Correct option is A
The question involves implementing a common bus for 8 registers, each of 16 bits, using multiplexers. The solution can be broken into the following steps:
1. Understanding the Problem:
· Each register has 16 bits.
· There are 8 registers.
· A multiplexer (MUX) allows multiple inputs (registers in this case) to be selected and sent to a common output (the bus).
2. Requirements of the Multiplexers:
· Size of MUX:
· Since there are 8 registers, the MUX needs to select 1 out of 8 inputs for each bit of the bus.
· Therefore, the required MUX size is 8 × 1.
· Number of MUX:
· Each register has 16 bits, and all bits need to connect to the bus.
· Therefore, 16 multiplexers are required (one for each bit of the bus).
3. Conclusion: The correct configuration is 16 multiplexers, each of size 8 × 1.
Information Booster
1. Common Bus System: A common bus is used to transfer data between multiple registers and other components of a computer system.
2. Multiplexer: A digital device that selects one input from multiple inputs and forwards it to the output, controlled by select lines.
3. Bus Width: The width of the bus is equal to the number of bits in a register (16 bits in this case).
4. Select Lines: For an 8 × 1 multiplexer, log2 8 = 3 select lines are needed.
Additional Knowledge
· A common bus system minimizes the number of connections needed between components in a digital system.
· 8 × 1 MUX: Used for selecting data from 8 sources.
· 16 Multiplexers: Required because each of the 16 bits from the registers needs a dedicated multiplexer for data transmission through the bus.

