Solution

Solution

Total instruction size is:

$16$ bits

Opcode uses:

$4$ bits

So, remaining bits for memory address are:

$16-4=12$ bits

With $12$ address bits, maximum addressable memory locations are:

$2^{12}=4096$

Solution

Opcode stands for Operation Code.

It tells the CPU which operation has to be performed, such as addition, subtraction, data movement, comparison, etc.

Operand tells the data or address on which the operation is performed.

Register is a small storage location inside the CPU.

Immediate address represents data given directly in the instruction.