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Learn cmos logic gates circuits test MCQs: 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) for online masters degree. Practice assessment test for scholarships, online learning complex gate quiz questions for competitive assessment in engineering majors.

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