Question 1 |

The transfer function of a real system, H(s), is given as:

H(s)=\frac{As+B}{s^2+Cs+D}

where A, B, C and D are positive constants. This system cannot operate as

H(s)=\frac{As+B}{s^2+Cs+D}

where A, B, C and D are positive constants. This system cannot operate as

low pass filter. | |

high pass filter | |

band pass filter. | |

an integrator. |

Question 1 Explanation:

Put s=0, H(0)=\frac{A \times 0+B}{0+C \times 0+D}=\frac{B}{D}

So, the system pass low frequency component. Put s=\infty , H(\infty )=0

For high pass filter, high frequency component should be non zero. Hence this system cannot be operated as high pass filter.

So, the system pass low frequency component. Put s=\infty , H(\infty )=0

For high pass filter, high frequency component should be non zero. Hence this system cannot be operated as high pass filter.

Question 2 |

For an ideal MOSFET biased in saturation, the magnitude of the small signal
current gain for a common drain amplifier is

0 | |

1 | |

100 | |

infinite |

Question 2 Explanation:

For ideal MOSFET, i_G=0

Therefore, Current gain, A_I=\frac{i_s}{i_G}=\infty

Therefore, Current gain, A_I=\frac{i_s}{i_G}=\infty

Question 3 |

The most commonly used relay, for the protection of an alternator against loss of
excitation, is

offset Mho relay. | |

over current relay. | |

differential relay | |

Buchholz relay. |

Question 4 |

The geometric mean radius of a conductor, having four equal strands with each strand
of radius 'r', as shown in the figure below, is

4r | |

1.414r | |

2r | |

1.723r |

Question 4 Explanation:

Redraw the configuration:

\therefore \; GMR=(r' \times 2r\times 2r\times 2\sqrt{2}r)^{1/4}

Where, r'=0.7788r

Hence, GMR=1.723r

\therefore \; GMR=(r' \times 2r\times 2r\times 2\sqrt{2}r)^{1/4}

Where, r'=0.7788r

Hence, GMR=1.723r

Question 5 |

The valid positive, negative and zero sequence impedances (in p.u.), respectively, for
a 220 kV, fully transposed three-phase transmission line, from the given choices are

1.1, 0.15 and 0.08 | |

0.15, 0.15 and 0.35 | |

0.2, 0.2 and 0.2 | |

0.1, 0.3 and 0.1 |

Question 5 Explanation:

We have,

X_0 \gt X_1=X_2

(for 3-\phi transposed transmission line)

X_0 \gt X_1=X_2

(for 3-\phi transposed transmission line)

Question 6 |

The steady state output (V_{out}), of the circuit shown below, will

saturate to +V_{DD} | |

saturate to -V_{EE} | |

become equal to 0.1 V | |

become equal to -0.1 V |

Question 6 Explanation:

Redraw the circuit:

From circuit,

\begin{aligned} V_{out} &=-\frac{1}{C_1}\int I\cdot dt \\ &= -\frac{1}{R_1C_1}\int 0 \cdot 1dt \\ \\ &=-\frac{0.1}{R_1C_1}\int dt \\ \\ &= -\frac{0.1}{R_1C_1}t \end{aligned}

Hence, V_{out}=-V_{EE}

From circuit,

\begin{aligned} V_{out} &=-\frac{1}{C_1}\int I\cdot dt \\ &= -\frac{1}{R_1C_1}\int 0 \cdot 1dt \\ \\ &=-\frac{0.1}{R_1C_1}\int dt \\ \\ &= -\frac{0.1}{R_1C_1}t \end{aligned}

Hence, V_{out}=-V_{EE}

Question 7 |

The Bode magnitude plot of a first order stable system is constant with frequency.
The asymptotic value of the high frequency phase, for the system, is -180^{\circ}. This
system has

one LHP pole and one RHP zero at the same frequency | |

one LHP pole and one LHP zero at the same frequency | |

two LHP poles and one RHP zero | |

two RHP poles and one LHP zero. |

Question 7 Explanation:

The given system is non-minimum phase system
Therefore, transfer function, T.F=\frac{s-1}{s+1}

Hence, one LHP pole and one RHP zero at the same frequency.

Hence, one LHP pole and one RHP zero at the same frequency.

Question 8 |

A balanced Wheatstone bridge ABCD has the following arm resistances:

R_{AB}=1k\Omega \pm 2.1%; R_{BC}=100\Omega \pm 0.5%, R_{CD} is an unknown resistance; R_{DA}=300\Omega \pm 0.4%; . The value of R_{CD} and its accuracy is

R_{AB}=1k\Omega \pm 2.1%; R_{BC}=100\Omega \pm 0.5%, R_{CD} is an unknown resistance; R_{DA}=300\Omega \pm 0.4%; . The value of R_{CD} and its accuracy is

30\Omega \pm 3\Omega | |

30\Omega \pm 0.9\Omega | |

3000\Omega \pm 90\Omega | |

3000\Omega \pm 3\Omega |

Question 8 Explanation:

The condition for balanced bridge

\begin{aligned} R_{AB}R_{CD}&=R_{DA}R_{BC} \\ R_{CD} &=\frac{300 \times 100}{1000}=30\Omega \\ %Error &=\pm (2.1+0.5+0.4)=\pm 3% \\ \therefore \; R_{CD}&=30\pm 30 \times \frac{3}{100}=30\pm 0.9\Omega \end{aligned}

\begin{aligned} R_{AB}R_{CD}&=R_{DA}R_{BC} \\ R_{CD} &=\frac{300 \times 100}{1000}=30\Omega \\ %Error &=\pm (2.1+0.5+0.4)=\pm 3% \\ \therefore \; R_{CD}&=30\pm 30 \times \frac{3}{100}=30\pm 0.9\Omega \end{aligned}

Question 9 |

The open loop transfer function of a unity gain negative feedback system is given by G(s)=\frac{k}{s^2+4s-5}.

The range of k for which the system is stable, is

The range of k for which the system is stable, is

k \gt 3 | |

k \lt 3 | |

k \gt 5 | |

k \lt 5 |

Question 9 Explanation:

Characteristic equation:

\begin{aligned} 1+G(s)H(s)&=0\\ 1+\frac{k}{s^2+4s-5}&=0\\ s^2+4s+k-5&=0 \end{aligned}

R-H criteria:

\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k-5\\ 4 & 0\\ k-5 & \end{matrix}

Hence, for stable system,

k-5 \gt 0 \;\; \Rightarrow \; k \gt 5

\begin{aligned} 1+G(s)H(s)&=0\\ 1+\frac{k}{s^2+4s-5}&=0\\ s^2+4s+k-5&=0 \end{aligned}

R-H criteria:

\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k-5\\ 4 & 0\\ k-5 & \end{matrix}

Hence, for stable system,

k-5 \gt 0 \;\; \Rightarrow \; k \gt 5

Question 10 |

Consider a 3 x 3 matrix A whose (i,j)-th element, a_{i,j}=(i-j)^3. Then the matrix A will be

symmetric. | |

skew-symmetric. | |

unitary | |

null. |

Question 10 Explanation:

for \; i=j\Rightarrow a_{ij}=(i-i)^3=0\forall i

for \; i\neq j\Rightarrow a_{ij}=(i-j)^3=(-(j-i))^3=-(j-i)^3=-a_{ji}

\therefore \; A_{3 \times 3 } is skew symmetric matrix.

for \; i\neq j\Rightarrow a_{ij}=(i-j)^3=(-(j-i))^3=-(j-i)^3=-a_{ji}

\therefore \; A_{3 \times 3 } is skew symmetric matrix.

There are 10 questions to complete.