Question 1 |

The value of Z in figure, which is most appropriate to cause parallel resonance
at 500 Hz is

125.00 mH | |

304.20 \mu F | |

2.0 \mu F | |

0.05 \mu F |

Question 1 Explanation:

Atresonance, the circuit should be in unity power factor.

Hence, 'Z' should be capacitive.

Admittance of the parallel circuit.

Y=\frac{1}{jL\omega }+\frac{1}{1/jC\omega }=0

\frac{-1}{L\omega }+C\omega =0

\therefore \;\; C=\frac{1}{L\omega ^2}=\frac{1}{2(2 \pi \times 500)^2} =0.05\mu F

Hence, 'Z' should be capacitive.

Admittance of the parallel circuit.

Y=\frac{1}{jL\omega }+\frac{1}{1/jC\omega }=0

\frac{-1}{L\omega }+C\omega =0

\therefore \;\; C=\frac{1}{L\omega ^2}=\frac{1}{2(2 \pi \times 500)^2} =0.05\mu F

Question 2 |

A parallel plate capacitor is shown in figure. It is made two square metal plates
of 400 mm side. The 14 mm space between the plates is filled with two layers of
dielectrics of \varepsilon _r= 4, 6 mm thick and \varepsilon _r= 2, 8 mm thick. Neglecting fringing of
fields at the edge the capacitance is

1298 pF | |

944 pF | |

354 pF | |

257pF |

Question 2 Explanation:

When two capacitor formed by two layer of dielectics are connected in series, then equivalent capacitance

\begin{aligned} C_{eq} &= \frac{C_1 \times C_2}{C_1+C_2}\\ &=\frac{\frac{\varepsilon _{r1} \varepsilon _0 A}{d_1} \times \frac{\varepsilon _{r2} \varepsilon _0 A}{d_2}}{\frac{\varepsilon _{r1} \varepsilon _0 A}{d_1} + \frac{\varepsilon _{r2} \varepsilon _0 A}{d_2}} \\ &= \frac{(8.85 \times 10^{-12})(400 \times 10^{-3})^2 \times 4 \times 2}{4 \times 8 \times 10^{-3}+6 \times 2 \times 10^{-3}} \\ & =257 \; \text{pF} \end{aligned}

\begin{aligned} C_{eq} &= \frac{C_1 \times C_2}{C_1+C_2}\\ &=\frac{\frac{\varepsilon _{r1} \varepsilon _0 A}{d_1} \times \frac{\varepsilon _{r2} \varepsilon _0 A}{d_2}}{\frac{\varepsilon _{r1} \varepsilon _0 A}{d_1} + \frac{\varepsilon _{r2} \varepsilon _0 A}{d_2}} \\ &= \frac{(8.85 \times 10^{-12})(400 \times 10^{-3})^2 \times 4 \times 2}{4 \times 8 \times 10^{-3}+6 \times 2 \times 10^{-3}} \\ & =257 \; \text{pF} \end{aligned}

Question 3 |

The inductance of a long solenoid of length 1000 mm wound uniformly with
3000 turns on a cylindrical paper tube of 60 mm diameter is

3.2 \mu H | |

3.2 mH | |

32.0 mH | |

3.2 H |

Question 3 Explanation:

Let, B=\mu _0nI

Then, \phi =B.S=\mu _0nSI

Where 'S' is the cross sectional area of solenoid flux linkage,

a'=n\phi =\mu _0n^2SI

Hence, Inductance length =\mu _0n^2S

For,

l=1000mm=1m

\therefore \;\;L=4 \pi \times 10^{-7} \times 3000^2 \times \pi \times (30 \times 10^{-3})^2

\;\;\;=32mH

Then, \phi =B.S=\mu _0nSI

Where 'S' is the cross sectional area of solenoid flux linkage,

a'=n\phi =\mu _0n^2SI

Hence, Inductance length =\mu _0n^2S

For,

l=1000mm=1m

\therefore \;\;L=4 \pi \times 10^{-7} \times 3000^2 \times \pi \times (30 \times 10^{-3})^2

\;\;\;=32mH

Question 4 |

Total instantaneous power supplied by a 3-phase ac supply to a balanced R-L
load is

zero | |

constant | |

pulsating with zero average | |

pulsating with the non-zero average |

Question 4 Explanation:

Impedance of the load

=Z_L=R+j\omega L=|Z|\angle \theta _L

where, \theta _L=\tan ^{-1}\left ( \frac{\omega L}{R} \right )

Voltages of 3-\phi supply

\begin{aligned} V_a&=V_m \sin \omega t \\ V_b&=V_m \sin (\omega t-120^{\circ}) \\ V_c &=V_m \sin (\omega t+120^{\circ}) \\ I_a&=\frac{V_a}{Z_L}=\frac{V_m \sin \omega t}{|Z|\angle \theta _L} \\ &= I_M \sin (\omega t-\theta _L)\\ \text{where, } I_m&=\frac{V_m}{|Z|}\\ \text{Similarly,}&\\ I_b&=I_m \sin (\omega t-120-\theta _L)\\ I_c&=I_m \sin (\omega t+120-\theta _L)\\ & \text{Instantaneous power}\\ &=P=V_aI_a+V_bI_b+V_cI_c\\ P&=V_mI_m[\sin \omega t\cdot \sin (\omega t-\theta _L)\\ &+\sin (\omega t-120^{\circ}) \cdot \sin (\omega t-120-\theta _L)\\ &+\sin (\omega t+120^{\circ})\cdot \sin (\omega t+120-\theta _L)]\\ &=\frac{V_mI_m}{2}[(\cos \theta _L -\cos(2\omega t-\theta _L) ) \\ &+(\cos \theta _L -\cos(2\omega t-240-\theta _L) )\\ &+(\cos \theta _L -\cos(2\omega t+240-\theta _L) )]\\ P&=\frac{3V_mI_m}{2}\cos \phi = \text{constant} \end{aligned}

=Z_L=R+j\omega L=|Z|\angle \theta _L

where, \theta _L=\tan ^{-1}\left ( \frac{\omega L}{R} \right )

Voltages of 3-\phi supply

\begin{aligned} V_a&=V_m \sin \omega t \\ V_b&=V_m \sin (\omega t-120^{\circ}) \\ V_c &=V_m \sin (\omega t+120^{\circ}) \\ I_a&=\frac{V_a}{Z_L}=\frac{V_m \sin \omega t}{|Z|\angle \theta _L} \\ &= I_M \sin (\omega t-\theta _L)\\ \text{where, } I_m&=\frac{V_m}{|Z|}\\ \text{Similarly,}&\\ I_b&=I_m \sin (\omega t-120-\theta _L)\\ I_c&=I_m \sin (\omega t+120-\theta _L)\\ & \text{Instantaneous power}\\ &=P=V_aI_a+V_bI_b+V_cI_c\\ P&=V_mI_m[\sin \omega t\cdot \sin (\omega t-\theta _L)\\ &+\sin (\omega t-120^{\circ}) \cdot \sin (\omega t-120-\theta _L)\\ &+\sin (\omega t+120^{\circ})\cdot \sin (\omega t+120-\theta _L)]\\ &=\frac{V_mI_m}{2}[(\cos \theta _L -\cos(2\omega t-\theta _L) ) \\ &+(\cos \theta _L -\cos(2\omega t-240-\theta _L) )\\ &+(\cos \theta _L -\cos(2\omega t+240-\theta _L) )]\\ P&=\frac{3V_mI_m}{2}\cos \phi = \text{constant} \end{aligned}

Question 5 |

A 500 kVA, 3-phase transformer has iron losses of 300 W and full load copper
losses of 600 W. The percentage load at which the transformer is expected to
have maximum efficiency is

50.00% | |

70.70% | |

141.40% | |

200.00% |

Question 5 Explanation:

Copper loass at any load

\begin{aligned} P_{cu} &=x^2 \times [\text{Full load copper loss}] \\ &= x^2 \times P_{fl,cu}\\ \text{where,}\;\; x&= \text{fraction of laod} \end{aligned}

Maximum occurs when

Copper loss-Iron loss

x^2 \times P_{fl,cu}=P_i

x^2 \times 600=300 x=0.707 =70.7\%

\begin{aligned} P_{cu} &=x^2 \times [\text{Full load copper loss}] \\ &= x^2 \times P_{fl,cu}\\ \text{where,}\;\; x&= \text{fraction of laod} \end{aligned}

Maximum occurs when

Copper loss-Iron loss

x^2 \times P_{fl,cu}=P_i

x^2 \times 600=300 x=0.707 =70.7\%

Question 6 |

For a given stepper motor, the following torque has the highest numerical value

Detent torque | |

Pull-in torque | |

Pull-out torque | |

Holding torque |

Question 6 Explanation:

Detent torque: It is amount of torque that the motor produces when it is not energized. No current is slowing through the winding.

Holding Torque: It is amount of torque that the motor produces when it has rated current flowing through the winding but motor is at rest..

Pull-in Torque: Shows the maximum value of torque at given speeds that the motor can start, stop, or reverse in synchronism with the input pulses.

Pull-out Torque: Shows the maximum value of torque at given speeds that the motor can generate while running in synchronism. If the motor is run outside of this curve, it will stall.

Holding Torque: It is amount of torque that the motor produces when it has rated current flowing through the winding but motor is at rest..

Pull-in Torque: Shows the maximum value of torque at given speeds that the motor can start, stop, or reverse in synchronism with the input pulses.

Pull-out Torque: Shows the maximum value of torque at given speeds that the motor can generate while running in synchronism. If the motor is run outside of this curve, it will stall.

Question 7 |

The following motor definitely has a permanent magnet rotor

DC commutator motor | |

Brushless dc motor | |

Stepper motor | |

Reluctance motor |

Question 7 Explanation:

Brushless dc motor definitely has a permanent magnet.

Question 8 |

The type of single-phase induction motor having the highest power factor at full
load is

shaded pole type | |

split-phase type | |

capacitor-start type | |

capacitor-run type |

Question 8 Explanation:

Capacitor start type motor has high staring torque and therefore is used for hard starting loads such as compressor, conveyors, pumps et.

Capacitor-run type motor has better running power factor and efficiency and a quiter and smoother operation. It is used for both easy and hard to start loads. Moder applications of such motors are ceiling fans, blower etc.

Capacitor-run type motor has better running power factor and efficiency and a quiter and smoother operation. It is used for both easy and hard to start loads. Moder applications of such motors are ceiling fans, blower etc.

Question 9 |

The direction of rotation of a 3-phase induction motor is clockwise when it is
supplied with 3-phase sinusoidal voltage having phase sequence A-B-C. For
counter clockwise rotation of the motor, the phase sequence of the power supply
should be

B-C-A | |

C-A-B | |

A-C-B | |

B-C-A or C-A-B |

Question 9 Explanation:

To reverse the direction of rotation, phase sequence of the supply has to be reversed. For clockwise direction, the phase sequence was A-B-C. For counter-clockwise direction, the phase sequence has to be A-C-B.

Question 10 |

For a linear electromagnetic circuit, the following statement is true

Field energy is equal to the co-energy | |

Field energy is greater than the co-energy | |

Field energy is lesser than the co-energy | |

Co-energy is zero |

Question 10 Explanation:

Where, \lambda =N\phi =Flux linkage

Field energy is the energy absorbed by the magnetic system to establish flux \phi.

For a linear electromagnetic circuit

Field energy = Co-energy =\frac{1}{2}\lambda i

There are 10 questions to complete.