Question

If 8085 microprocessor adds 87H and 79H, then :

• A. Both CARRY and ZERO flags will be set to 0
• B. CARRY flag will be set to 0, ZERO flag to 1
• C. CARRY flag will be set to 1, ZERO flag to 0
• D. Both CARRY and ZERO flags will be set to 1

Hint:

If addition exceeds 8 bits: \[ CY=1 \] If final 8-bit result becomes: \[ 00H \] then: \[ Z=1 \]

Answer 1

The Correct Option is C

Concept: In 8085:
• Carry flag becomes 1 if addition produces carry beyond 8 bits.
• Zero flag becomes 1 if result equals zero.

Step 1: Convert hexadecimal numbers into binary. \[ 87H = 10000111_2 \] \[ 79H = 01111001_2 \]

Step 2: Perform hexadecimal addition. Adding: \[ 87H+79H \] Hexadecimal calculation: \[ 87H+79H=100H \] Now: \[ 100H = 1\ 00000000_2 \]

Step 3: Determine Carry flag. The result exceeds 8 bits. A carry is generated beyond D7 bit. Therefore: \[ \text{Carry flag}=1 \]

Step 4: Determine Zero flag. The lower 8-bit result stored becomes: \[ 00H \] Since the result equals zero: \[ \text{Zero flag}=1 \]

Step 5: Write the final answer. Thus: \[ \text{Carry}=1,\quad \text{Zero}=1 \] Hence the correct option is: \[ \boxed{(D)\ \text{Both CARRY and ZERO flags will be set to 1}} \]