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

Assume that the debug utility initially shows the 8086 register/memory contents to be as follows: A X=0123 BX=0456 CX=0000 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000 DS=12EB ES=12EB SS=12EB CS=12EBIP=0203 NV UP EIPL 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.200 b.204 c.205 d.206

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