The following instructions were executed on 8085 microprocessor :

MVI A, 33H

MVI B, 78H A

ADD B

CMA

The Accumulator value after the execution of the fourth instruction is 

Explanation:

Let's analyze the instructions executed on the 8085 microprocessor step by step to understand how the accumulator value is determined after the fourth instruction.

Instructions:

1. MVI A, 33H: This instruction moves the immediate value 33H into the Accumulator (A). Therefore, after this instruction, the Accumulator (A) will hold the value 33H.
2. MVI B, 78H: This instruction moves the immediate value 78H into the register B. So, register B will hold the value 78H after this instruction.
3. ADD B: This instruction adds the content of register B to the Accumulator (A) and stores the result in the Accumulator. The values before the addition are:

• Accumulator (A) = 33H
• Register B = 78H

Now, performing the addition:

33H + 78H = ABH

After this instruction, the Accumulator (A) will hold the value ABH.

4. CMA: This instruction complements (inverts) the contents of the Accumulator. In other words, it changes all 1s to 0s and all 0s to 1s in the binary representation of the Accumulator's value.

Let's first convert ABH to binary:

ABH = 1010 1011 (in binary)

Complementing this value (changing all bits):

1010 1011 becomes 0101 0100 (in binary)

Converting 0101 0100 back to hexadecimal:

0101 0100 = 54H

Therefore, after the CMA instruction, the Accumulator (A) will hold the value 54H.

Correct Option Analysis:

The correct option is:

Option 3: 54H

This option correctly represents the value of the Accumulator after the execution of the fourth instruction (CMA).