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Digital electronics practice test MCQ: by de morgan's law y=bar(a)+bar(b+cd) is equals to with choices bar(a)bar(b+cd), bar(a)+bar(b)(cd), bar(a)+bar(b)bar(cd) and bar(a)bar(b)bar(cd) with online engineering degrees test preparation for communication and electronics engineering students. Free study guide is for online learning complex gate quiz with MCQs to practice test questions with answers.

MCQ: By De Morgan's Law Y=bar(A)+bar(B+CD) is equals to

1. bar(A)bar(B+CD)
2. bar(A)+bar(B)(CD)
3. bar(A)+bar(B)bar(CD)
4. bar(A)bar(B)bar(CD)

C

MCQ: By DE Morgan's Law Y=bar(A(B+CD)) is equals to

1. bar(A)bar(B+CD)
2. bar(A)+bar(B+CD)
3. A+bar(B+CD)
4. bar(A)+(B+CD)

B

MCQ: System in which Y is high for A input is high and simultaneously either B input is high or C and D both are high is represented as

1. Y=A(B+CD)
2. bar(Y)=A(B+CD)
3. bar(Y)=A(2+CD)
4. bar(Y)=AB+CD

A

MCQ: By De Morgan's Law Y=bar(A)+bar(B)bar(CD) is equals to

1. bar(A)bar(B+CD)
2. bar(A)+bar(B)(bar(C)+D)
3. bar(A)+bar(B)(C+D)
4. bar(A)+bar(B)(bar(C)+bar(D))

D

MCQ: System in which Y will be high if A is low or B and C are both low is represented as

1. Y=A+BC
2. Y=ABC
3. Y=A+bar(BC)
4. Y=bar(A)+bar(BC)

D