8085 Microprocessor RAL Instruction: Detailed Solution

The question asks us to determine the final number stored in the 8085 microprocessor's accumulator after executing the RAL instruction twice on an initial value of decimal 8. This problem requires understanding how the RAL (Rotate Accumulator Left through Carry) instruction works at the bit level.

RAL Instruction Overview

The RAL instruction in the 8085 microprocessor performs a left rotation of the accumulator's 8-bit contents through the Carry Flag (CY). Here's how it operates:

• The content of the accumulator is rotated left by one bit position.
• The most significant bit (D7) of the accumulator is moved to the Carry Flag (CY).
• The existing value of the Carry Flag (CY) is moved into the least significant bit (D0) of the accumulator.
• All other bits shift one position to the left (D0 moves to D1, D1 to D2, ..., D6 to D7).

Initial Decimal Value Setup

First, let's convert the initial decimal number 8 into its 8-bit binary representation. The decimal number 8 is represented as $\text{0000 1000}_2$ in binary. For the purpose of this calculation, we assume the initial state of the Carry Flag (CY) is 0, as is common when no prior operations affecting flags are mentioned.

First RAL Execution Step

Now, let's apply the RAL instruction for the first time:

• The D7 bit (which is 0) moves to the Carry Flag. So, CY becomes 0.
• The original Carry Flag value (which was 0) moves to the D0 bit of the accumulator. So, D0 becomes 0.
• The remaining bits (D0-D6) shift one position to the left.

After the first RAL instruction, the accumulator holds $\text{0001 0000}_2$, which is decimal 16.

Second RAL Execution Step

Next, we execute the RAL instruction for the second time, using the current state of the accumulator and Carry Flag:

• The current D7 bit (which is 0) moves to the Carry Flag. So, CY remains 0.
• The current Carry Flag value (which is 0) moves to the D0 bit of the accumulator. So, D0 remains 0.
• The remaining bits (D0-D6) shift one position to the left.

Final Number Calculation

After the second RAL instruction, the accumulator holds the binary value $\text{0010 0000}_2$. To find the final decimal number, we convert this binary value back to decimal:

$\text{0010 0000}_2 = (0 \times 2^7) + (0 \times 2^6) + (1 \times 2^5) + (0 \times 2^4) + (0 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (0 \times 2^0)$

$= 0 + 0 + 32 + 0 + 0 + 0 + 0 + 0$

$= 32_{10}$

Therefore, the final number stored in the accumulator is decimal 32.