Question 1 |

Which of the following is true?

A finite signal is always time bounded | |

A time bounded signal always possesses finite energy | |

A bounded signal is always zero outside the interval [-t_0,t_0
] for some t_0
| |

A time bounded signal is always finite |

Question 2 |

x(t) is a real valued function of a real variable with period T. Its trigonometric
Fourier Series expansion contains no terms of frequency \omega = 2\pi(2k)/T; k=1,2... Also, no sine terms are present. Then x(t) satisfies the equation

x(t)=-x(t-T) | |

x(t)=x(T-t)=-x(-t) | |

x(t)=x(T-t)=-x(t-T/2) | |

x(t)=x(t-T)=x(t-T/2) |

Question 2 Explanation:

Since trigonometric fourier series has no sine terms and has only cosine terms therefore this will be an even signal i.e. it will satisfy.

x(t)=x(-t)

or, we can write,

x(t-T)=x(-t+T)

but signal is periodic with period T.

therefore x(t-T)=x(t)

therefore, x(t)=x(T-t)\;\;...(i)

Now, since signal contains only odd harmonics i.e. no terms of frequency

\omega =\frac{2 \pi \times 2k}{T},\;\;\;k=1,2,3,4

i.e. no even harmonics.

This means signal contains half wave symmetry

this implies that,

x(t)=-x\left ( t-\frac{T}{2} \right )\;\;...(ii)

From eq. (i) and (ii),

x(t)=x(T-t)=-x\left ( t-\frac{T}{2} \right )

x(t)=x(-t)

or, we can write,

x(t-T)=x(-t+T)

but signal is periodic with period T.

therefore x(t-T)=x(t)

therefore, x(t)=x(T-t)\;\;...(i)

Now, since signal contains only odd harmonics i.e. no terms of frequency

\omega =\frac{2 \pi \times 2k}{T},\;\;\;k=1,2,3,4

i.e. no even harmonics.

This means signal contains half wave symmetry

this implies that,

x(t)=-x\left ( t-\frac{T}{2} \right )\;\;...(ii)

From eq. (i) and (ii),

x(t)=x(T-t)=-x\left ( t-\frac{T}{2} \right )

Question 3 |

In the figure the current source is 1\angle 0 A, R=1\Omega, the impedances are Z_{C} =-j \Omega and Z_{L}= 2j \Omega. The Thevenin equivalent looking into the circuit across X-Y is

\sqrt{2}\angle 0V , (1+2j)\Omega | |

2\angle 45^{\circ}V , (1-2j)\Omega | |

2\angle 45^{\circ}V , (1+j)\Omega | |

\sqrt{2}\angle 45^{\circ}V , (1+j)\Omega |

Question 3 Explanation:

To calculate Thevenin's impedance, current-souce is open-circuited

Z_{th}=R+Z_L+Z_c

\;\;=1+2j-j

\;\;=1+j \Omega

Open-circuit voltage at terminals X-Y

=I \times Z_{th}

\;\;=1\angle 0 \times (1+j)

\;\;=\sqrt{2}\angle 15^{\circ} volts

Z_{th}=R+Z_L+Z_c

\;\;=1+2j-j

\;\;=1+j \Omega

Open-circuit voltage at terminals X-Y

=I \times Z_{th}

\;\;=1\angle 0 \times (1+j)

\;\;=\sqrt{2}\angle 15^{\circ} volts

Question 4 |

The three limbed non ideal core shown in the figure has three windings with
nominal inductances L each when measured individually with a single phase AC
source. The inductance of the windings as connected will be

very low | |

L/3 | |

3L | |

very high |

Question 5 |

Which of the following statement holds for the divergence of electric and magnetic flux densities ?

Both are zero | |

These are zero for static densities but non zero for time varying densities. | |

It is zero for the electric flux density | |

It is zero for the magnetic flux density |

Question 5 Explanation:

The divergence of magnetic field is always zzero because magnetic flux makes always a closed path.

So, \bigtriangledown \cdot B=0 (Maxweel's equation)

while divergence of electric field,

\bigtriangledown \cdot \vec{E}=\frac{\rho _v}{\in}

So, \bigtriangledown \cdot B=0 (Maxweel's equation)

while divergence of electric field,

\bigtriangledown \cdot \vec{E}=\frac{\rho _v}{\in}

Question 6 |

In transformers, which of the following statements is valid ?

In an open circuit test, copper losses are obtained while in short circuit
test, core losses are obtained | |

In an open circuit test, current is drawn at high power factor | |

In a short circuit test, current is drawn at zero power factor | |

In an open circuit test, current is drawn at low power factor |

Question 6 Explanation:

Circuit model in open-circuit test:

In open-circuit test, the transformer draws only exciting current. The exciting current is only magnetizing in nature and is proportional to the sinusoidal flux and in phase with it. This is represnted by I_m lagging the induced emf by 90^{\circ}. However, the presence of eddy-current, and hysteresis, both demad the flow of active power into the system and as a consequence the exciting current I_0 has another component I_i in phase with E_1. Thus the exciting current lags the induced emf by an angle slightly less than 90^{\circ} making power factor very low.

Question 7 |

For a single phase capacitor start induction motor, which of the following
statements is valid ?

The capacitor is used for power factor improvement | |

The direction of rotation can be changed by reversing the main winding
terminals | |

The direction of rotation cannot be changed | |

The direction of rotation can be changed by interchanging the supply
terminals |

Question 7 Explanation:

Start capacitor is used to provide 90^{\circ} phase difference between I_m and I_a.

If supply terminals are interchanged, I_a and I_m will flow in the opposite direction. So, toque will act in the smae direction. Therefore, the direction of rotation will remainsame.

Question 8 |

In a DC machine, which of the following statements is true ?

Compensating winding is used for neutralizing armature reaction while
interpole winding is used for producing residual flux | |

Compensating winding is used for neutralizing armature reaction while
interpole winding is used for improving commutation | |

Compensating winding is used for improving commutation while interpole
winding is used for neutralizing armature reaction | |

Compensation winding is used for improving commutation while interpole
winding is used for producing residual flux |

Question 8 Explanation:

Compensating winding is placed in slots cut out in pole faces such that the axis of this winding coincides with the brish axis. The compensating winding neutralizes the armature mmf directly under the pole while in the interpolar region, there is incomplete neutralization.

To speed up the commutation process, the reactance voltage must be neutralized by injecting a suitable polarity dynamical (speed) voltage into the commutating coil. In rder that this injection is restricted to commutating coils, narrow iterpolar are provided in the interpolar region.

To speed up the commutation process, the reactance voltage must be neutralized by injecting a suitable polarity dynamical (speed) voltage into the commutating coil. In rder that this injection is restricted to commutating coils, narrow iterpolar are provided in the interpolar region.

Question 9 |

The concept of an electrically short, medium and long line is primarily based on
the

nominal voltage of the line | |

physical length of the line | |

wavelength of the line | |

power transmitted over the line |

Question 9 Explanation:

L= length of line

If L \leq 80 km

The line id short line.

If 80km \lt L \leq 250km

The line is medium line.

If L \gt 250km

The line is long line.

If L \leq 80 km

The line id short line.

If 80km \lt L \leq 250km

The line is medium line.

If L \gt 250km

The line is long line.

Question 10 |

Keeping in view the cost and overall effectiveness, the following circuit breaker is best suited for capacitor bank switching

vacuum | |

air blast | |

SF_{6} | |

oil |

Question 10 Explanation:

The arc time constant is the least (a few \mu s) in vaccum circuit breakers as compared to other breaker types. The rapid building up of dielectric strength after final arc extinction (20kV/\mu s) is unique feature of VCBs. These arc, therefore, ideally suited for capacitor switching.

There are 10 questions to complete.