) \(\mathrm{y}=1\) \(\mathrm{y}=\mathrm{A}\)Because the volume of \(\mathrm{A}=1\)
Karnataka CET-2018
Semiconductor Electronics Material Devices and Simple Circuits
151349
Identify the logic operation carried out by the following circuit :
1 AND
2 NAND
3 NOR
4 OR
Explanation:
D Given, The circuit combination will be Then, Boolean expression will be \(\mathrm{y} =\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}}\) \(\mathrm{y} =\overline{\overline{\mathrm{A}}}+\overline{\overline{\mathrm{B}}}\) \(\mathrm{y} =\mathrm{A}+\mathrm{B}\)The above expression represents OR gate,
Karnataka CET-2016
Semiconductor Electronics Material Devices and Simple Circuits
151351
The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as :
1 AND gate
2 NOT gate
3 NAND gate
4 NOR gate
Explanation:
D Given that the output of an OR gate is connected to both the input of a NAND gate then Output of NAND gate is, \(\mathrm{Z} =\overline{\mathrm{Y} \cdot \mathrm{Y}}\) \(=\overline{(\mathrm{A}+\mathrm{B}) \cdot(\mathrm{A}+\mathrm{B})}\) \(=\overline{\mathrm{A}+\mathrm{B}}\) \(\mathrm{y} =\overline{\mathrm{A}+\mathrm{B}}\)which represents NOR gate.
Karnataka CET-2013
Semiconductor Electronics Material Devices and Simple Circuits
151352
The following truth table with \(A\) and \(B\) as inputs is for ....... gate : {|l|l|l|} | { A } {c|}{ B } Output\)| |---|---| \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \(
1 NOR
2 AND
3 OR
4 XOR
Explanation:
D For an XOR gate, the output is high when one of them input is high. The Boolean expression is, \(\mathrm{y}=\overline{\mathrm{A}} \mathrm{B}+\overline{\mathrm{B}} \mathrm{A}\)Hence, the truth table with \(A\) and \(B\) as input is for XOR gate.
Semiconductor Electronics Material Devices and Simple Circuits
151347
If \(A=1\) and \(B=0\), then in terms of Boolean algebra, \(\mathbf{A}+\overline{\mathbf{B}}=\)
1 \(\mathrm{B}\)
2 \(\overline{\mathrm{B}}\)
3 \(\mathrm{A}\)
4 \(\overline{\mathrm{A}}\)
Explanation:
C Given that, \(\mathrm{A}=1 \text { and } \mathrm{B}=0\) Then the value of Boolean expression, \(\mathrm{y}=\mathrm{A}+\overline{\mathrm{B}}\) \(\mathrm{y}=1+\overline{0}\) \(\mathrm{y}=1+1\) \(\mathrm{y}=1\) \(\mathrm{y}=\mathrm{A}\)Because the volume of \(\mathrm{A}=1\)
Karnataka CET-2018
Semiconductor Electronics Material Devices and Simple Circuits
151349
Identify the logic operation carried out by the following circuit :
1 AND
2 NAND
3 NOR
4 OR
Explanation:
D Given, The circuit combination will be Then, Boolean expression will be \(\mathrm{y} =\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}}\) \(\mathrm{y} =\overline{\overline{\mathrm{A}}}+\overline{\overline{\mathrm{B}}}\) \(\mathrm{y} =\mathrm{A}+\mathrm{B}\)The above expression represents OR gate,
Karnataka CET-2016
Semiconductor Electronics Material Devices and Simple Circuits
151351
The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as :
1 AND gate
2 NOT gate
3 NAND gate
4 NOR gate
Explanation:
D Given that the output of an OR gate is connected to both the input of a NAND gate then Output of NAND gate is, \(\mathrm{Z} =\overline{\mathrm{Y} \cdot \mathrm{Y}}\) \(=\overline{(\mathrm{A}+\mathrm{B}) \cdot(\mathrm{A}+\mathrm{B})}\) \(=\overline{\mathrm{A}+\mathrm{B}}\) \(\mathrm{y} =\overline{\mathrm{A}+\mathrm{B}}\)which represents NOR gate.
Karnataka CET-2013
Semiconductor Electronics Material Devices and Simple Circuits
151352
The following truth table with \(A\) and \(B\) as inputs is for ....... gate : {|l|l|l|} | { A } {c|}{ B } Output\)| |---|---| \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \(
1 NOR
2 AND
3 OR
4 XOR
Explanation:
D For an XOR gate, the output is high when one of them input is high. The Boolean expression is, \(\mathrm{y}=\overline{\mathrm{A}} \mathrm{B}+\overline{\mathrm{B}} \mathrm{A}\)Hence, the truth table with \(A\) and \(B\) as input is for XOR gate.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Semiconductor Electronics Material Devices and Simple Circuits
151347
If \(A=1\) and \(B=0\), then in terms of Boolean algebra, \(\mathbf{A}+\overline{\mathbf{B}}=\)
1 \(\mathrm{B}\)
2 \(\overline{\mathrm{B}}\)
3 \(\mathrm{A}\)
4 \(\overline{\mathrm{A}}\)
Explanation:
C Given that, \(\mathrm{A}=1 \text { and } \mathrm{B}=0\) Then the value of Boolean expression, \(\mathrm{y}=\mathrm{A}+\overline{\mathrm{B}}\) \(\mathrm{y}=1+\overline{0}\) \(\mathrm{y}=1+1\) \(\mathrm{y}=1\) \(\mathrm{y}=\mathrm{A}\)Because the volume of \(\mathrm{A}=1\)
Karnataka CET-2018
Semiconductor Electronics Material Devices and Simple Circuits
151349
Identify the logic operation carried out by the following circuit :
1 AND
2 NAND
3 NOR
4 OR
Explanation:
D Given, The circuit combination will be Then, Boolean expression will be \(\mathrm{y} =\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}}\) \(\mathrm{y} =\overline{\overline{\mathrm{A}}}+\overline{\overline{\mathrm{B}}}\) \(\mathrm{y} =\mathrm{A}+\mathrm{B}\)The above expression represents OR gate,
Karnataka CET-2016
Semiconductor Electronics Material Devices and Simple Circuits
151351
The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as :
1 AND gate
2 NOT gate
3 NAND gate
4 NOR gate
Explanation:
D Given that the output of an OR gate is connected to both the input of a NAND gate then Output of NAND gate is, \(\mathrm{Z} =\overline{\mathrm{Y} \cdot \mathrm{Y}}\) \(=\overline{(\mathrm{A}+\mathrm{B}) \cdot(\mathrm{A}+\mathrm{B})}\) \(=\overline{\mathrm{A}+\mathrm{B}}\) \(\mathrm{y} =\overline{\mathrm{A}+\mathrm{B}}\)which represents NOR gate.
Karnataka CET-2013
Semiconductor Electronics Material Devices and Simple Circuits
151352
The following truth table with \(A\) and \(B\) as inputs is for ....... gate : {|l|l|l|} | { A } {c|}{ B } Output\)| |---|---| \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \(
1 NOR
2 AND
3 OR
4 XOR
Explanation:
D For an XOR gate, the output is high when one of them input is high. The Boolean expression is, \(\mathrm{y}=\overline{\mathrm{A}} \mathrm{B}+\overline{\mathrm{B}} \mathrm{A}\)Hence, the truth table with \(A\) and \(B\) as input is for XOR gate.
Semiconductor Electronics Material Devices and Simple Circuits
151347
If \(A=1\) and \(B=0\), then in terms of Boolean algebra, \(\mathbf{A}+\overline{\mathbf{B}}=\)
1 \(\mathrm{B}\)
2 \(\overline{\mathrm{B}}\)
3 \(\mathrm{A}\)
4 \(\overline{\mathrm{A}}\)
Explanation:
C Given that, \(\mathrm{A}=1 \text { and } \mathrm{B}=0\) Then the value of Boolean expression, \(\mathrm{y}=\mathrm{A}+\overline{\mathrm{B}}\) \(\mathrm{y}=1+\overline{0}\) \(\mathrm{y}=1+1\) \(\mathrm{y}=1\) \(\mathrm{y}=\mathrm{A}\)Because the volume of \(\mathrm{A}=1\)
Karnataka CET-2018
Semiconductor Electronics Material Devices and Simple Circuits
151349
Identify the logic operation carried out by the following circuit :
1 AND
2 NAND
3 NOR
4 OR
Explanation:
D Given, The circuit combination will be Then, Boolean expression will be \(\mathrm{y} =\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}}\) \(\mathrm{y} =\overline{\overline{\mathrm{A}}}+\overline{\overline{\mathrm{B}}}\) \(\mathrm{y} =\mathrm{A}+\mathrm{B}\)The above expression represents OR gate,
Karnataka CET-2016
Semiconductor Electronics Material Devices and Simple Circuits
151351
The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as :
1 AND gate
2 NOT gate
3 NAND gate
4 NOR gate
Explanation:
D Given that the output of an OR gate is connected to both the input of a NAND gate then Output of NAND gate is, \(\mathrm{Z} =\overline{\mathrm{Y} \cdot \mathrm{Y}}\) \(=\overline{(\mathrm{A}+\mathrm{B}) \cdot(\mathrm{A}+\mathrm{B})}\) \(=\overline{\mathrm{A}+\mathrm{B}}\) \(\mathrm{y} =\overline{\mathrm{A}+\mathrm{B}}\)which represents NOR gate.
Karnataka CET-2013
Semiconductor Electronics Material Devices and Simple Circuits
151352
The following truth table with \(A\) and \(B\) as inputs is for ....... gate : {|l|l|l|} | { A } {c|}{ B } Output\)| |---|---| \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \(
1 NOR
2 AND
3 OR
4 XOR
Explanation:
D For an XOR gate, the output is high when one of them input is high. The Boolean expression is, \(\mathrm{y}=\overline{\mathrm{A}} \mathrm{B}+\overline{\mathrm{B}} \mathrm{A}\)Hence, the truth table with \(A\) and \(B\) as input is for XOR gate.
