Answer:
See the table attached for complete solution to the problem.
The answer to your question is,
True.
-Mabel <3
Answer:
Total Memory= 4 KB = 4096 bytes = 32768 bits
Explanation:
<em><u>1. Data lines are 8 From D0 to D7</u></em>
so
Total memory at single address locations is 8 bits.
<em><u>2. Address lines are 12 (A0 to A11)</u></em>
There are 12 address lines but 3 out 12 are for selction of chip or memory bank.
so only 9 pins are there to address the locations on one chip.
Total No. of address locations on single chip = 2^9 = 512 locations
as 1 location is 1 byte so total memory of single chip is 512 bytes.
<u><em>3. Total Memory Bank </em></u>
There are total 3 selection pins for memory bank.
so
Total chips = 2^3 = 8.
<em><u>4. Total Memory </u></em>
Total size of 1 chip = 512 bytes
Total size of 8 chip = 8x512 bytes = 4096 bytes = 4096/1024 kb = 4 kb
<em>So total memory of system is 4 Kb = 4096 bytes = 32768 bits</em>
