Electronics and Communication Engineering - Digital Electronics - Discussion

Discussion Forum : Digital Electronics - Section 3 (Q.No. 3)
3.
Decimal number 46 in excess 3 code =
1000 1001
0111 1001
0111 1111
1000 1111
Answer: Option
Explanation:

Decimal 46 in excess 3 code = 46 + 33 = 79 in decimal = 0111 1001 in 4 bit binary.

Discussion:
7 comments Page 1 of 1.

Krishna said:   4 years ago
Thanks @Rohit.

Rohit said:   5 years ago
0100 0110
+ 0011 0011 individual 3 binary code add then;
---------------------------
0111 1001
(1)

Jagdish sonagara said:   5 years ago
Right answer is 1001111.

Ch@ndu said:   6 years ago
Excess 3 is obtained by 3 to the individual digit.

Sunil said:   6 years ago
Why we can not write it 46+3=49?

Please tell me.

Ashok said:   7 years ago
You are right @Shreekar.

But we add 3 for individual digit.

46 = 4+3 and 6+3 so its 79.
(1)

Shreekar said:   7 years ago
The excess 3 code should be 3 more than decimal code not 33.

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