16KB, or 16384 bytes, can be addressed with 14 address lines. (214 = 16384)
Bytes:32768 Bits:262144
To calculate the number of address lines required for a 64 kB memory, first convert 64 kB into bytes: 64 kB = 64 × 1024 bytes = 65,536 bytes. The number of address lines needed can be determined using the formula (2^n = \text{total number of addresses}), where (n) is the number of address lines. Since 65,536 is (2^{16}), you need 16 address lines to address a 64 kB memory.
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.
The Intel 8088 microprocessor has 20 address lines. This allows it to address up to 1 MB (2^20 bytes) of memory. The address lines are labeled A0 to A19, enabling the microprocessor to access a wide range of memory locations.
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.
The PIC16F877A microcontroller has 13 address lines, which allows it to address 8K words of program memory. This enables it to access a range of memory locations for storing instructions. Additionally, it has 368 bytes of data RAM and 256 bytes of Electrically Erasable Programmable Read-Only Memory (EEPROM) for data storage.
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.
To determine the number of address lines required for 1 GB of memory, we can use the formula (2^n = \text{Memory Size}), where (n) is the number of address lines. Since 1 GB equals (2^{30}) bytes, we need (30) address lines to uniquely address each byte in 1 GB of memory. Therefore, (30) address lines are required for 1 GB.