The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be _____ A. 
Since the potential at \( A \) and \( B \) is the same, no current flows through the middle resistor (30 \( \Omega \)).
The two parallel branches consist of: \[ R_1 = 10 \Omega + 20 \Omega = 30 \Omega \] \[ R_2 = 40 \Omega \] The equivalent resistance: \[ R_{{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(30)(40)}{30 + 40} = \frac{1200}{70} = 17.14 \Omega \] The total current using Ohm’s law: \[ I = \frac{40V}{17.14} = 2.33 A \]
The Boolean expression $\mathrm{Y}=\mathrm{A} \overline{\mathrm{B}} \mathrm{C}+\overline{\mathrm{AC}}$ can be realised with which of the following gate configurations.
A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate
B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate
C. 3-input OR gate, 3 NOT gates and one 2-input AND gate
Choose the correct answer from the options given below:
In the digital circuit shown in the figure, for the given inputs the P and Q values are:

The truth table corresponding to the circuit given below is 
In the circuit shown, assuming the threshold voltage of the diode is negligibly small, then the voltage \( V_{AB} \) is correctly represented by:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
An organic compound (X) with molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ is not readily oxidised. On reduction it gives $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}(\mathrm{Y})\right.$ which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are: