Test: ESE Electrical - 8 - Electrical Engineering (EE) MCQ

Correct Answer :- B
Current due to voltage source is:

I' = 10/(64+36) 

= 0.1A.

Current due to current source is:

I = 0.5A.

Current ib = I' + I 

= 0.5+ 0.1

= 0.6A

• Since V is not proportional to R therefore, the element can’t be a resistor.
• At t = 5 ms, even if i ≠ 0, the element behaves as a short circuit therefore, the element can’t be a capacitor (since at t = 0 only capacitor behaves as short circuit).
• The current at t = 0 is zero and at t = 5 ms voltage across the element is zero therefore, the element must be an inductor (at t = 0, an inductor acts as open circuit and at t =∞ it behaves as short circuit).
• From the given voltage and current profile, we have:
  

From given figure:

• The equivalent combination of C₂ and C₃
  
• The equivalent combination of this 1 μF and  C₁ = 2μF is C₁+ 1 μF = 3μF
• Hence, the equivalent capacitance between terminals A and B is
  

Q = S sin θ 
⇒ 

In a two-port network, the short-circuit forward transfer admittance represents the relationship between the input current and the output voltage when the output port is short-circuited. It is denoted as y21.

The different parameters denoted by y in a two-port network are:

• y11: Input admittance with the output port open (Short-circuit reverse transfer admittance)
• y12: Forward transfer admittance with the input port open (Open-circuit reverse transfer admittance)
• y21: Forward transfer admittance with the output port short-circuited (Short-circuit forward transfer admittance)
• y22: Output admittance with the input port open (Open-circuit forward transfer admittance)

Among these parameters, the short-circuit forward transfer admittance is represented by y21.

Therefore, the correct answer is C: y21.

As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the B impedance is IB = VBR∠240⁰/Z∠Ø = (V/Z)∠-240-Ø.

The line current I₂ is the difference of I_Y and I_R. So the line current I₂ is I₂ = I_Y – I_R = (-27.32 + j10) A.

To calculate the force on a conductor in a magnetic field, we can use the formula:

Force (F) = I * L * B * sin(theta)

Where:

• I is the current flowing through the conductor
• L is the length of the conductor
• B is the magnetic flux density
• theta is the angle between the direction of the current and the magnetic field

In this case, the given values are:

• I = 8A
• L = 12cm = 0.12m
• B = 3.75 units
• theta = 30 degrees = 30 * (pi/180) radians = pi/6 radians

Plugging in these values into the formula:

F = 8 x 0.12 x  3.75 x  sin(pi/6)

Using a calculator, we can find:

F ≈ 1.8 units

Therefore, the correct answer is D: 1.8.

The force per unit length of two conductors is given by
F = μ I1xI2/2πD, where I1 = I2 = 5 and D = 0.2. Thus F = 4π x 10^-7 x 52/ 2π x 0.2 = 25 x 10^-6 units.

Since, causal system

h(t) = 0 for t < 0

Input y(t) = x(t) * h(t)

h(τ) = 0 for τ < 0

x ( t - τ) = 0, for t - τ < - 2

∴ τ > f + 2

Sampling rate conversion by a rational factor involves changing the sampling rate of a signal by a non-integer ratio. This process requires both interpolation and decimation, but specifically, interpolation is necessary in this case.

Interpolation is the process of increasing the sampling rate by inserting additional samples between existing samples. It involves creating new samples that approximate the values of the original signal at higher frequencies. Interpolation helps in achieving a higher sampling rate while preserving the signal's characteristics.

Decimation, on the other hand, is the process of decreasing the sampling rate by discarding some samples. It involves reducing the number of samples in a signal while maintaining the important information and characteristics of the signal.

In sampling rate conversion by a rational factor, interpolation is required to generate additional samples to match the desired higher sampling rate. After interpolation, decimation can be performed if necessary to achieve the final desired sampling rate.

Therefore, the correct answer is A: Interpolation.

In the diagram given, a interpolator is in cascade with a decimator which together performs the action of sampling rate conversion by a factor I/D.

x(t) = ( e^-t - 2te^-2t) u(t)

A = 10 , T =  5, X [k] =  2

The maximum frequency of the band-limited signal.
f_m = 5 kHz
According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist  frequency which is given as
f_N = 2 f_m = 2 × 5 = 10 kHz
So the sampling frequency f_s ≥ f_N
f_s ≥ 10 kHz
Only the option (A) does not satisfy the condition.
∴ 5 kHz is not a valid sampling frequency.

The decimation of the data sequence should be repeated again and again until the resulting sequences are reduced to one point sequences. For N=2^v, this decimation can be performed v=log₂N times. Thus the total number of complex multiplications is reduced to (N/2)log₂N.

• In frequency sampling realization, the system function H(z) is characterized by the set of frequency samples {H(k+ α)} instead of {h(n)}.
• We view this FIR filter realization as a cascade of two filters. One is an all-zero or a comb filter and the other consists of parallel bank of single pole filters with resonant frequencies.

• We are required to design a low pass Butterworth filter to meet the following frequency domain specifications.
• KP ≤ 20 log|H(jΩ)| ≤ 0 and 20 log|H(jΩ)| ≤ KS.

• In closed loop system there is the presence of a feedback path and this feedback path has connected from output of the system to the input of the system.
• The whole operation is controlled by the output of the system.

In a signal flow graph, the net gain between the input and output node represents the relationship between an input variable and an output variable. This net gain is calculated by multiplying the individual gains along the paths from the input node to the output node.

The overall gain of the system refers to the cumulative effect of these individual gains on the input-output relationship. It represents the ratio of the output variable to the input variable, taking into account all the intermediate gains in the signal flow graph.

Therefore, the correct answer is A: Overall gain of the system.

Given, s² + 2s + 8 = 0
Here, ω_n = √8 rad/s = 2√2 rad/sec

system is underdamped.

Putting s = (z - 1) in the given characteristic equation, we get:
(z - 1)³ + 3 (K + 1) (z - 1)² + (7K+ 5) (z - 1) + (4K + 7) = 0
or, z³ + 3 Kz² + (K + 2)z + 4 = 0
Routh’s array is:


i.e. 3K² + 6K - 4 > 0 and K > 0
Now, roots of 3K² + 6K - 4 = 0 are: K = -2.5, 0.53
Since K > 0 ∴ K = 0.53 (K = -2.5 neglected)
For K > 0.53, 3 K² + 6 K - 4 > 0
Hence, the range of value of K is
0.53 < K < ∞ or ∞ > K > 0.53

Given, 

1+ G(s) = 0
or, s (s²+ 6 s + 12) + K = 0
or, K = - (s³ + 6s² + 12s)
For B.A. points, 

or, s² + 4s + 4 = 0
or, (s + 2)² = 0
or, s = - 2 , - 2
For given O.L.T.F., P = 3, Z = 0, P - Z = 3

Poles are at s = 0 and s = -3 ± j√3 = -3 ± j1.732
Since entire -ve real axis is a part of root locus, therefore s = -2 is a valid breakaway point.

Given, G(s) = (1+sT)/(sT)
The inverse polar plot of G(jω) is the polar plot of 1/G(jω)

Thus, 

and

Hence, inverse polar plot will be as show below,