Question 1 |

In the circuit shown below, X and Y are digital inputs, and Z is a digital output. The equivalent circuit is a

NAND gate | |

NOR gate | |

XOR gate | |

XNOR gate |

Question 1 Explanation:

The Boolean expression for the output of the digital circuit is shown below

The above expression is of XOR gate.

The above expression is of XOR gate.

Question 2 |

In the logic circuit shown in the figure, Y is given by

Y = ABCD | |

Y = (A + B)(C + D) | |

Y = A + B + C + D | |

Y = AB + CD |

Question 2 Explanation:

Question 3 |

For a 3-input logic circuit shown below, the output Z can be expressed as

Q+\bar{R} | |

P\bar{Q}+R | |

\bar{Q}+R | |

P+\bar{Q}+R |

Question 3 Explanation:

Z=\overline{\overline{P\bar{Q}}\cdot Q\cdot \overline{Q\cdot R} }

\;\;=P\bar{Q}+\bar{Q}+Q R

\;\;=\bar{Q}(P+1)+QR

\;\;=\bar{Q}+QR

\;\;=(\bar{Q}+Q)(\bar{Q}+R)

\;\;=\bar{Q}+R

Question 4 |

Which of the following logic circuits is a realization of the function F whose
Karnaugh map is shown in figure

A | |

B | |

C | |

D |

Question 4 Explanation:

F=\bar{A}\bar{C}+BC

Question 5 |

A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles

and AND gate | |

an OR gate | |

an XOR gate | |

a NAND gate |

Question 5 Explanation:

Truth table of XOR gate is,

So, from the XOR gate truth table, it is clear that the bulb can be turned ON and also can be turned OFF by any one of the switched irrespective of the state of the other switch.

So, from the XOR gate truth table, it is clear that the bulb can be turned ON and also can be turned OFF by any one of the switched irrespective of the state of the other switch.

There are 5 questions to complete.