Semiconductor Electronics Material Devices and Simple Circuits
151307
Which logic gate is represented by the following logic gates?
1 NOR
2 NAND
3 AND
4 \(\mathrm{OR}\)
Explanation:
C Given circuit, Then Boolean expression - \(\mathrm{Y}=\overline{\overline{\mathrm{A}}+\overline{\mathrm{B}}}\) Apply de-Morgan theorem, \(\mathrm{Y}=\overline{\overline{\mathrm{A}}} \cdot \overline{\overline{\mathrm{B}}}\) \(\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}\)Which is AND gate. So, logic gate is represent AND gate.
BCECE-2010
Semiconductor Electronics Material Devices and Simple Circuits
151309
The output of a NOT gate when its input is 0
1 Is 1
2 Is 0
3 Can be 0 or 1
4 Is 0 or 1
Explanation:
B When the input is 0 of NOT gate then output will be opposite. Which is 1 Boolean expression of NOT gate \(\mathrm{Y}=\overline{\mathrm{A}} \quad(\text { Where } \mathrm{A}=0 \text { ) }\) Truth table of NOT gate is - \((\because \mathrm{Y}=\overline{0}=1)\) {|c|c|c|} | Input (A) Output \((})\)\)| |---| \( \(\) 0 1\) \( \(\) 1 0\) \(
COMEDK 2012
Semiconductor Electronics Material Devices and Simple Circuits
151310
The output from a NAND gate is divided into two in parallel and fed to another NAND gate. The resulting gate is a
1 AND gate
2 NOR gate
3 OR gate
4 NOT gate
Explanation:
B The output of the logic gate is given as, \(\mathrm{C}=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) Then the Boolean expression will be- \(\mathrm{C}=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) \(\mathrm{C}=\mathrm{A} \cdot \mathrm{B}\) Which is AND gate. Hence, the resulting gate is AND gate.
COMEDK 2013
Semiconductor Electronics Material Devices and Simple Circuits
151314
Truth table for system of four NAND gates as shown in figure is
B Given circuit combination, Then Boolean expression \(\mathrm{Y}=\overline{(\overline{\mathrm{A}}+\mathrm{AB}) \cdot(\overline{\mathrm{B}}+\mathrm{AB})}\) Using De Morgan theorem \(Y=A \cdot \overline{A B}+B \cdot \overline{A B}\) \(Y=A \cdot(\bar{A}+\bar{B})+B(\bar{A}+\bar{B})\) \(Y=A \bar{A}+A \bar{B}+\bar{A} B+B \bar{B}\) \(Y=A \bar{B}+\bar{A} B \quad\left\{\begin{array}{l} A \bar{A}=0\) \(B \bar{B}=0 \end{array}\right\}\) Which is the expression for EX-OR gate and truth table {|l|l|l|} | \(\) \(\) \(=} +} \)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 0\) \(
Semiconductor Electronics Material Devices and Simple Circuits
151307
Which logic gate is represented by the following logic gates?
1 NOR
2 NAND
3 AND
4 \(\mathrm{OR}\)
Explanation:
C Given circuit, Then Boolean expression - \(\mathrm{Y}=\overline{\overline{\mathrm{A}}+\overline{\mathrm{B}}}\) Apply de-Morgan theorem, \(\mathrm{Y}=\overline{\overline{\mathrm{A}}} \cdot \overline{\overline{\mathrm{B}}}\) \(\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}\)Which is AND gate. So, logic gate is represent AND gate.
BCECE-2010
Semiconductor Electronics Material Devices and Simple Circuits
151309
The output of a NOT gate when its input is 0
1 Is 1
2 Is 0
3 Can be 0 or 1
4 Is 0 or 1
Explanation:
B When the input is 0 of NOT gate then output will be opposite. Which is 1 Boolean expression of NOT gate \(\mathrm{Y}=\overline{\mathrm{A}} \quad(\text { Where } \mathrm{A}=0 \text { ) }\) Truth table of NOT gate is - \((\because \mathrm{Y}=\overline{0}=1)\) {|c|c|c|} | Input (A) Output \((})\)\)| |---| \( \(\) 0 1\) \( \(\) 1 0\) \(
COMEDK 2012
Semiconductor Electronics Material Devices and Simple Circuits
151310
The output from a NAND gate is divided into two in parallel and fed to another NAND gate. The resulting gate is a
1 AND gate
2 NOR gate
3 OR gate
4 NOT gate
Explanation:
B The output of the logic gate is given as, \(\mathrm{C}=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) Then the Boolean expression will be- \(\mathrm{C}=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) \(\mathrm{C}=\mathrm{A} \cdot \mathrm{B}\) Which is AND gate. Hence, the resulting gate is AND gate.
COMEDK 2013
Semiconductor Electronics Material Devices and Simple Circuits
151314
Truth table for system of four NAND gates as shown in figure is
B Given circuit combination, Then Boolean expression \(\mathrm{Y}=\overline{(\overline{\mathrm{A}}+\mathrm{AB}) \cdot(\overline{\mathrm{B}}+\mathrm{AB})}\) Using De Morgan theorem \(Y=A \cdot \overline{A B}+B \cdot \overline{A B}\) \(Y=A \cdot(\bar{A}+\bar{B})+B(\bar{A}+\bar{B})\) \(Y=A \bar{A}+A \bar{B}+\bar{A} B+B \bar{B}\) \(Y=A \bar{B}+\bar{A} B \quad\left\{\begin{array}{l} A \bar{A}=0\) \(B \bar{B}=0 \end{array}\right\}\) Which is the expression for EX-OR gate and truth table {|l|l|l|} | \(\) \(\) \(=} +} \)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 0\) \(
NEET Test Series from KOTA - 10 Papers In MS WORD
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Semiconductor Electronics Material Devices and Simple Circuits
151307
Which logic gate is represented by the following logic gates?
1 NOR
2 NAND
3 AND
4 \(\mathrm{OR}\)
Explanation:
C Given circuit, Then Boolean expression - \(\mathrm{Y}=\overline{\overline{\mathrm{A}}+\overline{\mathrm{B}}}\) Apply de-Morgan theorem, \(\mathrm{Y}=\overline{\overline{\mathrm{A}}} \cdot \overline{\overline{\mathrm{B}}}\) \(\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}\)Which is AND gate. So, logic gate is represent AND gate.
BCECE-2010
Semiconductor Electronics Material Devices and Simple Circuits
151309
The output of a NOT gate when its input is 0
1 Is 1
2 Is 0
3 Can be 0 or 1
4 Is 0 or 1
Explanation:
B When the input is 0 of NOT gate then output will be opposite. Which is 1 Boolean expression of NOT gate \(\mathrm{Y}=\overline{\mathrm{A}} \quad(\text { Where } \mathrm{A}=0 \text { ) }\) Truth table of NOT gate is - \((\because \mathrm{Y}=\overline{0}=1)\) {|c|c|c|} | Input (A) Output \((})\)\)| |---| \( \(\) 0 1\) \( \(\) 1 0\) \(
COMEDK 2012
Semiconductor Electronics Material Devices and Simple Circuits
151310
The output from a NAND gate is divided into two in parallel and fed to another NAND gate. The resulting gate is a
1 AND gate
2 NOR gate
3 OR gate
4 NOT gate
Explanation:
B The output of the logic gate is given as, \(\mathrm{C}=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) Then the Boolean expression will be- \(\mathrm{C}=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) \(\mathrm{C}=\mathrm{A} \cdot \mathrm{B}\) Which is AND gate. Hence, the resulting gate is AND gate.
COMEDK 2013
Semiconductor Electronics Material Devices and Simple Circuits
151314
Truth table for system of four NAND gates as shown in figure is
B Given circuit combination, Then Boolean expression \(\mathrm{Y}=\overline{(\overline{\mathrm{A}}+\mathrm{AB}) \cdot(\overline{\mathrm{B}}+\mathrm{AB})}\) Using De Morgan theorem \(Y=A \cdot \overline{A B}+B \cdot \overline{A B}\) \(Y=A \cdot(\bar{A}+\bar{B})+B(\bar{A}+\bar{B})\) \(Y=A \bar{A}+A \bar{B}+\bar{A} B+B \bar{B}\) \(Y=A \bar{B}+\bar{A} B \quad\left\{\begin{array}{l} A \bar{A}=0\) \(B \bar{B}=0 \end{array}\right\}\) Which is the expression for EX-OR gate and truth table {|l|l|l|} | \(\) \(\) \(=} +} \)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 0\) \(
Semiconductor Electronics Material Devices and Simple Circuits
151307
Which logic gate is represented by the following logic gates?
1 NOR
2 NAND
3 AND
4 \(\mathrm{OR}\)
Explanation:
C Given circuit, Then Boolean expression - \(\mathrm{Y}=\overline{\overline{\mathrm{A}}+\overline{\mathrm{B}}}\) Apply de-Morgan theorem, \(\mathrm{Y}=\overline{\overline{\mathrm{A}}} \cdot \overline{\overline{\mathrm{B}}}\) \(\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}\)Which is AND gate. So, logic gate is represent AND gate.
BCECE-2010
Semiconductor Electronics Material Devices and Simple Circuits
151309
The output of a NOT gate when its input is 0
1 Is 1
2 Is 0
3 Can be 0 or 1
4 Is 0 or 1
Explanation:
B When the input is 0 of NOT gate then output will be opposite. Which is 1 Boolean expression of NOT gate \(\mathrm{Y}=\overline{\mathrm{A}} \quad(\text { Where } \mathrm{A}=0 \text { ) }\) Truth table of NOT gate is - \((\because \mathrm{Y}=\overline{0}=1)\) {|c|c|c|} | Input (A) Output \((})\)\)| |---| \( \(\) 0 1\) \( \(\) 1 0\) \(
COMEDK 2012
Semiconductor Electronics Material Devices and Simple Circuits
151310
The output from a NAND gate is divided into two in parallel and fed to another NAND gate. The resulting gate is a
1 AND gate
2 NOR gate
3 OR gate
4 NOT gate
Explanation:
B The output of the logic gate is given as, \(\mathrm{C}=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) Then the Boolean expression will be- \(\mathrm{C}=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) \(\mathrm{C}=\mathrm{A} \cdot \mathrm{B}\) Which is AND gate. Hence, the resulting gate is AND gate.
COMEDK 2013
Semiconductor Electronics Material Devices and Simple Circuits
151314
Truth table for system of four NAND gates as shown in figure is
B Given circuit combination, Then Boolean expression \(\mathrm{Y}=\overline{(\overline{\mathrm{A}}+\mathrm{AB}) \cdot(\overline{\mathrm{B}}+\mathrm{AB})}\) Using De Morgan theorem \(Y=A \cdot \overline{A B}+B \cdot \overline{A B}\) \(Y=A \cdot(\bar{A}+\bar{B})+B(\bar{A}+\bar{B})\) \(Y=A \bar{A}+A \bar{B}+\bar{A} B+B \bar{B}\) \(Y=A \bar{B}+\bar{A} B \quad\left\{\begin{array}{l} A \bar{A}=0\) \(B \bar{B}=0 \end{array}\right\}\) Which is the expression for EX-OR gate and truth table {|l|l|l|} | \(\) \(\) \(=} +} \)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 0\) \(
