Semiconductor Electronics Material Devices and Simple Circuits
151297
Which logic gate is represented by the following combination of logic gates?
1 \(\mathrm{OR}\)
2 NAND
3 AND
4 NOR
Explanation:
C Given that in this circuit using two NOT gate and one NOR gate Then, the Boolean expression will be- \(\mathrm{Y}=\overline{(\overline{\mathrm{A}}+\overline{\mathrm{B}})}=\mathrm{A} \cdot \mathrm{B}\) Truth table of \(\mathrm{Y}-\) {|c|c|c|c|c|c|} | \(\) \(\) \(}\) \(}\) \(}+}\) \(}+}}\)\)| |---| \( 0 0 1 1 1 0\) \( 0 1 1 0 1 0\) \( 1 0 0 1 1 0\) \( 1 1 0 0 0 1\) \( Which is output expression of AND gate.
JCECE-2016
Semiconductor Electronics Material Devices and Simple Circuits
151298
The circuit is equivalent to
1 NOR gate
2 AND gate
3 NAND gate
4 OR gate
Explanation:
C Given Circuit, \(\overline{\mathrm{Y}}_2=(\overline{\overline{\mathrm{A} \cdot \mathrm{B}}+\overline{\mathrm{A} \cdot \mathrm{B}}})\) \(\overline{\mathrm{Y}}_2=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}} \cdot \overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) \(\overline{\mathrm{Y}}_2=(\mathrm{A} \cdot \mathrm{B}) \cdot(\mathrm{A} \cdot \mathrm{B})\) \(\overline{\mathrm{Y}}_2=\mathrm{AB}\) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\) (apply NOR gate) (apply NOT gate) So output, \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)Which is the expression of NAND gate.
JCECE-2015
Semiconductor Electronics Material Devices and Simple Circuits
151299
In the adjoining circuit of logic gate, the output \(Y\) becomes zero if the inputs are
1 \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\)
2 \(\mathrm{A}=0, \mathrm{~B}=0, \mathrm{C}=0\)
3 \(\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=1\)
4 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0\)
Explanation:
B Given, circuit output \(\mathrm{Y}=0\) then- The expression of given circuit, \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B} \cdot \overline{\mathrm{C}}} \quad\) (using De, Morgen theorems) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}+\overline{\overline{\mathrm{C}}} \quad(\overline{\overline{\mathrm{C}}}=\mathrm{C} \rightarrow \text { Involution theorem })\) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}+\mathrm{C}=\overline{\mathrm{A}}+\overline{\mathrm{B}}+\mathrm{C}\) Then check option (1), \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\) \(\mathrm{Y}=\overline{1}+\overline{1}+0\) \(\mathrm{Y}=0+0+0\) \(\mathrm{Y}=0\)This is possible when \(\mathrm{A}=1, \mathrm{~B}=1\) and \(\mathrm{C}=0\).
JCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151300
What is an AND gate?
1 It has not equivalence to switching circuit
2 It is equivalent to switching circuit
3 It is equivalent to parallel switching circuit
4 It is a mixture of series and parallel switching
Explanation:
B AND gate is used as a switching circuit, AND gate is only active when both inputs are high same as series. Switching circuit which is active only when both switches are closed. The truth table of AND gate will be - Boolean Expression, \(\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}\) {|l|l|l|} | \(\) \(\) \(=\)\)| |---| \( 0 0 0\) \(0 1 0\) \(1 0 0\) \(1 1 1\) \( When both input are high then we get high output
JCECE-2011
Semiconductor Electronics Material Devices and Simple Circuits
151301
To get an output 1 from the circuit shown in the figure, the input must be
1 \(\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=0\)
2 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0\)
3 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1\)
4 \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\)
Explanation:
C Given circuit when output is 1 . In this circuit using OR gate and AND gate. So Boolean expression- \(\mathrm{Y}=(\mathrm{A}+\mathrm{B}) \cdot \mathrm{C} \quad(\) Taking value \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1)\) \(\mathrm{Y}=(1+0) \cdot 1\) \(\mathrm{Y}=1 \cdot 1\) \(\mathrm{Y}=1\)
Semiconductor Electronics Material Devices and Simple Circuits
151297
Which logic gate is represented by the following combination of logic gates?
1 \(\mathrm{OR}\)
2 NAND
3 AND
4 NOR
Explanation:
C Given that in this circuit using two NOT gate and one NOR gate Then, the Boolean expression will be- \(\mathrm{Y}=\overline{(\overline{\mathrm{A}}+\overline{\mathrm{B}})}=\mathrm{A} \cdot \mathrm{B}\) Truth table of \(\mathrm{Y}-\) {|c|c|c|c|c|c|} | \(\) \(\) \(}\) \(}\) \(}+}\) \(}+}}\)\)| |---| \( 0 0 1 1 1 0\) \( 0 1 1 0 1 0\) \( 1 0 0 1 1 0\) \( 1 1 0 0 0 1\) \( Which is output expression of AND gate.
JCECE-2016
Semiconductor Electronics Material Devices and Simple Circuits
151298
The circuit is equivalent to
1 NOR gate
2 AND gate
3 NAND gate
4 OR gate
Explanation:
C Given Circuit, \(\overline{\mathrm{Y}}_2=(\overline{\overline{\mathrm{A} \cdot \mathrm{B}}+\overline{\mathrm{A} \cdot \mathrm{B}}})\) \(\overline{\mathrm{Y}}_2=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}} \cdot \overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) \(\overline{\mathrm{Y}}_2=(\mathrm{A} \cdot \mathrm{B}) \cdot(\mathrm{A} \cdot \mathrm{B})\) \(\overline{\mathrm{Y}}_2=\mathrm{AB}\) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\) (apply NOR gate) (apply NOT gate) So output, \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)Which is the expression of NAND gate.
JCECE-2015
Semiconductor Electronics Material Devices and Simple Circuits
151299
In the adjoining circuit of logic gate, the output \(Y\) becomes zero if the inputs are
1 \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\)
2 \(\mathrm{A}=0, \mathrm{~B}=0, \mathrm{C}=0\)
3 \(\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=1\)
4 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0\)
Explanation:
B Given, circuit output \(\mathrm{Y}=0\) then- The expression of given circuit, \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B} \cdot \overline{\mathrm{C}}} \quad\) (using De, Morgen theorems) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}+\overline{\overline{\mathrm{C}}} \quad(\overline{\overline{\mathrm{C}}}=\mathrm{C} \rightarrow \text { Involution theorem })\) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}+\mathrm{C}=\overline{\mathrm{A}}+\overline{\mathrm{B}}+\mathrm{C}\) Then check option (1), \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\) \(\mathrm{Y}=\overline{1}+\overline{1}+0\) \(\mathrm{Y}=0+0+0\) \(\mathrm{Y}=0\)This is possible when \(\mathrm{A}=1, \mathrm{~B}=1\) and \(\mathrm{C}=0\).
JCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151300
What is an AND gate?
1 It has not equivalence to switching circuit
2 It is equivalent to switching circuit
3 It is equivalent to parallel switching circuit
4 It is a mixture of series and parallel switching
Explanation:
B AND gate is used as a switching circuit, AND gate is only active when both inputs are high same as series. Switching circuit which is active only when both switches are closed. The truth table of AND gate will be - Boolean Expression, \(\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}\) {|l|l|l|} | \(\) \(\) \(=\)\)| |---| \( 0 0 0\) \(0 1 0\) \(1 0 0\) \(1 1 1\) \( When both input are high then we get high output
JCECE-2011
Semiconductor Electronics Material Devices and Simple Circuits
151301
To get an output 1 from the circuit shown in the figure, the input must be
1 \(\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=0\)
2 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0\)
3 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1\)
4 \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\)
Explanation:
C Given circuit when output is 1 . In this circuit using OR gate and AND gate. So Boolean expression- \(\mathrm{Y}=(\mathrm{A}+\mathrm{B}) \cdot \mathrm{C} \quad(\) Taking value \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1)\) \(\mathrm{Y}=(1+0) \cdot 1\) \(\mathrm{Y}=1 \cdot 1\) \(\mathrm{Y}=1\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Semiconductor Electronics Material Devices and Simple Circuits
151297
Which logic gate is represented by the following combination of logic gates?
1 \(\mathrm{OR}\)
2 NAND
3 AND
4 NOR
Explanation:
C Given that in this circuit using two NOT gate and one NOR gate Then, the Boolean expression will be- \(\mathrm{Y}=\overline{(\overline{\mathrm{A}}+\overline{\mathrm{B}})}=\mathrm{A} \cdot \mathrm{B}\) Truth table of \(\mathrm{Y}-\) {|c|c|c|c|c|c|} | \(\) \(\) \(}\) \(}\) \(}+}\) \(}+}}\)\)| |---| \( 0 0 1 1 1 0\) \( 0 1 1 0 1 0\) \( 1 0 0 1 1 0\) \( 1 1 0 0 0 1\) \( Which is output expression of AND gate.
JCECE-2016
Semiconductor Electronics Material Devices and Simple Circuits
151298
The circuit is equivalent to
1 NOR gate
2 AND gate
3 NAND gate
4 OR gate
Explanation:
C Given Circuit, \(\overline{\mathrm{Y}}_2=(\overline{\overline{\mathrm{A} \cdot \mathrm{B}}+\overline{\mathrm{A} \cdot \mathrm{B}}})\) \(\overline{\mathrm{Y}}_2=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}} \cdot \overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) \(\overline{\mathrm{Y}}_2=(\mathrm{A} \cdot \mathrm{B}) \cdot(\mathrm{A} \cdot \mathrm{B})\) \(\overline{\mathrm{Y}}_2=\mathrm{AB}\) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\) (apply NOR gate) (apply NOT gate) So output, \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)Which is the expression of NAND gate.
JCECE-2015
Semiconductor Electronics Material Devices and Simple Circuits
151299
In the adjoining circuit of logic gate, the output \(Y\) becomes zero if the inputs are
1 \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\)
2 \(\mathrm{A}=0, \mathrm{~B}=0, \mathrm{C}=0\)
3 \(\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=1\)
4 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0\)
Explanation:
B Given, circuit output \(\mathrm{Y}=0\) then- The expression of given circuit, \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B} \cdot \overline{\mathrm{C}}} \quad\) (using De, Morgen theorems) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}+\overline{\overline{\mathrm{C}}} \quad(\overline{\overline{\mathrm{C}}}=\mathrm{C} \rightarrow \text { Involution theorem })\) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}+\mathrm{C}=\overline{\mathrm{A}}+\overline{\mathrm{B}}+\mathrm{C}\) Then check option (1), \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\) \(\mathrm{Y}=\overline{1}+\overline{1}+0\) \(\mathrm{Y}=0+0+0\) \(\mathrm{Y}=0\)This is possible when \(\mathrm{A}=1, \mathrm{~B}=1\) and \(\mathrm{C}=0\).
JCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151300
What is an AND gate?
1 It has not equivalence to switching circuit
2 It is equivalent to switching circuit
3 It is equivalent to parallel switching circuit
4 It is a mixture of series and parallel switching
Explanation:
B AND gate is used as a switching circuit, AND gate is only active when both inputs are high same as series. Switching circuit which is active only when both switches are closed. The truth table of AND gate will be - Boolean Expression, \(\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}\) {|l|l|l|} | \(\) \(\) \(=\)\)| |---| \( 0 0 0\) \(0 1 0\) \(1 0 0\) \(1 1 1\) \( When both input are high then we get high output
JCECE-2011
Semiconductor Electronics Material Devices and Simple Circuits
151301
To get an output 1 from the circuit shown in the figure, the input must be
1 \(\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=0\)
2 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0\)
3 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1\)
4 \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\)
Explanation:
C Given circuit when output is 1 . In this circuit using OR gate and AND gate. So Boolean expression- \(\mathrm{Y}=(\mathrm{A}+\mathrm{B}) \cdot \mathrm{C} \quad(\) Taking value \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1)\) \(\mathrm{Y}=(1+0) \cdot 1\) \(\mathrm{Y}=1 \cdot 1\) \(\mathrm{Y}=1\)
Semiconductor Electronics Material Devices and Simple Circuits
151297
Which logic gate is represented by the following combination of logic gates?
1 \(\mathrm{OR}\)
2 NAND
3 AND
4 NOR
Explanation:
C Given that in this circuit using two NOT gate and one NOR gate Then, the Boolean expression will be- \(\mathrm{Y}=\overline{(\overline{\mathrm{A}}+\overline{\mathrm{B}})}=\mathrm{A} \cdot \mathrm{B}\) Truth table of \(\mathrm{Y}-\) {|c|c|c|c|c|c|} | \(\) \(\) \(}\) \(}\) \(}+}\) \(}+}}\)\)| |---| \( 0 0 1 1 1 0\) \( 0 1 1 0 1 0\) \( 1 0 0 1 1 0\) \( 1 1 0 0 0 1\) \( Which is output expression of AND gate.
JCECE-2016
Semiconductor Electronics Material Devices and Simple Circuits
151298
The circuit is equivalent to
1 NOR gate
2 AND gate
3 NAND gate
4 OR gate
Explanation:
C Given Circuit, \(\overline{\mathrm{Y}}_2=(\overline{\overline{\mathrm{A} \cdot \mathrm{B}}+\overline{\mathrm{A} \cdot \mathrm{B}}})\) \(\overline{\mathrm{Y}}_2=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}} \cdot \overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) \(\overline{\mathrm{Y}}_2=(\mathrm{A} \cdot \mathrm{B}) \cdot(\mathrm{A} \cdot \mathrm{B})\) \(\overline{\mathrm{Y}}_2=\mathrm{AB}\) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\) (apply NOR gate) (apply NOT gate) So output, \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)Which is the expression of NAND gate.
JCECE-2015
Semiconductor Electronics Material Devices and Simple Circuits
151299
In the adjoining circuit of logic gate, the output \(Y\) becomes zero if the inputs are
1 \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\)
2 \(\mathrm{A}=0, \mathrm{~B}=0, \mathrm{C}=0\)
3 \(\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=1\)
4 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0\)
Explanation:
B Given, circuit output \(\mathrm{Y}=0\) then- The expression of given circuit, \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B} \cdot \overline{\mathrm{C}}} \quad\) (using De, Morgen theorems) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}+\overline{\overline{\mathrm{C}}} \quad(\overline{\overline{\mathrm{C}}}=\mathrm{C} \rightarrow \text { Involution theorem })\) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}+\mathrm{C}=\overline{\mathrm{A}}+\overline{\mathrm{B}}+\mathrm{C}\) Then check option (1), \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\) \(\mathrm{Y}=\overline{1}+\overline{1}+0\) \(\mathrm{Y}=0+0+0\) \(\mathrm{Y}=0\)This is possible when \(\mathrm{A}=1, \mathrm{~B}=1\) and \(\mathrm{C}=0\).
JCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151300
What is an AND gate?
1 It has not equivalence to switching circuit
2 It is equivalent to switching circuit
3 It is equivalent to parallel switching circuit
4 It is a mixture of series and parallel switching
Explanation:
B AND gate is used as a switching circuit, AND gate is only active when both inputs are high same as series. Switching circuit which is active only when both switches are closed. The truth table of AND gate will be - Boolean Expression, \(\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}\) {|l|l|l|} | \(\) \(\) \(=\)\)| |---| \( 0 0 0\) \(0 1 0\) \(1 0 0\) \(1 1 1\) \( When both input are high then we get high output
JCECE-2011
Semiconductor Electronics Material Devices and Simple Circuits
151301
To get an output 1 from the circuit shown in the figure, the input must be
1 \(\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=0\)
2 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0\)
3 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1\)
4 \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\)
Explanation:
C Given circuit when output is 1 . In this circuit using OR gate and AND gate. So Boolean expression- \(\mathrm{Y}=(\mathrm{A}+\mathrm{B}) \cdot \mathrm{C} \quad(\) Taking value \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1)\) \(\mathrm{Y}=(1+0) \cdot 1\) \(\mathrm{Y}=1 \cdot 1\) \(\mathrm{Y}=1\)
Semiconductor Electronics Material Devices and Simple Circuits
151297
Which logic gate is represented by the following combination of logic gates?
1 \(\mathrm{OR}\)
2 NAND
3 AND
4 NOR
Explanation:
C Given that in this circuit using two NOT gate and one NOR gate Then, the Boolean expression will be- \(\mathrm{Y}=\overline{(\overline{\mathrm{A}}+\overline{\mathrm{B}})}=\mathrm{A} \cdot \mathrm{B}\) Truth table of \(\mathrm{Y}-\) {|c|c|c|c|c|c|} | \(\) \(\) \(}\) \(}\) \(}+}\) \(}+}}\)\)| |---| \( 0 0 1 1 1 0\) \( 0 1 1 0 1 0\) \( 1 0 0 1 1 0\) \( 1 1 0 0 0 1\) \( Which is output expression of AND gate.
JCECE-2016
Semiconductor Electronics Material Devices and Simple Circuits
151298
The circuit is equivalent to
1 NOR gate
2 AND gate
3 NAND gate
4 OR gate
Explanation:
C Given Circuit, \(\overline{\mathrm{Y}}_2=(\overline{\overline{\mathrm{A} \cdot \mathrm{B}}+\overline{\mathrm{A} \cdot \mathrm{B}}})\) \(\overline{\mathrm{Y}}_2=\overline{\overline{\mathrm{A} \cdot \mathrm{B}}} \cdot \overline{\overline{\mathrm{A} \cdot \mathrm{B}}}\) \(\overline{\mathrm{Y}}_2=(\mathrm{A} \cdot \mathrm{B}) \cdot(\mathrm{A} \cdot \mathrm{B})\) \(\overline{\mathrm{Y}}_2=\mathrm{AB}\) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\) (apply NOR gate) (apply NOT gate) So output, \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)Which is the expression of NAND gate.
JCECE-2015
Semiconductor Electronics Material Devices and Simple Circuits
151299
In the adjoining circuit of logic gate, the output \(Y\) becomes zero if the inputs are
1 \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\)
2 \(\mathrm{A}=0, \mathrm{~B}=0, \mathrm{C}=0\)
3 \(\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=1\)
4 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0\)
Explanation:
B Given, circuit output \(\mathrm{Y}=0\) then- The expression of given circuit, \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B} \cdot \overline{\mathrm{C}}} \quad\) (using De, Morgen theorems) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}+\overline{\overline{\mathrm{C}}} \quad(\overline{\overline{\mathrm{C}}}=\mathrm{C} \rightarrow \text { Involution theorem })\) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}+\mathrm{C}=\overline{\mathrm{A}}+\overline{\mathrm{B}}+\mathrm{C}\) Then check option (1), \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\) \(\mathrm{Y}=\overline{1}+\overline{1}+0\) \(\mathrm{Y}=0+0+0\) \(\mathrm{Y}=0\)This is possible when \(\mathrm{A}=1, \mathrm{~B}=1\) and \(\mathrm{C}=0\).
JCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151300
What is an AND gate?
1 It has not equivalence to switching circuit
2 It is equivalent to switching circuit
3 It is equivalent to parallel switching circuit
4 It is a mixture of series and parallel switching
Explanation:
B AND gate is used as a switching circuit, AND gate is only active when both inputs are high same as series. Switching circuit which is active only when both switches are closed. The truth table of AND gate will be - Boolean Expression, \(\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}\) {|l|l|l|} | \(\) \(\) \(=\)\)| |---| \( 0 0 0\) \(0 1 0\) \(1 0 0\) \(1 1 1\) \( When both input are high then we get high output
JCECE-2011
Semiconductor Electronics Material Devices and Simple Circuits
151301
To get an output 1 from the circuit shown in the figure, the input must be
1 \(\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=0\)
2 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0\)
3 \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1\)
4 \(\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0\)
Explanation:
C Given circuit when output is 1 . In this circuit using OR gate and AND gate. So Boolean expression- \(\mathrm{Y}=(\mathrm{A}+\mathrm{B}) \cdot \mathrm{C} \quad(\) Taking value \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1)\) \(\mathrm{Y}=(1+0) \cdot 1\) \(\mathrm{Y}=1 \cdot 1\) \(\mathrm{Y}=1\)
