Digital Electronics - Memory and Storage - Discussion

Discussion Forum : Memory and Storage - General Questions (Q.No. 1)
1.
How many address bits are needed to select all memory locations in the 2118 16K × 1 RAM?
8
10
14
16
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
21 comments Page 1 of 3.

Adarsh said:   2 months ago
what is meant by 2218?

@All.

it is just the model number of the given RAM, you need to check the chip size they have given it is 16k x 1.

Hence, we know that chip size is equal to a multiple of the depth size of the chip (16k) and the width of the chip (1bit).
Hence it is equal to 16k only, we know that 2^n, where n is the number of address lines used, 16k (16*1024) can be written as (2^4) + (2^10) after solving we get 2^ (4+10) equals to 2^14 after refer this to 2^n we get the n=14 as the answer.

Rudra prasad patra said:   2 years ago
How is it?

Anybody, Please explain in detail.

Aniket jhala said:   4 years ago
What about 2118? Explain in detail.

Chandu said:   5 years ago
It refers to Intel family model number.

Keshava said:   7 years ago
What is mean by 2118?

S Singh said:   8 years ago
What is mean by 2118?

Ajaj Ahmad said:   8 years ago
According to me, 10 is the right option.

Vaishnavi atram said:   9 years ago
Ram is only 2gb.

How to increase the RAM?

Abshir abdirizak said:   9 years ago
The answer is 10 + 4 = 14.

Dhanaji said:   9 years ago
Sir but how to use this 10+4.


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