accumulator of 8085 contains data F9h and register C contains 48h. what will be contents of accumulator after exceution of SUBC*
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E: ADD E with Accumulator
Before execution : [A] = 71H
instruction: ADD E
71 H = 01110001
ADD 39 H = 00111001
10101010 = AA H
After execution [A] = AA H
[Cy] = 00H
(ii) ORA E Logically OR with Acc Before execution : [A] = 71H
Instruction: ORA E
71 H 01110001
ORA 39 H 00111001
011110 01 = 79 H
After execution : [A] = 79H
RRC: Rotate accumulator right by one
bit.
Before execution : [A] = 71H01 1100
01Instruction: RRC
After execution [A] = 1 0 1 1 1 0 0 0 = B8
H
[A] = B8H [Cy] = 01H
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