Accumulator Contents After RAL Instruction

The question asks to determine the final contents of the accumulator after the RAL instruction is executed twice. We are given the initial accumulator content as AAH and the Carry flag (CY) as 0. The RAL instruction stands for "Rotate Accumulator Left through Carry".

Understanding the RAL Instruction

The RAL instruction performs a left rotation of the bits in the accumulator, including the Carry flag. Here's how it works:

• The most significant bit (MSB) of the accumulator moves into the Carry flag (CY).
• The content of the Carry flag (CY) moves into the least significant bit (LSB) of the accumulator.
• All other bits of the accumulator shift one position to the left.

Initial State of Accumulator and Carry

Let's start with the given initial values:

• Accumulator (A) = AAH
• Carry Flag (CY) = 0

First, we convert the hexadecimal value AAH into its 8-bit binary equivalent:

AAH \(= 1010 \ 1010_2\)

First RAL Instruction Execution

Now, we will execute the RAL instruction for the first time.

After the first RAL instruction:

• The MSB of the original accumulator (1) moves to CY. So, CY becomes 1.
• The original CY (0) moves to the LSB of the accumulator.
• The remaining bits (\(010 \ 1010\)) shift left by one position.

So, the new state after the first RAL is:

• New Accumulator (A) = \(0101 \ 0100_2\)
• New Carry Flag (CY) = \(1\)

Converting the new accumulator value to hexadecimal:

\(0101 \ 0100_2 = 54H\)

Second RAL Instruction Execution

Now, we execute the RAL instruction again, using the results from the first execution as our new initial state.

• Current Accumulator (A) = \(0101 \ 0100_2\)
• Current Carry Flag (CY) = \(1\)

After the second RAL instruction:

• The MSB of the current accumulator (0) moves to CY. So, CY becomes 0.
• The current CY (1) moves to the LSB of the accumulator.
• The remaining bits (\(101 \ 0100\)) shift left by one position.

So, the final state after the second RAL is:

• Final Accumulator (A) = \(1010 \ 1001_2\)
• Final Carry Flag (CY) = \(0\)

Converting the final accumulator value to hexadecimal:

\(1010 \ 1001_2 = A9H\)

Conclusion

After executing the RAL instruction twice, the final contents of the accumulator are A9H.