Most kinds of data is organised into 8-bit blocks, so a processor of any bit width (4, 8, 16, 32, 64, etc) will need to be able to handle 8-bit (byte-wise) data.
based on the size of the data bus they determine whether it is a 8 bit or 16 bit . here in 8086 it has 16 bit data bus hence it is called as 16 bit microprocessor
Bit rate = 8 / (16 * 10-9) bits/second
Most 8-bit ISA cards should work in a 16-bit slot (unless they are hardwired to an incompatible IRQ or have a skirt on them). Some 16-bit ISA cards can operate in an 8-bit slot, but most will not.
the uc is 8 bit or 16 is depended on ALU and speed of process and seeing the ports
8 bit registers cannot be used as 16 bit registers. The reverse works, however, as the 16 bit general purpose registers of the 8086 and 8088 can be used as pairs of 8 bit registers. AX is divided into AH (high 8 bits) and AL (low 8 bits), and BX, CX, and DX are similarly divided.Operations on 16 bit and operations on 8 bit registers are similar. So you can do add ah, bl, just as you could do add ax, bx.
The 8085 is an 8-bit microprocessor. Even though there are some 16-bit registers (BC, DE, HL, SP, PC), with some 16-bit operations that can be performed on them, and a 16-bit address bus, the accumulator (A), the arithmetic logic unit (ALU), and the data bus are 8-bits in size, making the 8085 an 8-bit computer.
8/16
Because it is an 8-bit microprocessor.
The ISA slot started as 8-bit, and then evolved to 16-bits.
first you need to find the common bottom bit e.g 9/8 +6/16. first you need to do 8 times _ equals 16 and the answer is 2 so now whatever you do to the bottom you have to do to the top so the question is now 18/16 + 6/16 witch is 24/16
a. 8 bit b. 16 bit c.32 bit d.64 bit