⇒Semiconductor Devices

Correct Option is (C)

When the two inputs of a NAND gate are shorted, the resulting gate behaves as a NOT gate. This is because a NAND gate outputs the opposite of the AND operation. When both inputs are the same (due to shorting), the NAND gate essentially inverts that single input signal.

To illustrate :

• If the input is 0 (0 AND 0), a regular AND gate would output 0, but a NAND gate outputs the opposite, which is 1.


• If the input is 1 (1 AND 1), a regular AND gate would output 1, but a NAND gate outputs the opposite, which is 0.

So, with both inputs shorted, the NAND gate effectively becomes a NOT gate, inverting whatever single input it receives.

Therefore, the correct answer is :

Option C : a NOT gate.

Correct Option is (D)

$\therefore$ There are two NOR gates, one NOT gate, and one NAND gate.

$\therefore$ Output of NAND gate: $\overline{A\cdot B}$

Output of NOT and NOR Gate: $\overline{\overline{A}+B}$

Final output: $\stackrel{―}{(\overline{\mathrm{A}\cdot \mathrm{B}})+(\overline{\stackrel{―}{\mathrm{A}}+\mathrm{B}})}$

So, the output $\mathrm{Y}$ is 1 only if the input $\mathrm{A}$ and $\mathrm{B}$ is 1.

$\begin{array}{rl}& A=1\\ & B=1\\ & Y=\overline{\stackrel{―}{1\cdot 1}+\stackrel{―}{1}+1}\\ & Y=1\end{array}$

Correct Option is (C)

$\begin{array}{rl}& \therefore \mathrm{Y}=\overline{\mathrm{A}\cdot \mathrm{B}}\cdot (\mathrm{A}+\mathrm{B})\\ & \therefore \mathrm{Y}=(\bar{\mathrm{A}}+\bar{\mathrm{B}})\cdot (\mathrm{A}+\mathrm{B})\dots (\because \overline{\mathrm{A}\cdot \mathrm{B}}=\bar{\mathrm{A}}+\bar{\mathrm{B}})\\ & \therefore \mathrm{Y}=\bar{\mathrm{A}}\cdot \mathrm{B}+\bar{\mathrm{B}}\cdot \mathrm{A}\\ & \therefore \mathrm{Y}=\mathrm{A}\cdot \bar{\mathrm{B}}+\mathrm{B}\cdot \bar{\mathrm{A}}\end{array}$

This represents XOR gate.

Correct Option is (A)

[Note: The gate at the output in the circuit is AND gate. It requires minimum two inputs. As, there is only one input the gate is not operational.]

Correct Option is (C)

In the given circuit, there is AND, NOR and a NOT gate.

Output of AND gate: $\overline{\mathrm{A}\cdot \mathrm{B}}$

Output of NOT gate: $\bar{\mathrm{C}}$

$\therefore$ The boolean expression will be:

$\overline{\mathrm{A}\cdot \mathrm{B}}+\bar{\mathrm{C}}$