In the case of Intel 8085, what would be the address range for a memory system where A0-A11 are the only lines used for addressing, The memory address starts from A000HQuestion 5Select one:a.A00H-A600Hb.A000H-A800Hc.A000H-A500Hd.A800H-AFFFFH

The Intel 8085 is an 8-bit microprocessor. Its memory addressing is based on the binary system. If only A0-A11 lines are used for addressing, it means that 12 bits are used for addressing.

Step 1: Calculate the total number of addresses Since 12 bits are used for addressing, the total number of addresses that can be generated is 2^12 = 4096.

Step 2: Convert the starting address to decimal The given starting address is A000H. In decimal, this is 40960.

Step 3: Calculate the ending address The ending address would be the starting address plus the total number of addresses. So, 40960 + 4096 = 45056.

Step 4: Convert the ending address back to hexadecimal The ending address in hexadecimal is B000H.

So, the address range for the memory system would be A000H-B000H. However, this option is not given in your choices. There might be a mistake in the question or the provided options.