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What is the maximum directly addressable memory capacity of 32 bit microprocessor?
Assuming your memory card slot(s) permit, 2 to the power of 32 = 4294967296
which is apprx 4 GB
which is apprx 4 GB
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A microprocessor has a data bus with 64 lines 32 lines address bus.what will be the Maximum number of bits stored in memory?
2^32 is amount of blocks that address bus could locate. and each blocks is 64bit because data bus has 64 lines. then maximum number of bits stored in memory is (2^32)*64… bit. By: Mohammad Saghafi Email: email@example.com
What is the maximum memory address space that the processor can access directly if it is connected to a 16 bit memory considering a hypothetical microprocessor generating a 16-bit address?
If you assume that it has a 16-bit data bus, then it would be 128k so the microprocessor can access 2^16 points, which is 64k (from it being a 16bit address) 16bits = 2 by…tes (memory) so through a 16 bit memory, it can access 2*64k, which is 128k alternatively, if its 8bit memory, 8bits=1byte 1*64k = 64k I'm no expert, and i was searching for the answer myself, hope this helped
Maximum memory address space that the processor can access directly if it is connected to a 16-bit memory?
Max. memory address space= 216 X 2 bytes = 128 Kbytes
What is the maximum memory capacity that can be connected to a microprocessor data bus with 16 lines and the address bus has 20 lines?
You can address 2^20 (or 1,048,576) locations. Each location will consist of 16 bits (1 or 0) which is commonly called a Word (though that isn't always the case). I …think what you're really looking for is the answer in bytes. Each data bus line holds a bit, 8 bits is a byte, so you're data line is 2 bytes. Since each location holds two bytes, this would be 2,097,152 bytes, or 2MB
32 bit address bus can access more than 4 gigabytes (232) of memory. Sandeep Kr. Singh (MCA)
2 ki power 64 memory addressing capacity hogi
What is the What is the maximum memory address space that the processor can access directly if it is connected to a 8 bit memory considering a hypothetical microprocessor generating a 8 bit address?
microprocessor can access 2^8 points which is 256 then we have 8 bit memory = 1 bytes then 1*256 =256 bytes
for a 32-bit processor with 8-bit opcode: 2^(32-8) bits == 2^24 bits = about 16.8 MB
It depends on the processor and the operating system. In the 8086/8088, with a 20 bit address bus, you can address 1,048,576 bytes of memory.In the 8085, with a 16 bit addres…s bus, you can address 65,536 bytes of memory.In the IA-32 architecture (80386 and higher in 32 bit mode), you can address 4,294,967,296 bytes of memory (physical) minus about 805,306,368 bytes of memory for overhead (operating system).In the IA-64 architecture (80686 running in 64 bit mode on 64 bit Windows) you can address 206,158,430,208 bytes of memory.
32 bit address line can access 4GB of memory. As 2^10 -> 1KB; 2^20 -> 2MB; 2^30 -> 1GB and so on.... 32 bit gives (2^30) * (2^2) = 1GB * 4 = 4GB;
A microprocessor that uses 24 bit addressing, such as the Intel 80286, can address 224 or 16,777,216 memory locations. The IBM MainFrame, 360/44 or any modern version running …in AMODE=24 also has the same capacity.
The 8086/8088 can address a maximum of 220, or 1,048,576, or 1 MB of memory.
232 = 4,294,967,296, or 4 G
0x00000000 to 0xFFFFFFFF in hexadecimal 0 to 4294967295 in binary