Test: ESE Electrical - 2 - Electrical Engineering (EE) MCQ

100 = 65 + v₂
⇒ v₂ = 35 V

v₃ - 30 = v₂ 
⇒ v₃ = 65 V

105 + v₄ - v₃ - 65 = 0
⇒ v₄ = 25 V

v₄ + 15 - 55 + v₁ = 0
⇒ v₁ = 15 V 

Voltage across 60 Ω resistor = 30 V 
Current = 30 / 60 = 0.5 A 
Voltage across R1 is = (70 - 20) V = 50 V
R₁ = 50 / 0.5 = 100 Ω

If we go from +ve side of 1 kΩ through 7 V, 6 V and 5 V
We get, v₁ = 7 + 6 - 5 = 8 V

When current source i(t) connected across an inductor(L), then the voltage across the inductor is given as

When voltage source v(t) connected across a capacitor(C), then the current across the capacitor is given as

Calculation:

Given i(t) = (8e^-250t - 4e^-1000t) A for t ≥ 0, L = 0.5 mH

Then the voltage across the inductor is given as

At t = 0 its value is given as

V_L = 0.5 × 10^-3 × (- 2000 + 4000)

VL = 1 V

Using KVL,
12 - V₁ - 8 = 0.
V₁ =  4V.
8 - V₂ - 2=0.
V₂ = 6V.

Assuming i₁ and i₂ be the currents in loop 1 and 2 respectively. In loop 1, 4 + 2i1+3(i₁ - 17/3) + 4(i₁ - i₂) + 5 = 0
In loop 2, i₂(4 + 1 + 5)-4i₁ - 5 = 0
⇒ -4i₁ + 10i₂ = 5.
Solving these equations simultaneously i₂ = 1.041A and i₁ = 1.352A
V= i₂ x 5 = 5.21V.

10 sin (t + 30°) = 10 cos (t - 60°)

 

 ..........(i)

In a delta-connected load, the relation between line voltage and the phase voltage is line voltage = phase voltage.

The voltage V_YB is V_YB = 400 ∠ -120⁰V. The impedance Z₂ is Z₂ = 40 ∠ 60⁰Ω

⇒ I_Y = (400 ∠ -120^o)/(40 ∠ 60^o)=(-10 + j0)A.

Total flux leaving the entire surface is, ψ = 4πr2D from Gauss law. Ψ = 4π(1/16π2) X 16π = 4.

The standard unit of charge is Coulomb. One coulomb is defined as the 1 Newton of force applied on 1 unit of electric field

The autocorrelation function of a signal is defined as the measure of similarity or coherence between a signal and its time delayed version. Thus, autocorrelation is the correlation of a signal with itself.

Property of autocorrelation function:

• The autocorrelation function of energy signals exhibits complex conjugate symmetry, which means the real part of autocorrelation function R(τ) is an even function of delay parameter ( τ) and the imaginary part of R(τ) is an odd function of the parameter τ. Thus,

R(τ)=R∗(−τ)

All gives the same z transform with different ROC. So all are the solution.

x₁[n] is right-sided signal

With N even

Concept:
The sampling theorem states that “a signal can be exactly reproduced if it is sampled at the rate f_s which is greater than twice the maximum frequency W in the modulating signal."
If the sampled frequency is less than the Nyquist frequency, overlapping of lower and upper sidebands known as aliasing takes place.
The main reason for aliasing is under-sampling, i.e. sampling the signal at a frequency less than twice the Nyquist rate.
f_s < 2f_m
f_s = Sampling frequency
f_m = Modulating frequency
Application:
The minimum sampling rate to avoid aliasing will be:
f_s = 2 × 2f_m = 2 × 4 kHz
f_s = 8 kHz
∴ For f_s ≥ 8 kHz, the sampling frequency will satisfy the Nyquist rate and no aliasing will take place.
And for both 7 kHz and 5 kHz sampling rates, the Aliasing effect will be seen in the reconstructed signal as the sampling frequency is less than the Nyquist Rate.

Aliasing is explained with the help of the spectrum as shown:

Explanation: We know that, for a symmetric h(n), the number of filter coefficients that specify the frequency response is (M+1)/2 when M is odd and M/2 when M is even.

We know that, for a anti-symmetric h(n) h(M-1/2)=0 and thus the number of filter coefficients that specify the frequency response is (M-1)/2 when M is odd and M/2 when M is even.

On shifting the take-off point beyond the block G₃ and combining blocks G₂ and G₃, the given block diagram is reduced as shown below.

The block diagram is further reduced as shown below.

Taking Laplace transform on both sides of the given differential equation, we get (s² + 2s + 2) Y(s) = 3

Since, ξ < 1, therefore given system is underdamped.

Both assertion and reason are individually true. The correct reason for assertion is BIBO stability criteria,

Number of open loop poies = P = 3
and number of open loop zeros = Z = 1
Poles are at: s = -1 ± j, 0 and zero is at s = -2


= 0
Here, P - Z = 2
So, angle of asymptotes are given by