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
A 16 bit address bus can access 64 k, or 65536 objects of memory.
We generally don't talk about "virtual" memory at this small a scale, because most (though not all) 16 bit computers do not provide a protected virtual memory interface. The only computer I recall that did do so was the Modcomp Classic II, but that was a 1970's era minicomputer. The answer provided above still applied in that case.
If an address bus for a given computer has 16 lines, what is the maximum amount of memory it can access?
You can address 65,536 (216) memory addresses with a 16 bit address bus.
16: 216 = 65536 = 64K
32: 232 = 4294967296 = 4G
656 kilobytes.
4GB of RAM
31 bit addressing is a way of accessing the virtual memory. Based on the memory available the bit in the address can change. Or viceversa. for 31 bit the memory available is gb and for 24 bit its 16mb.
no just a bit sweg
However much is assigned to it by the virtual machine. Virtual or not, the 32-bit version of WinXP will not be able to address more than 4GB of memory in any combination of video and system ram. Dial back to 4GB instead of allocating excess...you gain nothing and just take away memory from the host. This presumes that your host is a 64-bit OS. If the host is 32-bit, also, you want to limit your virtual machine to half the available memory (2GB) or your host will suffer performance issues.
An 80286 has a 24 bit address bus. As such, it can address 224, or 16,777,216, or 16 MB of memory.
Real memory uses Physical addresses.These are the members that the memory chips react to on the bus. Virtual addresses are the logical addresses that nrefer to a process' address space. Thus, a machine with a 16-bit word can generate virtual addresses upto 64K, regardless of whether the machine has more or less memory than 64 KB
The 8086/8088 is a 16 bit computer running on a 20 bit address bus. Processes use a segmented memory architecture to access one of four 64kb memory segments from a physical space of 1mb.
Virtual memory does not exist physically but it is available in the systemThe procedure of fetching the chosen pgm. Segments or data from the secondary storage into the physical memory is called swappingSo it can address 1GBPhysical address calculation in PVAMIt uses 16-bit content of a segment register as a selector to address a descriptor stored in physical memoryThe descriptor is a block of contiguous memory locations containing information of a segmentPrivilege levels prevent unauthorized accessesMaximum segment size will be of 64kb
A memory with a 16 bit address bus can address 216 or 65536 distinct items. If each item is 32 bits in size, then the item is 4 bytes. The size of this memory is then 262144 bytes. (256Kb)
Even though the 8085 is an 8 bit microprocessor, it can address 64K memory, because it has a 16 bit address bus.
4GB
in 8086 there is 20 bit address bus,so it can address 1,048,576 address. At each address we can store 8 bit address (1-byte)but if want to write a word(16-bit)into a memory segment to store data in byte form then we write the data in two consecutive memory address which are even(low) and odd(high) memory.
A 32-bit OS can only address up to 4GB of memory.