Analyse the given program for 8085 and answer the question that follows.
MVI B, 06h
MVI A, F2H
ADD B
What is the content of Accumulator Register after the execution of the given program?

Let's trace the execution of the 8085 program step by step to determine the final content of the Accumulator (Register A).

Initial State:

• Register B contains an unknown value.
• Register A contains an unknown value.

Instruction 1: MVI B, 06h

• MVI stands for "Move Immediate".
• This instruction loads the immediate 8-bit value 06h into Register B.
• After this instruction:
  • Register B = 06_h

Instruction 2: MVI A, F2H

• This instruction loads the immediate 8-bit value F2h into the Accumulator (Register A).
• After this instruction:
  • Register A = F2_h

Instruction 3: ADD B

• ADD adds the content of the specified register (in this case, Register B) to the content of the Accumulator (Register A). The result is stored in the Accumulator.
• Before the addition:
  • Register A = F2_h
  • Register B = 06_h

Now, let's perform the hexadecimal addition:

   F2  (Hexadecimal)
+  06  (Hexadecimal)
-----

Starting from the rightmost digit:

• 2 + 6 = 8

Moving to the leftmost digit:

• F + 0 = F

Therefore, the result of the addition F2_h + 06_h is F8_h.

After the execution of the ADD B instruction, the content of the Accumulator (Register A) will be F8_h.