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TANGEDCO AE EE Full Test - 2 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - TANGEDCO AE EE Full Test - 2

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TANGEDCO AE EE Full Test - 2 - Question 1

If A = [2 0 1], then rank of AAT :

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 1

Concept:

AT is Transpose of a matrix which is obtained by changing rows to columns and columns to rows.

Rank of the matrix is Number of non- zero rows in the matrix.

Calculations:

Given A = [2 0 1]

AT is Transpose of a matrix which is obtained by changing rows to columns and columns to rows.

AT =

Consider, AAT = [2 0 1]

⇒AAT = [4 + 0 + 1] = [5]

We know that the Rank of the matrix is Number of non- zero rows in a matrix.

In AAT = [5], Number of non- zero rows in matrix is 1.

Hence, the Rank of the matrix AAT is one.

TANGEDCO AE EE Full Test - 2 - Question 2

In the Laurent expansion of valid in the region 1 < |z| < 2, the co-efficient of is

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 2

Concept:
Laurent series of f(z) is given by:


Calculation:
Given:

The given region is 1 < |z| < 2






Co-efficient of is -1

TANGEDCO AE EE Full Test - 2 - Question 3

Let X and Y be a bivariate random variable with correlation coefficient 1/2, and standard deviation 2 and 3 respectively, then Cov (X,Y) is

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 3

Concept:

Where r(x, y) is the correlation coefficient.
Calculation:

Given:

Correlation coefficient = 0.5

The variance of x = (standard deviation of x)2

(2)2 = 4

The variance of y = (standard deviation of y)2

(3)2 = 9

Now,

Correlation coefficient =

∴ Cov (x, y) = 0.5 × 6

Cov (x, y) = 3

TANGEDCO AE EE Full Test - 2 - Question 4

When an alternating current passes through an ohmic resistance, the electric power converted into heat is

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 4

Resistor:

  • A resistor limits the electrical current that flows through a circuit.
  • Resistance is the restriction of current.
  • The resistor dissipates the power into heat in Watts.
  • The power taken by the reactance is called reactive power.

Formula:

Apparent power S is

S = P + jQ

S = VIcos∅ + jVIsin∅

True power P = VIcos∅ Watts

Reactive power Q = VIsin∅ Var

The resistor dissipates electrical power = I2. R in watts into heat in the form of True power.

Important points

  • An inductor stores the energy in the form of magnetic energy(Kinetic energy)
  • A Capacitor stores the energy into electrical energy(potential energy).
TANGEDCO AE EE Full Test - 2 - Question 5

A doubly excited salient pole motor will have

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 5

Salient Pole Synchronous Motor:
For Salient pole synchronous motor armature is exited with A.C. and field is excited with D.C.
Hence it is called a doubly excited machine.
Mutual torque:

  • The torque which is produced by electric and magnetic field mutually is called Mutual torque
  • It is also known as Electromagnetic torque.
  • The electromagnetic torque will be in the same direction of rotor rotation.

For the production of unidirectional torque.

  • The relative speed between stator and rotor fields must be zero
  • The number of stator poles must be equal to the number of stator poles or an integer multiple.

Reluctance torque:

  • When the motor is rotating and then any one of the fields is de-excited or cutoff then reluctance torque will come into play.
  • The reluctance torque will be in the same direction of rotor rotation.
  • It will have the same speed as the synchronous speed of the motor.
  • The motor cannot be started with reluctance torque as the motor will have magnetic locking of rotor and stator.
  • So, special stator and rotor construction are needed to start with reluctance torque.

Hence, a doubly excited salient pole motor will have mutual torque and reluctance torque at the same instant.

TANGEDCO AE EE Full Test - 2 - Question 6

The presence of earth in case of overhead lines

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 6

As the earth is considered to be an infinite source of electrons, the presence of overhead lines causes charge carrier to be separated by a dielectric(air), causing an increase in capacitance.

TANGEDCO AE EE Full Test - 2 - Question 7

Three insulating materials with breakdown strength 250 kV/cm, 200 kV/cm, 150 kV/cm and permittivities of 2.5, 3.0 and 3.5 are used in a single core cable. If the factor of safety for the materials is 5, the location of the materials with respect to the core of the cable will be

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 7

Potential Distribution around the cable:

  • The volage and electrostatic or voltage stress in a single core cable has a maximum value (gmax) at the conductor surface and goes on decreasing as we move towards the sheath.
  • The maximum voltage that can be safely applied to a cable depends upon gmax i.e., electrostatic stress at the conductor surface.
  • For safe working of a cable having homogeneous dielectric, the strength of dielectric must be more than gmax.
  • If a dielectric of high strength is used for a cable, it is useful only near the conductor where stress is maximum. But as we move away from the conductor, the electrostatic stress decreases, so the dielectric will be unnecessarily overstrong.
  • The Material with the highest product of breakdown strength and permittivity should be placed nearer to the conductor.

The layer of cable and their voltage stress can be drawn as shown,

According to the above concept, the graph for potential distribution can be drawn as,

It will be possible only when ε1 < ε2 < ε3

Hence, the location of the materials with respect to the core of the cable will be 2.5, 3.0, 3.5.

In that case,

V1 = 250 × 2.5 = 625 V

V2 = 200 × 3 = 600 V

V3 = 150 × 3.5 = 525 V

Other Related Points
The unequal stress distribution in a cable is undesirable for two reasons. Firstly, insulation of greater thickness is required which increases the cable size. Secondly, it may lead to a breakdown of insulation.

Methods to achieve uniform stress distribution:

The process of achieving uniform electrostatic stress in the dielectric of cables is known as the grading of cables.

In order to overcome the above disadvantages, it is necessary to have a uniform stress distribution in cables. This can be achieved by distributing the stress in such a way that its value is increased in the outer layers of the dielectric. This is known as the grading of cables. The following are the two main methods of grading of cables:

  • Capacitance grading
  • Inter-sheath grading
TANGEDCO AE EE Full Test - 2 - Question 8

If a projectile is thrown with velocity 'v' and makes an angle 'θ' with the x-axis then the horizontal range of the projectile is given by the formula ______________.

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 8

Concept:

  • Projectile motion: Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
  • Initial Velocity: The initial velocity can be given as x components and y components.
  • Component of initial velocity in x-direction, (ux) = ucosθ
  • Component of initial velocity in the y-direction, (uy) = usinθ
  • In the case of projectile motion, we can see a free-fall motion of a body on a parabolic path with constant velocity.
  • If a body is thrown at a certain angle then during its movement, we get two components of velocity as given below.

  • And thus, the range of a projectile is the displacement of a particle along the x-axis and can be given as:

The range of the projectile,

  • Whereas the time of flight is the total time for which projectile stayed in the air.

Time of flight for the projectile, (R) = ucosθ × t 
The angle of projection = θ
Initial velocity = u
Gravitational acceleration = g
Time of flight = t
Range of projectile = R
Explanation:
As given above,
v = initial velocity of projectile
θ = angle with x-axis
Time of flight for the projectile,

And the range of the projectile,
Thus, by comparing the above two equation range of projectile can also be modified as

But (By using trigonometric relation)

TANGEDCO AE EE Full Test - 2 - Question 9

Behaviour of conductors, semiconductors and insulators is explained on the basis of

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 9
  • Behaviour of conductors, semiconductors and insulators is explained on the basis of energy band structure.
  • Energy band gap is measured in eV or electron volt and it is the energy separation between the valence band and conduction band; The valence band is below the conduction band.
  • The forbidden gap between the valence band and conduction band is very large in insulators. The energy gap of insulator is approximately equal to 15 electron volts (eV).
  • In a conductor, valence band and conduction band overlap each other, Therefore, there is no forbidden gap in a conductor.
  • In semiconductors, the forbidden gap between valence band and conduction band is very small. It has a forbidden gap of about 1 electron volt (eV).

Explanation

  • In conductor the conduction band is partially filled and the valanced band is partially empty or when the conduction and valance bands overlap. When there is overlap electrons from valence band can easily move into the conduction band.

  • In insulator, there exists a large band gap between conduction band and valence band Eg (Eg > 3 eV).
  • The valance band is fully filled with electrons.
  • There are no electrons in the conduction band, and therefore no electrical conduction is possible.

  • In a semiconductor, there exists a finite but small band gap between the conduction band and valence band (Eg < 3 eV). Because of the small bandgap, at room temperature, some electrons from the valence band can acquire enough energy to cross the energy gap and enter the conduction band.
TANGEDCO AE EE Full Test - 2 - Question 10

A DC shunt generator delivers 480 A at 200 V. Its armature and field winding resistance are 0.02 Ω and 40 Ω respectively. What will be the value of generated emf?

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 10

Concept:
EMF generated in the shunt generator is given by
Eg = V + IaRa
Where, Ia = armature current
V = Terminal voltage
Ra = armature resistance
Rsh = Shunt resistance
Field current in shunt generator is given by

Calculation:
Given -
V = 200 volt
Load current (I) = 480 A
Rsh = 40 Ω
Ra= 0.02 Ω
According to the question, the circuit diagram of DC shunt generator can be drawn as

∴ Field current,
∴ Armature current, Ia = If + I = 480 + 5
Ia = 485 A
Now generated emf is calculated as
Eg = 200 + 485 × 0.02 = 209.7 V

TANGEDCO AE EE Full Test - 2 - Question 11

A d.c. series generator has a no load induced e.m.f. of 10 volts. If its speed is doubled, the no load induced e.m.f. is

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 11

Concept:
Emf equation of DC generator:

Eg = PϕZN/60A
Where
P = Pole
ϕ = Field flux
Z = No. of conductors
N = Speed
A = No. of parallel path
So, emf directly proportional to field flux and the speed.
Calculation:
No load induced emf = 10 V
Speed is doubled. As the induced emf is directly proportional to speed, the induced emf will get doubled.
Therefore, at increased speed no load induced emf = 20 V

TANGEDCO AE EE Full Test - 2 - Question 12

The reactors are used in transmission line to

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 12

Fault current is inversely proportional to the reactance of the network. Therefore with the series reactor, the fault current magnitude is reduced. It is preferred in the generating stations in some special applications for limiting the current flowing through the low MVA rating circuit breakers.
Note:

  • Series reactors are used as current limiting reactors to increase the impedance of a system. They are also used to limit the starting currents of synchronous electric motors and to compensate reactive power in order to improve the transmission capacity of power lines.
  • A shunt reactor is an absorber of reactive power, thus increasing the energy efficiency of the system.
  • Whenever an inductive load is connected to the transmission line, power factor lags because of lagging load current. To compensate this, a shunt capacitor is connected which draws current leading the source voltage. The power factor can be improved.
  • Series capacitors are used to compensate for the inductance of the transmission line. They will increase the transmission capacity and the stability of the line. These are also used to share the load between parallel lines.
TANGEDCO AE EE Full Test - 2 - Question 13

Bounded-input bounded-output stability implies asymptotic stability for

1. Completely controllable system

2. Completely observable system

3. Uncontrollable system

4. Unobservable system

Which of the above statements are correct?
Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 13

BIBO stability: A linear system is said to be BIBO stable if the output is bounded for an arbitrary bounded input.

Asymptotic stability: It is the same as BIBO stability, except pole-zero cancellation should not be there.

If a system is asymptotic stable, then the system is BIBO stable but not vice versa.

Asymptotic stability → BIBO stability

BIBO stability + no pole-zero cancellation Asymptotic stability

To have no pole-zero cancellation, the state space representation is minimal, i.e. both controllable and observable.

BIBO stability + controllable and observable system → Asymptotic stability

Therefore, bounded-input bounded-output stability implies asymptotic stability for a completely controllable and observable system.

This also implies that a marginally stable system with minimal realization is not BIBO stable.

Important Points:

Controllability:

  • The concept of controllability of a system which is related to the transfer of any initial state of the system to any other desired state, in a finite length of time by application of proper inputs.
  • A system is said to completely state controllable if it is possible to transfer the system state from any initial state x(t0) to any other desired state X(tf) in a specified time interval (tf) by a control vector u(t).
  • A system is said to completely output controllable if it is possible to construct an unconstrained input vector u(t) which will transfer any given initial output y(t0) to any final output Y(tf) in a finite time interval t0 ≤ t ≤ tf.

Observability:

  • The observability is related to the problem of determining the system state by measuring the output for a finite length of time.
  • A system is said to be completely observable if every state X(t0) can be completely identified by measurements of the outputs Y(t) over a finite time interval.
  • If the system is not completely observable means that few of its state variables are not practically measurable and are shielded from the observation.
TANGEDCO AE EE Full Test - 2 - Question 14

A uniform plane wave with an intensity of magnetic field 6 A/m is travelling in a lossless non magnetic medium(ϵ = 36ϵ0). The magnitude of associated electric field is:

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 14

Concept:
Intrinsic impedance:
It is the ratio of the electric field intensity to the corresponding magnetic field
intensity for an electromagnetic wave.
It is denoted by ‘η’ and is mathematically calculated as:

For lossless nonmagnetic medium,

Calculation:
H = 6 A/m

E = ηH


E = 120π = 377 V/m = 0.377 kV/m

TANGEDCO AE EE Full Test - 2 - Question 15
By increasing the gain K of type 1 system, for a step input, the steady state error
Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 15

Concept:

Ka = 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.

Application:

For a type 1 system, for step input, the steady state error is independent of systems gain K and it is always zero. Therefore, by increasing the gain K, the steady state error remains zero.

Important Points:

As the type of the system increases, the steady-state error decreases.

The steady-state error is inversely proportional to the gain. Therefore, it can be reduced by increasing the system gain.

TANGEDCO AE EE Full Test - 2 - Question 16
If it is possible that every state X(t0) can be completely identified by measurements of the outputs Y(t) over a finite time interval, then the system is said to be
Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 16

Controllability:

  • The concept of controllability of a system which is related to the transfer of any initial state of the system to any other desired state, in a finite length of time by application of proper inputs.
  • A system is said to be completely state controllable if it is possible to transfer the system state from any initial state x(t0) to any other desired state X(tf) in a specified time interval (tf) by a control vector u(t).
  • A system is said to be completely output controllable if it is possible to construct an unconstrained input vector u(t) which will transfer any given initial output y(t0) to any final output Y(tf) in a finite time interval t0 ≤ t ≤ tf.

Observability:

  • The observability is related to the problem of determining the system state by measuring the output for finite length of time.
  • A system is said to be completely observable if every state X(t0) can be completely identified by measurements of the outputs Y(t) over a finite time interval.
  • If the system is not completely observable means that few of its state variables are not practically measurable and are shielded from the observation.
TANGEDCO AE EE Full Test - 2 - Question 17

An electromagnetic wave is transmitted into a conducting medium of conductivity σ. The depth of penetration is

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 17

Derivation:
The depth of penetration δ of a plane electromagnetic wave incident normally on a good conductor is mathematically defined as:

α is the attenuation constant given by:

For a conducting medium σ >>1. The above expression for the attenuation constant can, therefore, be approximated as:

∴ The attenuation constant becomes:

Thus, the skin depth becomes:

Observation:

  • The skin depth is inversely proportional to the square root of frequency.
  • It is inversely proportional to the square root of the conductivity of the medium.
  • It is inversely proportional to the square root of the permeability of the medium.
TANGEDCO AE EE Full Test - 2 - Question 18

The rotor power output of 3-phase induction motor is 15 kW. The rotor copper losses at a slip of 4% will be -

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 18

Concept:
Power flow in the Induction motor is as shown below.


Rotor input or air gap power
Rotor copper losses
Gross mechanical power output 
The relation between the rotor air gap power, rotor copper losses and gross mechanical power output is,

Calculation:
Given that, slip (s) = 4% = 0.04
Rotor power output (Pg) = 15 kW
Rotor copper losses,

TANGEDCO AE EE Full Test - 2 - Question 19

To limit the line charging current, long distance EHV lines are connected with ________ at both ends.

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 19

Concept:

  • Technical improvements in the construction of EHV cables have made possible the installation of very long EHV AC cable lines.
  • With a sufficient degree of voltage control, such lines can retain a large part of their theoretical power transmission capability.
  • EHV cable lines over 25-30 km long require the compensation of the reactive power generated by the cable, by means of fixed or variable shunt devices.
  • Transmitting an active power up to 90% of the cable thermal limit could be operated by using only two shunt reactors at the cable terminals.
  • Fixed shunt reactors are adequate if the mixed line is practically symmetrical, i.e. the two stretches of the overhead line are of the same length.
  • On the other hand, if one of the overhead line sections is much longer than the other, or the terminal voltages of the mixed line are rather different, variable compensation allows to fully exploit the carrying capacity of the cable.
  • Shunt reactors contain efficiently temporary overvoltages due to load rejection and no-load energization.
TANGEDCO AE EE Full Test - 2 - Question 20
When is the peak inverse voltage highest in AC to DC converter?
Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 20

Peak inverse voltage :

Peak Inverse Voltage (PIV) is the maximum voltage that the diode can withstand during reverse bias condition. If a voltage is applied more than the PIV, the diode will be destroyed.

PIV's of different converters are given below

Here Vmax= peak value of supply voltage Vmph= peak voltage of phase voltage of a 3 phase supply

From the above table, we can observe that the single-phase midpoint full wave converter is having the highest PIV

TANGEDCO AE EE Full Test - 2 - Question 21

The minimum number of NAND gates required to implement is:

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 21

F = A + AB̅ + AB̅C
F = A + AB̅(1 + C)
F = A(1 + B̅)
F = A
Hence, the Zero Nand gate is required to implement this function.
Important Point

TANGEDCO AE EE Full Test - 2 - Question 22

What will be the order of the odd length even symmetric FIR impulse response if it is

known that there is a zero at and second zero at z = 1.

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 22

For the FIR filter to be symmetric root must occur in a reciprocal pair.


i.e., if z1 is the root of H(z) then 1/ z1 is also be a root.
One root H(z1) = 1

2nd root
(as complex roots occur in pair 

Then,


→ Now FIR filter will be symmetric
Thus the order of the filter is given by

The highest power of z in H(z) is 6
Thus the order of the filter is 6Important PointsMost of the FIR filters are linear-phase filters, i.e. when a linear-phase filter is desired, an FIR is usually used.
Since the length of the impulse response of a digital filter can be either even or odd, there are in total of four types of linear phase FIR.

  • Type 1: The impulse response has odd length ( N is odd) and is even symmetric about the midpoint, i.e.

h(n) = h(N -n -1)
The amplitude spectrum has even symmetry about ω = 0 and ω = π.

  • Type 2: The impulse response has even length and is even symmetric about its midpoint M., In this case, M is not an integer.

The amplitude spectrum has even symmetry about ω = 0 and odd about ω = π.

  • Type 3: The impulse response has odd length and odd symmetry about the midpoint. The amplitude spectrum is odd about ω = 0 and ω = π. It has a period of 2π.
  • Type 4: The impulse response has even length and odd symmetry about the midpoint. The amplitude spectrum is odd about ω = 0 and even about ω = π. It has a period of 4π.
TANGEDCO AE EE Full Test - 2 - Question 23

Determine the characteristic equation for the system matrix of a discrete system given below.

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 23

Characteristic equation of a matrix A is given by, |zI – A| = 0

Given matrix,

⇒ z (z + 5) + 3 = 0

⇒ z2 + 5z + 3 = 0
TANGEDCO AE EE Full Test - 2 - Question 24

The characteristic equation of a feedback control system is

2s4 + s3 + 2s2 + 5s + 10 = 0

The number of roots in the right half of the s-plane is
Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 24

Concept:

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

According to the Routh tabulation method,

The system is said to be stable if there are no sign changes in the first column of Routh array

The number of poles lie on the right half of s plane = number of sign changes

Calculation:

Characteristic equation: 2s4 + s3 + 2s2 + 4s + 15 = 0

By applying Routh tabulation method,

Number of roots lie on the right side = Number of sign changes = 2
TANGEDCO AE EE Full Test - 2 - Question 25

In the circuit shown, an ideal switch S is operated at 100 KHz with a duty ratio of 50%. Given that Δ icis 1.6 A peak to peak and Iois 5 A dc, the peak current in S is

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 25

Explanation
Input voltage, Vin = 24V
Operating frequency, f = 100 kHz
Duty ratio, D = 50%
Inductor peak-to-peak current ripple, ΔIL = 1.6 A
Average inductor current, ILavg = 5 A
The peak inductor current ILpeak is calculated as:
ILpeak = ILavg + (ΔIL / 2) = 5 A + (1.6 A / 2) = 5 A + 0.8 A = 5.8 A
Since the peak current in the switch S is the same as the peak inductor current during the on-state, the peak current in S is:
ISpeak = 5.8 A

TANGEDCO AE EE Full Test - 2 - Question 26
Rank of the matrix is
Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 26

Concept:

Rearrange the rows with leading zeroes on the left and convert into echelon form

Rank of the matrix = number of non-zeroes row in the echelon form

Calculation:

Given:

matrix =

Replace R1 with R3, R2 with R4, R3 with R2 and R4 with R1

Applying row transformations

Transformation 1:

Transformation 2:

Transformation 3:

R2 R2 - (R1/3)

Transformation 4:

R3 R3 + (3/2)R2

This is in the Echelon form and

Number of non zero rows = 2

So rank = 2

TANGEDCO AE EE Full Test - 2 - Question 27
If f (z) = u + iv is an analytic function, then
Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 27

Concept:

If a function f satisfies Laplace's equation 2f = 0, then f is said to be a harmonic function.

Calculation:

If f (z) = u + iv is an analytic function, then

Now

Both u and v are satisfying Laplace’s equation (∇2f = 0).

Both u and v are harmonic functions.

TANGEDCO AE EE Full Test - 2 - Question 28

If L-1 [f(s)] = f(t), then L-1 [f(s – a)] is

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 28

Concept:

If L-1{F(s)} = f(t)

then

L-1F(s – a) = eat.f(t) = eat. L-1{F(s)}

Other Related Points

L-1{F(s + a)} = e-at.f(t)

TANGEDCO AE EE Full Test - 2 - Question 29
In a closed system, volume changes from 1 m3 to 2 m3 and heat addition is 2000 kJ. The change in internal energy given the pressure volume relation as p = 1000 V kPa, Where V is volume in m3
Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 29

Concept:

Apply the first law of thermodynamics

dQ = dU + dW

Calculation:

Given:

V1 = 1 m3, V2 = 2 m3, dQ = 2000 kJ, P = 1000 V kPa ⇒ 1000 × 103 V Pa.

Heat is added to the system, so it is positive

We know work done

We know that

dU = dQ - dW

dU = 2000 - 1500

dU = 500 kJ.

TANGEDCO AE EE Full Test - 2 - Question 30

Armature reaction in a synchronous generator at rated voltage and zero power factor leading is

Detailed Solution for TANGEDCO AE EE Full Test - 2 - Question 30

Concept:
Armature Reaction: The effect of Armature (stator) flux on the flux produced by the rotor field poles is called Armature Reaction.
When the current flows through the armature winding of the alternator, a flux is produced by the resulting MMF.


This armature flux reacts with the main pole flux, causing the resultant flux to become either less than or more than the original main field flux.
Explanation:

  • Armature reaction of the alternator at zero pf lagging will be purely demagnetizing because at zero pf lagging and will be opposite to each other.

Zero pf lagging → purely demagnetizing

  • Armature reaction of the alternator at zero pf leading will be purely magnetizing because at zero pf leading and will be in the same direction to each other.

Zero pf leading → purely magnetizing

  • The generator supplies a leading load (as leading load takes the leading VAR) and in return gives lagging VAR (magnetic energy) to the generator.

In case of purely resistive load, the armature reaction is cross magnetizing only.
So, Armature reactance in a synchronous generator at rated voltage and zero power factor leading is only magnetizing.

Important Points

  • The armature reaction flux is constant in magnitude and rotates at synchronous speed.
  • The armature reaction is cross-magnetizing when the generator supplies a load at unity power factor.
  • When the generator supplies a load, at lagging power, the armature reaction is partly demagnetizing and partly cross-magnetizing.​
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