Question 1 |

Roots of the algebraic equation x^{3}+x^{2}+x+1=0 are

(+1, +j, -j) | |

(+1, -1, +1) | |

(0, 0, 0) | |

(-1, +j, -j) |

Question 1 Explanation:

-1 is one of the root since

(-1)^3+(-1)^2+(-1)+1=0

By polynomial division

\frac{x^3+x^2+x+1}{(x-(-1))}=x^2+1 \Rightarrow \;\; x^3+x^2+x+1=(x^2+1)(x+1)

So root are (-1,+j,-j)

(-1)^3+(-1)^2+(-1)+1=0

By polynomial division

\frac{x^3+x^2+x+1}{(x-(-1))}=x^2+1 \Rightarrow \;\; x^3+x^2+x+1=(x^2+1)(x+1)

So root are (-1,+j,-j)

Question 2 |

With K as a constant, the possible solution for the first order differential equation \frac{dy}{dx}=e^{-3x} is

-\frac{1}{3}e^{-3x}+K | |

-\frac{1}{3}e^{3x}+K | |

-\frac{1}{3}e^{-3x}+K | |

-3e^{-x}+K |

Question 2 Explanation:

\begin{aligned} \frac{dy}{dx}&=e^{-3x}\\ \int dy&=\int e^{-3x}dx\\ y&=\frac{e^{-3x}}{-3}+K\\ y&=-\frac{1}{3}e^{-3x}+K \end{aligned}

Question 3 |

The r.m.s value of the current i(t) in the circuit shown below is

\frac{1}{2}A | |

\frac{1}{\sqrt{2}}A | |

1A | |

\sqrt{2}A |

Question 3 Explanation:

V_s=1 \sin t\equiv V_m \sin \omega t

V_m=1 V and \omega =1 rad/sec

Impedance of the branch containing inductor and capacitor

Z=j(X_L-X_C)

\;\;=j\left ( \omega L-\frac{1}{\omega C} \right )

\;\;=j\left ( 1 times 1 -\frac{1}{1 times 1} \right )=0

So, this branch is short circuit and the whole current flow through it

i(t)=\frac{1.0 \sin t}{1}=1.0 \sin t

rms value of the current =\frac{1}{\sqrt{2}}A

V_m=1 V and \omega =1 rad/sec

Impedance of the branch containing inductor and capacitor

Z=j(X_L-X_C)

\;\;=j\left ( \omega L-\frac{1}{\omega C} \right )

\;\;=j\left ( 1 times 1 -\frac{1}{1 times 1} \right )=0

So, this branch is short circuit and the whole current flow through it

i(t)=\frac{1.0 \sin t}{1}=1.0 \sin t

rms value of the current =\frac{1}{\sqrt{2}}A

Question 4 |

The fourier series expansion f(t)=a_{0}+\sum_{n=1}^{\infty }a_{n}cosn\omega t+b_{n}sin n\omega t of the periodic signal shown below will contain the following nonzero terms

a_{0} \; and \; b_{n},n=1,3,5,...\infty | |

a_{0} \; and \; a_{n},n=1,2,3,...\infty | |

a_{0},a_{n} \; and \; b_{n},n=1,3,5,...\infty | |

a_{0} \; and \; a_{n},n=1,3,5,...\infty |

Question 4 Explanation:

Let, x(t)= Even and Hws

Fourier series expansion of x(t) contains cos terms with odd harmonics.

NOw, f(t)=1+x(t)

Fourier series of f(t) contains dc and cos terms with odd harmonics.

Fourier series expansion of x(t) contains cos terms with odd harmonics.

NOw, f(t)=1+x(t)

Fourier series of f(t) contains dc and cos terms with odd harmonics.

Question 5 |

A 4-point starter is used to start and control the speed of a

dc shunt motor with armature resistance control | |

dc shunt motor with field weakening control | |

dc series motor | |

dc compound motor |

Question 5 Explanation:

4-point starter is used to control the speed of shunt motor in field weakening region, i.e. above rates speeds.

In field weakening region field current will reduce in 3-point starter holding coil unable to hold the plunger in ON position.

In field weakening region field current will reduce in 3-point starter holding coil unable to hold the plunger in ON position.

Question 6 |

A three phase, salient pole synchronous motor is connected to an infinite bus. It
is operated at no load a normal excitation. The field excitation of the motor is
first reduced to zero and then increased in reverse direction gradually. Then the
armature current

Increases continuously | |

First increases and then decreases steeply | |

First decreases and then increases steeply | |

Remains constant |

Question 6 Explanation:

As field current reduces, the flux will start reducing, to keep this flux constant, the armature draws high current from bus.

Ar zero field current the motor acts as synchronous reluctance motor in this case the magnetizing current fully taken from bus. Ie we increase field current in reverse direction to keep flux constant, motor draws more current. During this process the load angle increases at one point the reversed field force dominates reluctance torque and rotor slip one pole pitch anf align to opposite pole. The instant align to opposite pole the flux will be very high, to reduce this flux current drops steeply t synchronous motor value. Here torque is reluctance + synchronous motor torque.

Ar zero field current the motor acts as synchronous reluctance motor in this case the magnetizing current fully taken from bus. Ie we increase field current in reverse direction to keep flux constant, motor draws more current. During this process the load angle increases at one point the reversed field force dominates reluctance torque and rotor slip one pole pitch anf align to opposite pole. The instant align to opposite pole the flux will be very high, to reduce this flux current drops steeply t synchronous motor value. Here torque is reluctance + synchronous motor torque.

Question 7 |

A nuclear power station of 500 MW capacity is located at 300 km away from a
load center. Select the most suitable power evacuation transmission configuration among the following options

A | |

B | |

C | |

D |

Question 7 Explanation:

For transmission of bulk power of very long distance high voltage (400 kV) is used.

To increase reliability, double circuit is used.

To increase reliability, double circuit is used.

Question 8 |

The frequency response of a linear system G(j \omega) is provided in the tubular form
below

Gain Margin and phase margin are

Gain Margin and phase margin are

6 dB and 30^{\circ} | |

6 dB and -30^{\circ} | |

-6 dB and 30^{\circ} | |

-6 dB and -30^{\circ} |

Question 8 Explanation:

At gain crossover frequency (\omega_{gc}), magnitude of G(j\omega) is 1.

|G(j\omega_{gc})|=1

Phase of G(j\omega )=\angle G(j\omega_{gc})=-150^{\circ}

Phase margin=180^{\circ}+\angle G(j\omega_{gc})

\;\;\; =180^{\circ}-150^{\circ}=30^{\circ}

At phase cross frequency (\omega_{pc}), phase of G(j\omega ) is -180^{\circ},

\angle G(j\omega_{pc})=-180^{\circ}

M=magnitude of G(j\omega ) at \omega _{pc}

=|G(j\omega_{pc})|=0.5

Gain margin

=20log\frac{1}{M}

=20 log \frac{1}{0.5}=6 dB

|G(j\omega_{gc})|=1

Phase of G(j\omega )=\angle G(j\omega_{gc})=-150^{\circ}

Phase margin=180^{\circ}+\angle G(j\omega_{gc})

\;\;\; =180^{\circ}-150^{\circ}=30^{\circ}

At phase cross frequency (\omega_{pc}), phase of G(j\omega ) is -180^{\circ},

\angle G(j\omega_{pc})=-180^{\circ}

M=magnitude of G(j\omega ) at \omega _{pc}

=|G(j\omega_{pc})|=0.5

Gain margin

=20log\frac{1}{M}

=20 log \frac{1}{0.5}=6 dB

Question 9 |

The steady state error of a unity feedback linear system for a unit step input is
0.1. The steady state error of the same system, for a pulse input r(t) having a
magnitude of 10 and a duration of one second, as shown in the figure is

0 | |

0.1 | |

1 | |

10 |

Question 9 Explanation:

Let the system is represnted as

\frac{Y(s)}{X(s)}=\frac{G(s)}{1+G(s)H(s)}

H(s)=1 (unity \; feedback)

Error=E(s)=\frac{X(s)}{1+G(s)}

Steady state error for X(s)=1/s, \; e_{ss}=0.1

e_{ss}=\lim_{s \to 0}sE(s)

\;\; =\lim_{s \to 0}\frac{sX(s)}{1+G(s)}

\Rightarrow \; 0.1=\lim_{s \to 0}\frac{s\cdot \frac{1}{s}}{1+G(s)}

\Rightarrow \; 0.1=\lim_{s \to 0}\frac{1}{1+G(s)}

When input,

X(t)=10[u(t)-u(t-1)]

X(s)=10\left ( \frac{1}{s}-\frac{e^{-s}}{s} \right )

\;\;=\frac{10}{s}(1-e^{-s})

e_{ss}=\lim_{s \to 0}sE(s)

\;\;=\lim_{s \to 0}\frac{sX(s)}{1+G(s)}

\;\;=\lim_{s \to 0}\frac{s \times \frac{10}{s}[1-e^{-s}]}{1+G(s)}

e_{ss}=\lim_{s \to 0}\frac{10(1-e^{-s})}{1+G(s)}=0

\frac{Y(s)}{X(s)}=\frac{G(s)}{1+G(s)H(s)}

H(s)=1 (unity \; feedback)

Error=E(s)=\frac{X(s)}{1+G(s)}

Steady state error for X(s)=1/s, \; e_{ss}=0.1

e_{ss}=\lim_{s \to 0}sE(s)

\;\; =\lim_{s \to 0}\frac{sX(s)}{1+G(s)}

\Rightarrow \; 0.1=\lim_{s \to 0}\frac{s\cdot \frac{1}{s}}{1+G(s)}

\Rightarrow \; 0.1=\lim_{s \to 0}\frac{1}{1+G(s)}

When input,

X(t)=10[u(t)-u(t-1)]

X(s)=10\left ( \frac{1}{s}-\frac{e^{-s}}{s} \right )

\;\;=\frac{10}{s}(1-e^{-s})

e_{ss}=\lim_{s \to 0}sE(s)

\;\;=\lim_{s \to 0}\frac{sX(s)}{1+G(s)}

\;\;=\lim_{s \to 0}\frac{s \times \frac{10}{s}[1-e^{-s}]}{1+G(s)}

e_{ss}=\lim_{s \to 0}\frac{10(1-e^{-s})}{1+G(s)}=0

Question 10 |

Consider the following statement

(1) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil.

(2) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit.

(1) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil.

(2) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit.

(1) is true but (2) is false | |

(1) is false but (2) is true | |

both (1) and (2) are true | |

both (1) and (2) are false |

Question 10 Explanation:

The current coil carries a current of I +I_P and produces a filed corresponding to this current. The compensating coil is connected in series with the pressure coil circuit and is made as nearly as possible identical and coincident with the current coil. It is so connected that it opposes the field of the current coil. The compensating coil carries a current I_P and produces a field corresponding to this current. This field acts as in opposition to the current coil field. Thus the resultant field is due to current I only. Hence the error caused by the pressure coil current flowing in the current coil is neutralized.

There are 10 questions to complete.