Digital Electronics And Logic Gates

Step 1: Understanding the Concept:
The question asks to convert a decimal number (base-10) to its binary equivalent (base-2). The standard method for this is the repeated division by 2.
Step 2: Key Formula or Approach:
To convert a decimal number to binary, we repeatedly divide the decimal number by 2 and record the remainders. The process continues until the quotient becomes 0. The binary number is then obtained by reading the remainders from the bottom up.
Step 3: Detailed Explanation:
Let's convert the decimal number 25 to binary:
\begin{itemize} \item Divide 25 by 2: Quotient = 12, Remainder = 1 \item Divide 12 by 2: Quotient = 6, Remainder = 0 \item Divide 6 by 2: Quotient = 3, Remainder = 0 \item Divide 3 by 2: Quotient = 1, Remainder = 1 \item Divide 1 by 2: Quotient = 0, Remainder = 1 \end{itemize} Reading the remainders from the bottom up gives us 11001.
So, .
Alternatively, we can express 25 as a sum of powers of 2:
The powers of 2 are ..., 32, 16, 8, 4, 2, 1.
25 can be written as:
The coefficients of the powers of 2 give the binary representation: 11001.
Step 4: Final Answer:
The binary equivalent of 25 is (11001) . This corresponds to option (C).

Step 1: Understanding the Concept:
A universal logic gate is a logic gate that can be used, by itself, to implement any other type of logic function or logic gate.
Step 2: Detailed Explanation:
The basic logic gates are AND, OR, and NOT. These can be combined to create any digital logic circuit.
However, it has been shown that either the NAND gate or the NOR gate can be used to construct all three basic logic gates:
- NOT gate from NAND: Connect both inputs of a NAND gate together.
- AND gate from NAND: Follow a NAND gate with a NOT gate (which is another NAND gate).
- OR gate from NAND: Use three NAND gates according to De Morgan's theorem.
Because any logic function can be built using only NAND gates, the NAND gate is considered a universal gate. The same is true for the NOR gate.
The basic gates OR, AND, and NOT are not universal because, for example, you cannot create an AND gate using only OR gates.
Step 3: Final Answer:
Out of the given options, the NAND gate is a universal logic gate. Therefore, option (D) is correct.