Electrical Engineering - Online Test

Q1. When the output Y in the circuit below is ‘1’, it implies that data has

Answer : Option A
Explaination / Solution:
No Explaination.


Q2. Match the inferences X, Y, and Z, about a system, to the corresponding properties of the elements of first column in Routh‟s Table of the system characteristic equation X: The system is stable … P: … when all elements are positive Y: The system is unstable … Q: … when any one element is zero Z: The test breaks down … R: … when there is a change in sign of coefficients
Answer : Option D
Explaination / Solution:
No Explaination.


Q3. Consider the feedback system shown in the figure. The Nyquist plot of G (s) is also shown. Which one of the following conclusions is correct ?

Answer : Option D
Explaination / Solution:

Given the feedback system and the Nyquist plot of G (s)is

For the given system, we have the open loop transfer function as
G (s) = KG (s)
Considering the open loop system G (s) is stable, we have no open loop poles in right half plane
P = 0
From Nyquist theorem, we know that
N = P - Z
Where N is the number of encirclements of (-1 + j0), P is number of open loop poles in right half plane, Z is number of closed loop poles in right half plane. For stability, we must have
Z = 0
N = 0,   if closed loop system is stable
≠ 0,   if closed loop system is unstable
observing the Nyquist plot, we conclude that the plot of KG(s) encircles (-1 + j0)
if  K> 1
Hence, N ≠ 0 for sufficient large and positive value of K . Thus, the closedsystem is unstable for sufficiently large and positive K .


Q4. The trigonometric Fourier series for the waveform f (t) shown below contains

Answer : Option C
Explaination / Solution:

For a function x(t) trigonometric fourier series is

For an even function x(t),Bn = 0
Since given function is even function so coefficient Bn = 0, only cosine and constant
terms are present in its fourier series representation.
Constant term :


Constant term is negative.


Q5. The transfer function 


Answer : Option D
Explaination / Solution:

For the given capacitance, in the circuit, we have the reactance.


Q6. The logic function implemented by the circuit below is (ground implies logic 0)

Answer : Option D
Explaination / Solution:



Q7. A first-order low-pass filter of time constant T is excited with different input signals (with zero initial conditions up to t = 0). Match the excitation signals X, Y, Z with the corresponding time responses for t ≥ 0: 
X: Impulse                           P: 1 – e-t/T 
Y: Unit step                          Q: t – T(1 – e(-t/T)
Z :Ramp                               R: e-t/T
Answer : Option C
Explaination / Solution:

In general the first order, L.P.F filter transfer function is  because G(0) = k and G(∞) = ∞ if we take this transfer function as reference and give different input such as s(t).r(t).u(t)



Q8. The feedback configuration and the pole-zero locations of are shown below. The root locus for negative values of k , i.e. for -3 < k < 0, has breakaway/break-in points and angle of departure at pole P (with respect to the
positive real axis) equal to

Answer : Option B
Explaination / Solution:

The characteristic equation is


Q9. A function n(x) satisfied the differential equation where L is a constant. The boundary conditions are :n(0) = K and n() = 0. The solution to this equation is
Answer : Option D
Explaination / Solution:

Given differential equation


Q10. A source has an internal impedance of If a purely resistive load connected to this source has to extract the maximum power out of the source, its value inΩ should be
Answer : Option C
Explaination / Solution:

For the purely resistive load, maximum average power is transferred when

where
is the equivalent thevinin (input) impedance of the circuit. Hence,
 we obtain