16KB, or 16384 bytes, can be addressed with 14 address lines. (214 = 16384)
Bytes:32768 Bits:262144
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
In case of IPv4, the address has 4 bytes. In case of IPv6, the address has 16 bytes.
If you are addressing bytes, then 512K words (16 bit words) requires 20 address lines.I gave that answer because the question was categorized 8086/8088. If you are addressing words, then the answer is 19 address lines.
10 bytes - 4 for the network, 6 for the MAC address.
The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
A Mac address is a 48bit addressing scheme (usually represented in HEX). There are 8 bits in a bytes therefore it is 6 bytes long.
IPv4 addresses are 4 bytes. IPv6 IP addresses are 16 bytes.
How many no of address lines required in 1MB memory 11,16,22 or 24 u haven't specified correct options! 20 address lines will be required because 1 MB is 1024 KB that is 1024*1024 Byte which is equivalent to (2^10)^2 bytes if ur memory is Byte addressable then address lines required will be 20.
32 bits or 4 bytes and an int is not an address, it is a primitive so it directly access the data without a reference.
There are 20 address lines and 16 data lines in the 8086 microprocessor. The low order 16 address lines are multiplexed with the data lines. Some of the high order address lines are multiplexed with status lines.