Question

A laptop computer has two internal signals: An unplugged signal, which is '1' if the laptop's power supply is connected, and '0' otherwise. A low battery signal, which is '0' if the laptop's battery has reached an almost empty state, and '1' otherwise. Suppose the laptop's power control system accepts a single hibernate signal which determines if the laptop should change its current operating state and shut down. If the laptop should shut down when the battery is low and its charger is unplugged, which gate could be used to produce the hibernate signal?

Answer 1

Answer:

The correct usage is a NOR gate which is indicated in the explanation.

Explanation:

The truth table for the given two signals, namely

p=unplugged signal

q=low battery signal

can  form a truth table of following form

Here p has 2 states

1 if the power supply is connected

0 otherwise

Similarly q has 2 states

0 if the battery has reached almost zero state

1 otherwise

As the condition for the Hibernate Signal is given as to only activate when the battery is low and the power supply is not connected. This indicate that the value of Hibernate signal should be 1 when both p and q are 0.

Using this condition, the truth table is formed as

unplugged signal | low battery signal |  Hibernate Signal

            0               |                 0             |             1

            0               |                 1              |             0

            1                |                 0             |             0

            1                |                 1              |             0

Now the truth table of NOR is given as

a    |     b    |  a or b   |   ~(a or b)

0    |     0    |      0      |        1

0    |     1     |      1       |        0

1     |     0    |      1       |        0

1     |     1     |      1       |        0

This indicates that the both truth tables are same thus the NOR gate is to be used for this purpose.