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Chapter 19: Problem 2619

For Boolean identities match the pair: (1) \(\underline{\underline{A}}\) (P) \(\underline{\mathrm{A}}+\underline{\mathrm{B}}\) (2) \(\underline{\mathrm{A}+\mathrm{B}}\) (Q) A \cdot B (3) \(\underline{\mathrm{A} \cdot \mathrm{B}}\) (R) \(\underline{A} \cdot \underline{B}\) (4) \(\mathrm{A} \cdot(\underline{\mathrm{A}}+\mathrm{B})\) (S) \(\mathrm{A}\) (A) \((1)-(\mathrm{S}),(2)-(\mathrm{P}),(3)-(\mathrm{Q}),(4)-(\mathrm{R})\) (B) \((1)-(\mathrm{S}),(2)-(\mathrm{R}),(3)-(\mathrm{Q}),(4)-(\mathrm{P})\) (C) (1) - (S), (2) - (Q), (3) - (P), (4) - (R) (D) \((1)-(\mathrm{S}),(2)-(\mathrm{R}),(3)-(\mathrm{P}),(4)-(\mathrm{Q})\)

Short Answer

Expert verified
The short answer is: (D) (1) - (S), (2) - (R), (3) - (P), (4) - (Q)

Step by step solution

01

1) Simplifying the given Boolean identities expressions

Let's start by simplifying or rearranging the expressions: (1) \(\underline{\underline{A}}\) : The complement of complement A is simply A. (2) \(\underline{\mathrm{A}+\mathrm{B}}\) : This is the complement of the sum of A and B. (3) \(\underline{\mathrm{A} \cdot \mathrm{B}}\) : This is the complement of the product of A and B. (4) \(\mathrm{A} \cdot(\underline{\mathrm{A}}+\mathrm{B})\) : Simplifying this expression using the distributive law of Boolean algebra, we have \(A \cdot \underline{A} + A \cdot B\). And since \(A\cdot\underline{A}=0\), it becomes \(0 + A \cdot B\), which is simply \(A \cdot B\). Now let's move to the equivalent expressions.
02

2) Identify the equivalent expressions

For each original expression (1, 2, 3, and 4), we want to associate it with the equivalent expression (P, Q, R, and S): (P) \(\underline{\mathrm{A}}+\underline{\mathrm{B}}\) : Using De Morgan's law, this simplifies to \(\underline{A \cdot B}\). (Q) A \(\cdot\) B : This is simply the product of A and B. (R) \(\underline{A} \cdot \underline{B}\) : Using De Morgan's law, this simplifies to \(\underline{A + B}\). (S) A : This is simply A.
03

3) Match the pairs

Now that we have simplified and/or identified the equivalent expressions for each Boolean identity, we can match the pairs as follows: (1) A -> (S) A (2) \(\underline{\mathrm{A}+\mathrm{B}}\) -> (R) \(\underline{A \cdot B}\) (3) \(\underline{\mathrm{A} \cdot \mathrm{B}}\) -> (P) \(\underline{A} + \underline{B}\) (4) \(A \cdot B\) -> (Q) A \(\cdot\) B So the correct choice is given by option (D): (D) (1) - (S), (2) - (R), (3) - (P), (4) - (Q)

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Most popular questions from this chapter

A N-P-N transistor conducts when collector is and emitter is with respect to base. (A) positive, negative (B) positive, positive (C) negative, negative (D) negative, positive

Direction for Assertion - Reason type questions (A) If both Assertion and Reason are true and reason is the correct explanation of assertion. (B) If both Assertion and Reason are true but Reason is not the correct explanation of assertion. (C) If Assertion is true but Reason is false (D) If both assertion and reason are false A : A transistor amplifier circuit in common emitter configuration has low input impedance \(\mathrm{R}\) : Base-emitter junction is forward biased (A) (B) (C) (D)

A n-p-n transistor is used in common emitter made in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by $2 \mathrm{~mA}\( and \)0.04 \mathrm{~V}$ in input voltage. An amplifier has voltage gain \(A_{V}=1000\). The voltage gain in \(\mathrm{dB}\) is (A) \(20 \mathrm{~dB}\) (B) \(30 \mathrm{~dB}\) (C) \(3 \mathrm{~dB}\) (D) \(60 \mathrm{~dB}\)

A n-p-n transistor is used in common emitter mode in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by $2 \mathrm{~m} \mathrm{~A}\( and \)0.04 \mathrm{~V}$ in input voltage. The input resistance is (A) \(1 \mathrm{k} \Omega\) (B) \(10 \Omega\) (C) \(10 \mathrm{k} \Omega\) (D) \(100 \Omega\)

A n-p-n transistor is used in common emitter made in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by $2 \mathrm{~mA}\( and \)0.04 \mathrm{~V}$ in input voltage. If a load of \(6 \mathrm{k} \Omega\) is used, then the voltage gain of the amplifier is (A) 100 (B) 200 (C) 300 (D) 400
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