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).
Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance \( R_p = 1 \, \Omega \) as shown in the figure. An external resistance of \( R_e = 2 \, \Omega \) is connected via the sliding contact.
The current \( i \) is :
In Bohr model of hydrogen atom, if the difference between the radii of \( n^{th} \) and\( (n+1)^{th} \)orbits is equal to the radius of the \( (n-1)^{th} \) orbit, then the value of \( n \) is:
Given the function:
\[ f(x) = \frac{2x - 3}{3x - 2} \]
and if \( f_n(x) = (f \circ f \circ \ldots \circ f)(x) \) is applied \( n \) times, find \( f_{32}(x) \).