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.

There are 5 questions to complete.