Question 1 |

The common emitter forward current gain of the transistor shown is \beta _{F}=100.

The transistor is operating in

The transistor is operating in

Saturation region | |

Cutoff region | |

Reverse active region | |

Forward active region |

Question 1 Explanation:

We assume BJT is in active region, applying KVL in base emitter circuit

10-0.7=1k\Omega \times I_c+270 \times I_b

=I_b(270+100k)

\therefore \; \; I_b=\frac{9.3}{370}mA

\therefore \;\; I_c=\frac{93}{37}mA

I_{c(sat)}=\frac{10-0.2}{2k}=4.9 mA

I_{c(sat)} \gt I_{c(active)}

\therefore \; BJT is in active region.

10-0.7=1k\Omega \times I_c+270 \times I_b

=I_b(270+100k)

\therefore \; \; I_b=\frac{9.3}{370}mA

\therefore \;\; I_c=\frac{93}{37}mA

I_{c(sat)}=\frac{10-0.2}{2k}=4.9 mA

I_{c(sat)} \gt I_{c(active)}

\therefore \; BJT is in active region.

Question 2 |

The three-terminal linear voltage regulator is connected to a 10 \Omega load resistor as shown in the figure. If V_{in} is 10 V, what is the power dissipated in the transistor ?

0.6W | |

2.4W | |

4.2W | |

5.4W |

Question 2 Explanation:

V_0=V_Z-V_{BE}

\;\; =6.6V-0.7V=5.9V

I_L=\frac{V_0}{R_L}=\frac{5.9}{10}=0.59A=I_C

V_{CE}-V_i-V_0

\;\; =10-5.9=4.1V

P_Q=V_{CE} \times I_C

\;\;=4.1 \times 0.59A=2.4 W

Question 3 |

Consider the transformer connections in a part of a power system shown in the figure. The nature of transformer connections and phase shifts are indicated for all but one transformer.

Which of the following connections, and the corresponding phase shift \theta, should be used for the transformer between A and B ?

Which of the following connections, and the corresponding phase shift \theta, should be used for the transformer between A and B ?

Star-Star (\theta=0^{\circ}) | |

Star-Delta (\theta=-30^{\circ} ) | |

Delta-Star (\theta=30^{\circ}) | |

Star-Zigzag (\theta=30^{\circ}) |

Question 3 Explanation:

Taking V_1 as the reference

V_1=220\angle 0^{\circ}kV

\angle \theta _1=0^{\circ}

Phase difference V_2 and V_1 is 0^{\circ}.

So, V_2=400\angle 0^{\circ}kV

V_2 leads V_3 by 30^{\circ}.

So, V_3=15\angle -30^{\circ}kV

\angle \theta _3=-30^{\circ}

V_3 lags V_4 by 30^{\circ}.

\angle \theta _3=\angle \theta _4-30^{\circ}

\angle \theta _4=\angle \theta _3+30^{\circ}

\angle \theta _4=-30^{\circ}+30^{\circ}=0^{\circ}

Phase difference between V_1 and V_2 is \theta

\theta =\angle \theta _1-\angle \theta _4=0-0=0^{\circ}

Question 4 |

The incremental cost curves in Rs/MWhr for two generators supplying a common load of 700 MW are shown in the figures. The maximum and minimum generation limits are also indicated. The optimum generation schedule is :

Generator A : 400 MW, Generator B : 300 MW | |

Generator A : 350 MW, Generator B : 350 MW | |

Generator A : 450 MW, Generator B : 250 MW | |

Generator A : 425 MW, Generator B : 275 MW |

Question 4 Explanation:

Maximum incremental cost in Rs/Mwhr for generator A =600 (at 450 MW)

Minimum incremental cost in Rs/Mwhr for generator B =650 (at 150 MW)

As maximum value of incremental cost of A is less than minimum value of B.

Therefore, generator 'A' will operate at its maximum (o/p) 450 MW and B at (700-450)=250 MW.

Minimum incremental cost in Rs/Mwhr for generator B =650 (at 150 MW)

As maximum value of incremental cost of A is less than minimum value of B.

Therefore, generator 'A' will operate at its maximum (o/p) 450 MW and B at (700-450)=250 MW.

Question 5 |

Two regional systems, each having several synchronous generators and loads are inter connected by an ac line and a HVDC link as shown in the figure. Which of the following statements is true in the steady state :

Both regions need not have the same frequency | |

The total power flow between the regions (P_{ac}+P_{dc}) can be changed by controlled the HDVC converters alone | |

The power sharing between the ac line and the HVDC link can be changed
by controlling the HDVC converters alone. | |

The directions of power flow in the HVDC link (P_{dc}) cannot be reversed |

Question 5 Explanation:

Both region are connected by HVDC link as well as AC line. So AC link is possible when both regions have frequency.

By changing fringe angle (\alpha) of converter, we can change the power sharing, between the AC line and HVDC link.

By changing fringe angle (\alpha) of converter, we can change the power sharing, between the AC line and HVDC link.

Question 6 |

Considered a bundled conductor of an overhead line consisting of three identical
sub-conductors placed at the corners of an equilateral triangle as shown in the
figure. If we neglect the charges on the other phase conductor and ground, and
assume that spacing between sub-conductors is much larger than their radius, the
maximum electric field intensity is experienced at

Point X | |

Point Y | |

Point Z | |

Point W |

Question 6 Explanation:

Electric field intensity at various points are shown as follow.

It is clear from the above diagrams that minimumcancellation of vector occurs at the point Y. Hence maximum electric field intensity.

It is clear from the above diagrams that minimumcancellation of vector occurs at the point Y. Hence maximum electric field intensity.

Question 7 |

The circuit shown in the figure is

a voltage source with voltage \frac{rV}{R_{1}\parallel R_{2}} | |

a voltage source with voltage \frac{r \parallel R_{2}}{R_{1}} V | |

a current source with current \frac{r \parallel R_{2}}{R_{1}+R_{2}} \frac{V}{r} | |

a current source with current \frac{R_{2}}{R_{1}+R_{2}} \frac{V}{r} |

Question 7 Explanation:

It behaves as current source because the output current I_o depends upon (V_{in}) and resistance only.

Where, I_0=\frac{\left ( \frac{V \times R_2}{R_1+R_2} \right )}{r}=\frac{R_2}{R_1+R_2}\cdot \frac{V}{r}

Where, I_0=\frac{\left ( \frac{V \times R_2}{R_1+R_2} \right )}{r}=\frac{R_2}{R_1+R_2}\cdot \frac{V}{r}

Question 8 |

The system shown in the figure is

Stable | |

Unstable | |

Conditionally stable | |

Stable for input u_1, but unstable for input u_2 |

Question 8 Explanation:

(T/F)_1=\frac{\frac{s-1}{s+2}}{1+\frac{s-1}{s+2} \times \frac{1}{s-1}}

\;\;=\frac{s-1}{s+3}

Pole is in LHS of s-plane, hence stable.

(T/F)_2=\frac{\frac{1}{s-1}}{1+\frac{1}{s-1} \times \frac{s-1}{s+2}}

\;\;=\frac{s+2}{(s-1)(s+3)}

Hence, unstable as it ha pole at right hand side of s-plane.

Question 9 |

Let a signal a_1 \sin(\omega _{1}t + \phi_1) be applied to a stable linear time invariant system. Let the corresponding steady state output be represented as a_2 F(\omega _{2}t + \phi _{2}). Then
which of the following statement is true?

F is not necessarily a "Sine" or "Cosine" function but must be periodic
with \omega _{1} = \omega _{2} | |

F must be a "Sine" or "Cosine" function with a_1=a_2 | |

F must be a "Sine" function with \omega _1=\omega _2 and \phi _{1} = \phi _{2} | |

F must be a "Sine" or "Cosine" function with \omega _{1} = \omega _{2} |

Question 9 Explanation:

The effect of linear time invariant systems is best understtod by considering that, at any given input frequency \omega _1, the output is sinusoidal at same frequency,as the input applied, with amplitude given by output = |H| x input and phase equal to \phi _2=\phi _H+\phi _1, where |H| is the magnitude to the frequency response and \phi _H is its phase angle. Both |H| and \phi _H are functions of frequency.

Question 10 |

The frequency spectrum of a signal is shown in the figure. If this is ideally sampled
at intervals of 1 ms, then the frequency spectrum of the sampled signal will be

A | |

B | |

C | |

D |

Question 10 Explanation:

Given that, sampling interval =1 msec

i.e. T_s=1 \; msec=10^{-3}\; sec

Therefore sampling frequency

f_s=\frac{1}{T_s}=\frac{1}{10^{-3}}=1 kHz

after sampling new signal in frequency domain

U_T(f)=\frac{1}{T_s}\sum_{n=-\infty }^{\infty }U(f-nf_s)

Therefore, spectrum of sampled signal will be

There are 10 questions to complete.