Question: Assume that the debug utility initially shows the 8 0 8 6 register / memory contents to be as follows: AX = 0 1 2

Assume that the debug utility initially shows the 8086 register/memory contents to be as follows:
AX=0123 BX=0456 CX=0000 DX=0000 SP=0000 BP=0000 SI=0000 DI=000
DS =12 E B ES =12 E B S S-12 E B CS 12 E B I P=0203 NV UP EI PL NZ NA PO NC 12EB:0203 B83412 MOV AX,1234. What is the new contents of the IP register immediately after the current instruction is executed? a.204 b.200 c.206 d.205

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