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.
It requires 30 address bits to address 1GB of RAM.230 = 1,073,741,824
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;
You need 30 address lines to access 1G of memory. 230 = 1,073,741,824. log2 (1,073,741,824) = 30.
The number of lines of text that can be read with 1GB of data depends on the length of each line. Assuming an average line of text contains about 100 characters, and considering that 1GB is approximately 1 billion bytes, you could fit around 10 million lines of text in 1GB. However, this is a rough estimate and can vary based on formatting and encoding.
You cannot address 1GB memory with the 8085 or the 8086/8088 without some kind of external demultiplexor that is software controlled. The address bus on the 8085 is 16 bits, giving addressibility of 64KB; while the address bus on the 8086/8088 is 20 bits, giving addressibility of 1MB. To address 1GB, you need a 30 bit address bus.
1000kb = 1mb 1000mb = 1gb 1,000,000kb = 1gb
In a 1GB memory space, the total number of bytes is (1 \times 2^{30} = 1,073,741,824) bytes. Since memory addresses typically start at 0, the memory address of the last byte would be (1,073,741,824 - 1 = 1,073,741,823). Therefore, the memory address of the last byte of a 1GB memory is 1,073,741,823.
On a Flip Mino HD 1GB = 30min.
1GB = 1024 MB 1MB= 1024 KB So 1GB= 1048576 KB
1Gb
1024mb (megabits) = 1gb (gigabits)
About 1000.