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8085 has 16-bit address bus which makes _____bytes of total memory accessible.
2┬░16
64KB
Both
None
Both
The Intel 8085 microprocessor utilizes a 16-bit address bus, which determines its maximum addressable memory space. Since each bit can represent two states (0 or 1), 16 bits provide 2┬░16 unique memory addresses, totaling 65,536 bytes of memory.
The Intel 8085 microprocessor utilizes a 16-bit address bus, which determines its maximum addressable memory space. Since each bit can represent two states (0 or 1), 16 bits provide 2┬░16 unique memory addresses, totaling 65,536 bytes of memory.
MemoryCapacity=2┬░n bytes where n is the number of address lines
1KB=2┬░10bytes=1024bytes
The number of address lines (n) directly determines the addressable memory capacity according to the formula 2┬░n bytes. For n=16, the calculation is 2┬░16=65,536. Since 1024 bytes are defined as 1 kilobyte (KB), the capacity is 102465536тАЛ=64 KB.
The 8085 address bus width is 16 bits.
The address space ranges from 0000H to FFFFH.
Total addressable memory is exactly 65,536 bytes.
65,536 bytes is equivalent to 64 KB.
Direct addressing of up to 64KB simplifies memory management.
Compatibility with byte-addressable memory architecture.
Limited address space compared to modern 32-bit or 64-bit processors.
Requires bank switching if memory needs exceed 64KB.
Embedded controller designs.
Educational hardware for microprocessor assembly language programming.
Legacy industrial control systems.
Option A is mathematically correct as 2┬░16 defines the address space.
Option B is the standard shorthand notation for 65,536 bytes in binary computer architecture.
Both A and B are equivalent representations, making C the most accurate choice.
C is correct тАФ Both 2┬░16 and 64KB are equivalent representations of the total memory addressable by a 16-bit address bus.
Remember that for any microprocessor, if the address bus has n lines, the total memory capacity is always 2┬░n bytes. This rule is fundamental for all CPU architecture questions.