Step 1: Resolution of 8-bit ADC.
For 8-bit:
\[
2^8 = 256 \, \text{levels}
\]
Input range: \(0 \to 5 \, V\).
Thus, step size:
\[
\Delta = \frac{5}{256} = 0.01953 \, V = 19.53 \, mV
\]
Step 2: Equivalent digital count.
For \(V_{in} = 1.9922 \, V\):
\[
N = \frac{V_{in}}{\Delta} = \frac{1.9922}{0.01953} \approx 102
\]
Step 3: Convert to hex.
Decimal 102 = Hex \(66\).
Step 4: Final result.
Thus, the ADC output is:
\[
\boxed{66H}
\]
A continuous time periodic signal \( x(t) \) is given by: \[ x(t) = 1 + 2\cos(2\pi t) + 2\cos(4\pi t) + 2\cos(6\pi t) \] If \( T \) is the period of \( x(t) \), then evaluate: \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt \quad {(round off to the nearest integer).} \]
The maximum percentage error in the equivalent resistance of two parallel connected resistors of 100 \( \Omega \) and 900 \( \Omega \), with each having a maximum 5% error, is: \[ {(round off to nearest integer value).} \]
Consider a distribution feeder, with \( R/X \) ratio of 5. At the receiving end, a 350 kVA load is connected. The maximum voltage drop will occur from the sending end to the receiving end, when the power factor of the load is: \[ {(round off to three decimal places).} \]
In the circuit with ideal devices, the power MOSFET is operated with a duty cycle of 0.4 in a switching cycle with \( I = 10 \, {A} \) and \( V = 15 \, {V} \). The power delivered by the current source, in W, is: \[ {(round off to the nearest integer).} \] 
The induced emf in a 3.3 kV, 4-pole, 3-phase star-connected synchronous motor is considered to be equal and in phase with the terminal voltage under no-load condition. On application of a mechanical load, the induced emf phasor is deflected by an angle of \( 2^\circ \) mechanical with respect to the terminal voltage phasor. If the synchronous reactance is \( 2 \, \Omega \), and stator resistance is negligible, then the motor armature current magnitude, in amperes, during loaded condition is closest to: \[ {(round off to two decimal places).} \]