Assume initial value of register A is 08 H, B is 07 H and carry flag is 0. What should be the sequence of 8085 code to make the value in B, thrice its original value? (Suffix H denotes hexadecimal number)

Concept:

In the 8085 microprocessor, we use instructions like MOV, ADD, and RLC to manipulate register values.

We are given:
A = 08H,
B = 07H,
Carry Flag = 0
Our goal is: Make B = 3 × original B = 3 × 07H = 15H

Step-by-Step Instruction Execution:

Instruction 1: MOV A, B
This copies the content of B (07H) into register A.
Now: A = 07H, B = 07H

Instruction 2: RLC (Rotate Accumulator Left)
This rotates A one bit to the left.
A = 07H = 0000 0111 (binary)
After RLC: 0000 1110 = 0EH (14 in decimal)
Now: A = 0EH

Instruction 3: ADD B
Add contents of B (07H) to A (0EH):
0EH + 07H = 15H (21 in decimal)
Now: A = 15H

Instruction 4: MOV B, A
Copy the final result in A to B.
So B = 15H (final answer)

Conclusion:

The value in register B becomes three times its original value (07H × 3 = 15H).
Thus, the correct sequence is:

MOV A, B
RLC
ADD B
MOV B, A

Answer:

Option 2: MOV A, B → RLC → ADD B → MOV B, A