Question

An 8-bit microcontroller has an external memory map from \(0000\text{H}\) to \(1FFF\text{H}\). The number of bytes this RAM can store is

• A. \(8193\)
• B. \(8191\)
• C. \(8192\)
• D. \(8000\)

Hint:

For an inclusive memory range, \[ \boxed{ \text{Memory Size} = (\text{Last Address}-\text{First Address})+1 } \] bytes.

Answer 1

The Correct Option is C

Step 1: Calculate the total number of memory locations. The memory range is \[ 0000_{\text{H}} \text{ to } 1FFF_{\text{H}}. \] Hence, \[ \text{Number of locations} = (1FFF_{\text{H}}-0000_{\text{H}})+1. \]

Step 2: Convert the hexadecimal value. \[ 1FFF_{\text{H}} = 8191_{10}. \] Therefore, \[ 8191+1 = 8192. \] Since each address stores one byte, \[ \boxed{8192\ \text{bytes}} \] can be stored. Therefore, \[ \boxed{(C)} \] is the correct answer.