# Digital Electronics - Code Converters and Multiplexers - Discussion

### Discussion :: Code Converters and Multiplexers - General Questions (Q.No.5)

5.

From the following list of input conditions, determine the state of the five output leads on a 74148 octal-to-binary encoder.
 I0 = 1 I3 = 1 I6 = 1 I1 = 1 I4 = 0 I7 = 1 I2 = 1 I5 = 1 EI = 0

 [A]. GS = L, A0 = L, A1 = L, A2 = H, EO = H [B]. GS = L, A0 = H, A1 = L, A2 = L, EO = H [C]. GS = L, A0 = L, A1 = H, A2 = L, EO = H [D]. GS = L, A0 = H, A1 = H, A2 = L, EO = H

Answer: Option D

Explanation:

No answer description available for this question.

 Tariku Kajela said: (Dec 22, 2010) explain it

 Chinnu said: (Apr 27, 2012) If any one knows please tell the answer, I try to solve but I don't know how to solve this.

 Ashish said: (Oct 14, 2012) I guess the answer should be a) as if I4=0 then output in binary is 100, NOTE- inputs of 74148 are low active hence giving a zero to I4 gives binary equivalent of 4 i.e 100, A2=1,A1=0,A0=0.

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