1 votes 1 votes In the following circuit, $\text{X}$ is given by $\text{X}=\text{A}\; \overline{\text{B}}\; \overline{\text{C}}+\overline{\text{A}} \;\text{B}\; \overline{\text{C}}+\overline{\text{A}}\; \overline{\text{B}} \;\text{C}+\text{A B C}$ $\text{X}=\overline{\text{A}} \;\text{B C + A} \;\overline{\text{B}}\; \text{C + A B} \;\overline{\text{C}}+\overline{\text{A}}\; \overline{\text{B}}\; \overline{\text{C}}$ $\text{X = A B + B C + A C}$ $\text{X}=\overline{\text{A}} \;\overline{\text{B}}+\overline{\text{B}}\; \overline{\text{C}}+\overline{\text{A}} \;\overline{\text{C}}$ Digital Logic goclasses2024-iiith-mock-5 goclasses digital-logic combinational-circuit multiplexer 1-mark + – GO Classes asked Apr 30, 2023 • retagged Apr 29 by Lakshman Bhaiya GO Classes 102 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes $\mathrm{Y}$ is $\text{A} \oplus \text{B}$ So, $\mathrm{X}=(\text{A} \oplus \text{B}) \oplus \text{C}$ $\mathrm{X}$ is $1$ iff Odd number of inputs are $1.$ So, the answer is Option A. GO Classes answered Apr 30, 2023 GO Classes comment Share Follow See all 0 reply Please log in or register to add a comment.
