Question 1 |

A single-phase 100 kVA, 1000 V / 100 V, 50 Hz transformer has a voltage drop of 5% across its series impedance at full load. Of this, 3% is due to resistance. The percentage regulation of the transformer at full load with 0.8 lagging power factor is

4.8 | |

6.8 | |

8.8 | |

10.8 |

Question 1 Explanation:

Percent voltage regulation

=\left ( \frac{I_2R_{02}}{E_2} \times 100 \right )\cos \phi _2\pm \left ( \frac{I_2X_{02}}{E_2} \times 100 \right ) \sin \phi _2

\% \; VR.=V_r \cos \phi _2\pm V_x \sin \phi _2

(where, '+' lag p.f. and '-' lead p.f.)

At full load:

Given, V_r=3\%

Impedance drop, V_z=5\%

\therefore \; Reluctance drop, V_x=\sqrt{5^2-3^2}=4\%

Voltage regulation at full load at 0.8 p.f. lagging

V.R.=3(0.8)+4(0.6)=4.8\%

=\left ( \frac{I_2R_{02}}{E_2} \times 100 \right )\cos \phi _2\pm \left ( \frac{I_2X_{02}}{E_2} \times 100 \right ) \sin \phi _2

\% \; VR.=V_r \cos \phi _2\pm V_x \sin \phi _2

(where, '+' lag p.f. and '-' lead p.f.)

At full load:

Given, V_r=3\%

Impedance drop, V_z=5\%

\therefore \; Reluctance drop, V_x=\sqrt{5^2-3^2}=4\%

Voltage regulation at full load at 0.8 p.f. lagging

V.R.=3(0.8)+4(0.6)=4.8\%

Question 2 |

In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when the load angle in electrical degrees is

0 | |

45 | |

60 | |

90 |

Question 2 Explanation:

Salient pole synchronous motor power and torque relations per phase:

P=\frac{E_bV}{X_t}\sin \delta +\frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \;Watts

T=\frac{60}{2 \pi N_S}\left [ \frac{E_bV}{X_d} \sin \delta + \frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \right ]\; N-m

The second term is reluctance power or reluctance torque, which is directly proportional to \sin 2\delta.

Therefore, reluctance torque will be maximum.

When, \delta =45^{\circ}

\because \;\;\sin2 (45^{\circ})\Rightarrow \sin 90^{\circ}=1 (Maximum)

P=\frac{E_bV}{X_t}\sin \delta +\frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \;Watts

T=\frac{60}{2 \pi N_S}\left [ \frac{E_bV}{X_d} \sin \delta + \frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \right ]\; N-m

The second term is reluctance power or reluctance torque, which is directly proportional to \sin 2\delta.

Therefore, reluctance torque will be maximum.

When, \delta =45^{\circ}

\because \;\;\sin2 (45^{\circ})\Rightarrow \sin 90^{\circ}=1 (Maximum)

Question 3 |

A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load voltage and current?

v_{0}\geq 0 \text{ and } i_{0} \lt 0 | |

v_{0} \lt 0 \text{ and } i_{0}\lt 0 | |

v_{0}\geq 0 \text{ and } i_{0}\geq 0 | |

v_{0} \lt 0 \text{ and } i_{0}\geq 0 |

Question 4 |

Four power semiconductor devices are shown in the figure along with their relevant
terminals. The device(s) that can carry dc current continuously in the direction shown when
gated appropriately is (are)

Triac only | |

Triac and MOSFET | |

Triac and GTO | |

Thyristor and Triac |

Question 5 |

Two wattmeter method is used for measurement of power in a balanced three-phase load
supplied from a balanced three-phase system. If one of the wattmeters reads half of the
other (both positive), then the power factor of the load is

0.532 | |

0.632 | |

0.707 | |

0.866 |

Question 5 Explanation:

In two wattmeter method,

\tan \phi =\frac{\sqrt{3}(W_1-W_2)}{W_1+W_2}

\begin{aligned} \text{Given}, \; W_2 &=\frac{W_1}{2} \\ \tan \phi &=\frac{\sqrt{3}\left (W_1-\frac{W_1}{2} \right ) }{\left (W_1+\frac{W_1}{2} \right )} \\ \phi &= 30^{\circ}\\ \cos \phi &=\cos 30^{\circ}=0.866 \end{aligned}

\tan \phi =\frac{\sqrt{3}(W_1-W_2)}{W_1+W_2}

\begin{aligned} \text{Given}, \; W_2 &=\frac{W_1}{2} \\ \tan \phi &=\frac{\sqrt{3}\left (W_1-\frac{W_1}{2} \right ) }{\left (W_1+\frac{W_1}{2} \right )} \\ \phi &= 30^{\circ}\\ \cos \phi &=\cos 30^{\circ}=0.866 \end{aligned}

There are 5 questions to complete.