To get output 1 for the following logic circuit, the correct choice of the inputs is:

(1) A = 1, B = 1, C = 0
(2) A = 0, B = 1, C = 0
(3) A = 1, B = 0, C = 1
(4) A = 0, B = 0, C = 1
The logic circuit shown involves a combination of NAND gates. Let's break down the behavior of the circuit based on the inputs:
1. The first gate is a NAND gate. A NAND gate outputs 1 unless both its inputs are 1.
For inputs A = 1, B = 0 at this gate, the output will be 1 (since \(1 \cdot 0 = 0\), and the NAND of 0 is 1).
2. The second gate is another NAND gate. The input to this gate comes from the output of the first gate and the input C.
Now, the input to the second gate will be \(1 \text{ (from the first NAND gate)}\) and \(C = 1\).
Since the NAND of \(1 \cdot 1 = 1\), the output of the second gate will be 0.
Thus, the correct inputs that produce the output 1 are:
A = 1, B = 0, C = 1
Hence, the correct choice is Option (3).
Assertion (A): We cannot form a p-n junction diode by taking a slab of a p-type semiconductor and physically joining it to another slab of an n-type semiconductor.
Reason (R): In a p-type semiconductor, \( n_e \gg n_h \) while in an n-type semiconductor \( n_h \gg n_e \).
The graph shows the variation of current with voltage for a p-n junction diode. Estimate the dynamic resistance of the diode at \( V = -0.6 \) V.


