Q. 8

# Truth table for the given circuit (Fig. 14.6) is

A.

A

B

E

0

0

1

0

1

0

1

0

1

1

1

0

B.

A

B

E

0

0

1

0

1

0

1

0

1

1

1

0

A | B | E |

0 | 0 | 1 |

0 | 1 | 0 |

1 | 0 | 0 |

1 | 1 | 1 |

C.

A | B | E |

0 | 0 | 0 |

0 | 1 | 1 |

1 | 0 | 0 |

1 | 1 | 1 |

D.

A | B | E |

0 | 0 | 0 |

0 | 1 | 1 |

1 | 0 | 1 |

1 | 1 | 0 |

Answer :

In the given network of gates, the output at C is due to A and B through the **AND** gate. The output at D is due to the signal of A through the **NOT** gate and B. Therefore,

and

The output at E is due to C and D through an **OR** gate. therefore

Therefore, the truth table can be constructed as follows,

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