Question

Combination of NAND gates is shown in the figure. It is equivalent to

Choose the correct answer from the options given below

• A. AND gate
• B. NOR gate
• C. OR gate
• D. X-OR gate

Hint:

Memorize this standard shorthand combination for entrance exams:
NOTs before a NAND $\implies$ OR gate.
NOTs before a NOR $\implies$ AND gate.
Recognizing these standard gate configurations saves you from drawing full truth tables under time pressure!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question presents a digital logic circuit constructed from three NAND gates. We need to analyze how binary signals propagate through this configuration to find its single gate equivalent.

Step 2: Key Formula or Approach:
The Boolean output of a standard two-input NAND gate is $\overline{A \cdot B}$.
If the two inputs of a NAND gate are tied together, its Boolean expression simplifies to: $$Y = \overline{A \cdot A} = \overline{A}$$ This means a shorted-input NAND gate acts exactly like a

NOT gate (inverter).
We can analyze the total combination using

De Morgan's Laws: $$\overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B$$

Step 3: Detailed Explanation:
Let's analyze the circuit network step-by-step from left to right: 1. The top-left NAND gate has its inputs shorted together and receives input $A$. Its output is: $$Y_1 = \overline{A}$$ 2. The bottom-left NAND gate has its inputs shorted together and receives input $B$. Its output is: $$Y_2 = \overline{B}$$ 3. The final NAND gate on the right receives these two inverted signals ($Y_1$ and $Y_2$) as its inputs. Its output $Y$ is: $$Y = \overline{Y_1 \cdot Y_2}$$ Substitute the intermediate expressions for $Y_1$ and $Y_2$ into the final output equation: $$Y = \overline{\overline{A} \cdot \overline{B}}$$ Apply De Morgan's Law to break the outer inversion bar and change the multiplication operator to addition: $$Y = \overline{\overline{A}} + \overline{\overline{B}}$$ Since a double negation cancels out ($\overline{\overline{A}} = A$), the expression simplifies to: $$Y = A + B$$ The expression $A + B$ is the exact Boolean function of an

OR gate.

Step 4: Final Answer:
The combination is equivalent to an OR gate, matching option (C).