TANGEDCO AE EE Full Test - 4 - Electrical Engineering (EE) MCQ

Given that,
z(t) = x(t) * y(t)
Apply Laplace transform,
Z(s) = X(s) . Y(s)
Now apply time scaling property,

Apply inverse Laplace
z(ct) = c x(ct) * y(ct)

Let

It is a ∞° form. To evaluate such limits use log concept.

Taking log on both sides

Now applying L-H rule,

Concept:
Cauchy’s Integral Formula - If f(z) is analytic within and on the closed curve C and if z_o is any point inside C then

Cauchy’s Residue Theorem – If f(z) is analytic in a closed curve C except at a finite number of singular point lies inside C then,

If f(z) has a simple pole at z = a then,

Calculation:

∴ Sum of residue must be equal to 2

∴ Z = 1 must lies inside C, Z = 2 lies outside C.

Concept:


g(x) = ϕ(x-2) = |x-2|
Shift the graph to right on X-axis by 2 units.

F(x) = g(x) + 3 = 3 + |x-2|
Shift the above graph 3 units on the y-axis in the upward direction.

As the graph is continuous on the given point (2, 3). So the given function is continuous at the point (2, 3). But there is a sharp edge at point (2, 3), so the limiting value of the function in the left and right proximity of the point will be different. Hence the function is not differentiable at the point (2, 3).

In limit calculation problems, the first thing we should try is to simply plug in the values and see if we can compute the limit.
In the question, we can simply put the values of x and y in the function to get:

= 1 + 4 = 5
Important Point
L' Hospital's rule:
If by plugging the values the limit is coming to be 0/0 or , all we need to do is differentiate the numerator and differentiate the denominator and then take the limit. We can keep on differentiating until the limit can be estimated.
For example,

Putting x → 0, we are getting 0/0

Differentiating the numerator and denominator we get,

Putting x → 0, we are getting 0/0
Differentiating the numerator and denominator again we get,

Putting x → 0, we are getting 0/0

Differentiating the numerator and denominator again we get,

=

Concept:

• Wavelength (λ): It is equal to the distance traveled by the wave during the time in which any one particle of the medium completes one vibration about its mean position. It is the length of one wave.
• Frequency (f) : It is defined as the number of vibrations completed by the particle in one second.
  • It is the number of complete wavelengths traversed by the wave in one second.

The relation between velocity, frequency, and wavelength is given by
c = f × λ
where, c = velocity of light, f = frequency Hz, λ = wavelength
Calculation:
Given that, f = 108 MHz, c = 3 × 10⁸ m/s (standard value)
Putting all the values in
c = f × λ
λ = c/f = 3 × 10⁸/ 108 × 10⁶ = 2.7 m
The correct option is 2.7 m.

int add (int a, int b) // formal argument
{
return a + b;
}
main()
{
int c;
c = add (10, 20);
printf("c = %d", c);
}

In this Program Add function is following the call by value mechanism so that value cannot be called outside the function So the answer will be a+b = 10+ 20 = 30

Therefore Option B is correct

Concept:

• The same adsorbent (charcoal in this case) at the same temperature will adsorb different gases to a different extent. The extent to which gases are adsorbed is proportional to the critical temperature of the gas.

while Loop:
Both while loops and do-while loops are condition-controlled, meaning that they continue to loop until some condition is met. While loops check for the stopping condition first, and may not execute the body of the loop at all if the condition is initially false.

where the body can be either a single statement or a block of statements within { curly braces }.
do-while Loops:
do-while loops are exactly like while loops, except that the test is performed at the end of the loop rather than the beginning. This guarantees that the loop will be performed at least once.

In theory, the body can be either a single statement or a block of statements within { curly braces }, but in practice, the curly braces are almost always used with do-whiles.
for Loops:
for-loops are counter-controlled, meaning that they are normally used whenever the number of iterations is known in advance.

Concept:

• ​The distinction between a solid and a fluid is made on the basis of the substance’s ability to resist an applied shear (tangential) stress that tends to change its shape.
• A solid can resist an applied shear by deforming its shape whereas a fluid deforms continuously under the influence of shear stress, no matter how small its shape is.
• A fluid can be defined as a substance which is capable of flowing and changing its shape according to the surrounding without offering internal resistance. Liquid offers no resistance to change of shape and it conforms to the shape of its surrounding.
• The fluid is a substance that continues to deform under the action of shear forces. If shear force is absent, fluid will be at rest.

The given network has a pole at s = -1 and a zero at s = -2.

The transfer function of the network can be represented as

TF = s+2 / s+1

The phase angle of the above system is,

The above phase angle is negative for all the frequencies. Therefore, the above system provides a lagging angle. This network is an example of lag compensator.

Therefore, if this network is excited by sinusoidal input, the output lags the input.

Note:

If the network provides a leading angle, then the output will be leading the input.

If the network provides zero angle, then the output will be in phase with the input.

Concept:
Voltage regulation is the change in secondary terminal voltage from no load to full load at a specific power factor of load and the change is expressed in percentage.
E₂ = no-load secondary voltage
V₂ = full load secondary voltage
Voltage regulation for the transformer is given by the ratio of change in secondary terminal voltage from no load to full load to no load secondary voltage.

Voltage regulation
It can also be expressed as,

Regulation

Regulation = %R cos ϕ ± %X sin ϕ
+ sign is used for lagging loads
- ve sign is used for leading loads
Where %R = ohmic drop
%X = reactance drop
Calculation:
Ohmic loss (%R) = 1%
Reactance drop (%X) = 5%
Power factor = cos ϕ = 0.8 leading
∴ sin ϕ = 0.6
Regulation = %R cos ϕ ± %X sin ϕ
= 1 (0.8) - 5 (0.6) = -2.2%

In a Bode magnitude plot,

• For a pole at the origin, the initial slope is -20 dB/decade
• For a zero at the origin, the initial slope is 20 dB/decade
• The slope of magnitude plot changes at each corner frequency
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The final slope of Bode magnitude plot = (Z – P) × 20 dB/decade

Type-1 System:

A low-frequency slope of the magnitude plot of -20 dB/decade indicates a single pole of G(s)H(s) at s = 0 and therefore a type 1 system.

Concept:

For the Step-Down, DC chopper-

V0 = α Vs

Where V0 = Average output Voltage

Vs = Input Voltage

And the duty ratio is α.

Ripple factor is given by-

Voltage Ripple factor

Where form factor (FF) =

So, V.R.F. =

Calculation:

Vs = 200 Volt

V0 = (1/3) × 200 = 66.66 Volt

V.R.F. = ≈ 1.5

The correct answer is At the network.

Packet-filtering firewalls operate at the network layer (Layer 3) of the OSI model. Packet-filtering firewalls make processing decisions based on network addresses, ports, or protocols.

Transport Layer :

• It performs the same functions as that of the transport layer present in the OSI model. Here are the key points regarding the transport layer:
  1. It uses TCP and UDP protocol for an end to end transmission.
  2. TCP is a reliable and connection-oriented protocol.
  3. TCP also handles flow control.
  4. The UDP is not reliable and a connectionless protocol also does not perform flow control.
• Protocols utilized in this layer: TCP/IP and UDP protocols are deployed in this layer.

Internet Layer :

• The responsibility of this layer is to permit the host to insert packets into the network and then make them travel independently to the destination. However, the sequence of receiving the packet may vary from the sequence they were sent.
• Protocols used: Internet Protocol (IP) is deployed in the Internet layer.

Chopper circuits are known as DC to DC converters. Similar to the transformers of the AC circuit, choppers are used to step up and step down the DC power. They change the fixed DC power to variable DC power.

Applications of Choppers:

Load Commutated Chopper:

In this type of commutation, the current through the SCR is reduced below the holding current value by resonating the load. In this type, the load circuit is so designed that even though the supply voltage is positive, an oscillating current tends to flow and when the current through the SCR reaches zero, the device turns off.

This type of chopper is used in Series Inverter circuits.

Current Commutated Chopper:

This is a type of commutation in which an LC series circuit is connected across the SCR. Since the commutation circuit has negligible resistance it is always under-damped i.e., the current in the LC circuit tends to oscillate whenever the SCR is on.

Voltage Commutated Chopper:

Dividing, we get

…1)

We know that X(z)

Comparing 1) and 2) we get

Concept:
Let's consider 3-phase balanced voltage phasors as
V_R = V_m sin (ωt)
V_Y = V_m sin (ωt - 120°)
V_B = V_m sin (ωt -240°)
Where
V_R, V_Y, V_B are phase voltages of the balanced 3 phase system
V_m is the maximum value
Calculation:
Given that
at particular instant V_R = + 30 V, V_Y = -90 V
V_R = V_m sin (ωt) = +30
V_Y = V_m sin (ωt - 120°) = -90
Divide V_R by V_Y
We will get, sin (ωt - 120°) = -3 sin (ωt)By solving wt =19.1°
And V_R = 30 = V_m sin (ωt)
⇒ V_m = 91.68 V
Now substitute the value of wt and V_m in the V_B equation,
V_B = V_m sin (ωt +120°)
= 91.68 sin( 19.1°+120°)
=60 V
Therefore the value of B phase V_B = +60 V
Alternative method:
At any instant of time sum of all the three phasors of three phase balanced system is equal to zero
so, V_R + V_Y + V_B =0
Given that at some instant V_R = +30 V , V_Y = -90 V
From the above equation +30 + (-90) + V_B = 0
⇒ V_B = +60 V

The differential relay actually compares between primary current and secondary current of power transformer, if any unbalance found in betweenprimary and secondary currents the relay will actuate and inter trip both the primary and secondary circuit breaker of the transformer.

For differential protection of power transformer, CT s must be connected in star in the delta side of the transformer and in delta in the star side of the transformer.

Concept:

• The number of infinite radii half circles will be equal to the number of poles at the origin.
• The infinite radius half-circle will start at the point where the mirror image of the polar plot ends. And this infinite radius half-circle will end at the point where the polar plot starts.

Analysis:

The above plot is not having ‘∞’ radius semi-circle. It means the system is a Type-0 system.

In a differentially compound DC generator,

Flux due to shunt field ϕ_rh = 80% of ϕ

Flux due to series field ϕ_re = 20% of ϕ

Since all the resistances are ignored, the voltage drop is zero.

⇒ V_t = E_g

As, V_t = E_g - I_aR_a

V_t = terminal voltage

E_g = generated voltage

The generated voltage is,

⇒ E_g ∝ ϕ

⇒ V_t ∝ ϕ ⇒ V_t ∝ (ϕ_rh – ϕ_re)

Now, the series field is suddenly short-circuited

⇒ ϕ_re = 0

Percentage change in terminal voltage = 100 × (V_tnew - V_told) / V_told

⇒ % change in V_t = 100 × (0.8 ϕ - 0.6 ϕ) / 0.6 ϕ

⇒ % change in V_t = 100 × 1 / 3 = 33.33%

∴ Terminal voltage increases by 33.33%

Concept:

KP = position error constant =

Kv = velocity error constant =

Ka = acceleration error constant =

Steady state error for different inputs is given by

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input and steady-state error for parabolic-input.

Calculation:

Given transfer function is

Acceleration error constant

Steady state error

Concept
Harmonic Analysis on AC side of the converter
The Fourier series for the source current waveform is

where,
θ_n = -nα

Here only odd harmonics are present, even harmonics are canceled (or) eliminated
The RMS value of n^th harmonic input current,
RMS value of fundamental current,
RMS value of total input current,

Also, θ₁ = -α (Negative sign for θ₁ indicates that fundamental current lags the source voltage).

Fundamental Displacement Factor (or) Displacement Factor

FDF = cos θ₁ = cos (-α) = cos α

Current distortion factor (CDF) or (g)

Harmonic Factor (or) Total Harmonic Distortion

Harmonic estimated by the shape of the waveform e.g. square wave case, THD = 48.34% does not depend upon the magnitude of voltage (or) current waveform.

Power factor

Harmonics on DC side of the converter

Voltage ripple factor (VRF) measure the harmonics on DC side of the converter

VRF = 48.3426%
At α = 0
i.e. ripple is minimum
At α = 90°, ripple is maximum
Active power Input (P)
Only the RMS fundamental component of contributes to delivered useful power in the AC side of the converter but in
case of dc, the average value is responsible to provide useful power
Active power input P = V_sr I_s1 cos α

P= V₀I₀
Reactive power input (Q):

Q = P tan α
Q = V₀ I₀ tan α
Calculation
Input power factor = 0.9 cos α
= 0.9 × cos 30° = 0.78

A discrete-time signal is periodic if there is a non-zero integer N ∈ discrete time such that for all n ∈ discrete time, x(n + N) = x(n).

The smallest value of N is known as the fundamental period.

The signal repeats after every N value.

x(n) = (-1)ⁿ

= 1, for n = 0, ±2, ±4, ±6, ±8, …

= 0, for n = ±1, ±3, ±5, …

The given signal repeats after every two values.

Therefore, the fundamental period = 2

Alternate method:

Given discrete-time signal is, x(n) = (-1)ⁿ

= (e^iπ)ⁿ = e^iπn

= cos nπ + sin nπ

= conπs

Fundamental frequency, ω = π

Fundamental time period

Concept:
Convolution of signals:
Convolution is a mathematical way of combining two signals to form a third signal.
Convolution methods:

1. Sum by column method or Convolution summation method
2. Periodic or Circular convolution method
3. Matrix method.

Let, no. of terms in x₁(n) = m,
No. of terms in x₂(n)= n
then no. of terms in output signal y(n) is =m + n - 1.
y(n) = x₁(n) * x₂(n)
Calculation:
Given signals are
x₁(n) = x₂(n) = {1, 1, 1}
m = 3, n = 3
By using Sum by column method.

y(n) = x₁(n) * x₂(n)

Output y(n) = {1, 2, 3, 2, 1}

Concept:
When a carrier is modulated by different waves having different modulation indexes, the overall modulation index is given by:

Calculation:
Given μ₁ = 0.4 and μ₂ = 0.3, the overall modulation index will be:

μ_eff = 0.5
Important Point
We can calculate the total power of the modulated signal for the given overall modulation index as:

P_c = Carrier Power
μ_eff = Effective modulation Index

Concept:

If two lines are perpendicular, then their dot product will be zero.

Calculation:

Let the point on the plane be (x, y, z)

Given that plane passes through the point (3, -1, 1), then the line vector passing through these two points is

since it is perpendicular to their dot product becomes zero

4x + 2y - z = 9

Concept:

The correlation coefficient is given by

Cov(X,Y) = E(XY) – E(X) E(Y)

Where cov(X, Y) = covariance of X and Y; σ = standard deviation; E = expectation;

Calculation:

Given X ∼ N (0, 1) ⇒ μ_x = 0 & σ_x² = 1 ⇒ σ_x = 1;

And the probability density function will be

μ_x = 0 ⇒ E(X) = 0;

var(X) = σ_X² = E (X²) – [E (X)]² = 1 ⇒ E (X²) = 1;

Given Y = X²; therefore Y is a chi-square distribution with 1 degree of freedom.

For a chi-square distribution with degree of freedom (n) = 1,

Expectation = E(Y) = n = 1; variance = σ_y² = 2 × n = 2;

Now

E(XY) = E (X³) =

Hence, the function inside the integral is odd, we can say it is symmetric w.r.t zero.

⇒ E(X³) = 0;

Now

Cov(X, X²) = E (X³) – E(X) E(X²) = 0 – 0 × 1 = 0;

A starter is necessary to start a DC motor because it restricts initial high armature current that exists, as the value of starting back-EMF is zero.
A resistor is used in the starter for a DC shunt motor

• When the connected dc motor is to be started, the lever is turned gradually to the right
• When the lever touches point 1, the field winding gets directly connected across the supply, and the armature winding gets connected with resistances R1 to R5 in series
• During starting, full resistance is added in series with the armature winding
• Then, as the lever is moved further, the resistance is gradually is cut out from the armature circuit
• Now, as the lever reaches to position 6, all the resistance is cut out from the armature circuit and armature gets directly connected across the supply

So, the purpose of including an external resistance at the time of starting a DC motor is to reduce the starting current.