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Question 171

UPPCL Technician 2021• Domain Knowledge

Which of the following types of materials lacks permanent magnetic dipoles?

  1. AOption (a)

    paramagnetic

  2. BOption (b)

    ferrimagnetic

  3. COption (c)

    diamagnetic

  4. DOption (d)

    ferromagnetic

View answer & explanation

Correct answer

Answer: (c)C.

diamagnetic

Why this answer is correct

Types of magnetic materials

1.) Diamagnetic material

  • The materials are repelled by a magnet such as zinc. mercury, lead, sulfur, copper, silver, bismuth, wood, etc., are known as diamagnetic materials.
  • They are slightly magnetized when placed in a very strong magnetic field and act in the direction opposite to that of the applied magnetic field.
  • Diamagnetic materials lack permanent magnetic dipoles.
  • Their permeability is slightly less than one.

2.) Paramagnetic material

  • The materials which are not strongly attracted to a magnet are known as paramagnetic materials. For example aluminum, tin magnesium, etc.
  • Such materials are magnetized only when placed on a super strong magnetic field and act in the direction of the magnetic field.
  • Their relative permeability is small but positive.

3.) Ferromagnetic material

  • The materials which are strongly attracted by a magnetic field or magnet is known as ferromagnetic material for eg: iron, steel, nickel, cobalt, etc.
  • The permeability of these materials is very very high.

4.) Ferrimagnetic material

  • In these materials, magnetic moments are aligned in unequal numbers in parallel and antiparallel directions, resulting in a net moment.
  • Ferrimagnetism comes in an oxide of iron called magnetite, with the chemical formula Fe3O4.

Question 172

UPPCL Technician 2021• Domain Knowledge

In which of the following systems is a battery NOT required?

  1. AOption (a)

    Solar inverter (standalone)

  2. BOption (b)

    Electric vehicle

  3. COption (c)

    UPS

  4. DOption (d)

    Electronics ballast

View answer & explanation

Correct answer

Answer: (d)D.

Electronics ballast

Why this answer is correct

Correct option 4

For electronic ballast system battery is not required

Concept:

UPS Block Diagram:

A static transfer switch makes an online UPS different from offline UPS. 

The block diagrams of both the UPS are shown in the figure.

Electric vehicle :

Electric vehicles contain AC_DC converter, Battery pack, DC-DC converter, motor, mech reactions, and block diagram is shown below

Solar Inverter:

Solar inverter always has a battery backup it as shown below

Electronic ballast:

An electronic ballast (or electrical ballast) is a device that controls the starting voltage and the operating currents of lighting devices.

Question 173

UPPCL Technician 2021• Domain Knowledge

Open circuit test is conducted on a transformer to determine:

  1. AOption (a)

    copper loss

  2. BOption (b)

    thermal loss

  3. COption (c)

    iron loss

  4. DOption (d)

    both iron and copper loss

View answer & explanation

Correct answer

Answer: (c)C.

iron loss

Why this answer is correct

Testing of Transformer

1.) Open Circuit Test

  • An open circuit test is conducted on a transformer to determine iron loss or constant loss.
  • This test is conducted on the LV side keeping the HV side open circuit.
  • This test is performed at rated voltage and rated frequency.
  • The voltage at rated frequency is applied to that LV side with the help of a variac of variable ratio auto transformer.
  • Now with the help of variac, the applied voltage gets slowly increased until the voltmeter gives a reading equal to the rated voltage of the LV side.

2.) Short Circuit Test

  • A short circuit test is conducted on a transformer to determine copper loss.
  • This test is conducted on the HV side keeping the LV side short circuit.
  • This test is performed at the rated current.
  • A low voltage of around 5-10% is applied to that HV side with the help of a variac.
  • Now with the help of variac applied voltage is slowly increased until the wattmeter, and an ammeter gives a reading equal to the rated current of the HV side.

Question 174

UPPCL Technician 2021• Domain Knowledge

Fuse is a safety device that is used to protect a circuit from-

  1. AOption (a)

    Under voltage

  2. BOption (b)

    Excess current

  3. COption (c)

    Under current

  4. DOption (d)

    Leakage current

View answer & explanation

Correct answer

Answer: (b)B.

Excess current

Why this answer is correct

The correct answer is option 2

Concept:

  • A fuse is an electrical safety device used to protect an electrical or electronic device from Excess-current
  • When the current above the rated value of the fuse flows through the fuse, the heat generated will cause the fuse wire to melt, thereby breaking the connection.
  • An electric fuse is a protective device that protects electrical equipment in the circuit by breaking the circuit when there is a short circuit.
  • It contains a fuse wire which has a low melting point, it allows normal current to pass through it safely but if there is any short circuit fuse wire can’t withstand the heat produced by that current (infinite current) so it melts and breaks the circuit
  • Whereas the fuse rating of an electrical fuse is the maximum amount of current that can pass through it without melting the fuse wire
  • Hence if an electrical safety fuse is rated (marked) as 16 A, it means that the fuse wire will be blown (break) if the current exceeds 16 A to prevent overload or short-circuit

Question 175

UPPCL Technician 2021• Domain Knowledge

In a shell type transformer, the leakage current is reduced by:

  1. AOption (a)

    opening HV and LV windings

  2. BOption (b)

    placing a magnet between HV and LV windings

  3. COption (c)

    sandwiching HV and LV windings

  4. DOption (d)

    providing air cooling between HV and LV windings

View answer & explanation

Correct answer

Answer: (c)C.

sandwiching HV and LV windings

Why this answer is correct

Correct option 3

Concept:

Shell type transformer

  • ​Shell-type transformers provide a shorter magnetic path compared to core type transformer
  • In shell type transformer leakage flux and magnetizing current are lesser as compared to the core type transformer. Therefore power transfer capability is more this is due to sandwiching HV and LV windings
  • It provides better support against electromagnetic force.
Core type transformerShell types transformer
1   It has one yoke and two limb1 . It has two yoke tree limb
2   leakage flux is more power transfer capability is less2 leakage flux is less  and power transfer capability is more
3. more amount of copper is required for winding3 less amount of copper is required for winding

Question 176

UPPCL Technician 2021• Domain Knowledge

Which of the following factors will NOT affect the tariff of electricity consumption?

  1. AOption (a)

    Number of energy meters connected

  2. BOption (b)

    Power factor of the load

  3. COption (c)

    Maximum demand

  4. DOption (d)

    Type of load

View answer & explanation

Correct answer

Answer: (a)A.

Number of energy meters connected

Why this answer is correct

The number of energy meters connected does not affect the tariff of electricity consumption.

Tariff

  • The electricity tariff is defined as the rate at which electrical energy is sold to a consumer.
  • The tariff should include the total cost of producing and supplying electrical energy plus the profit to the power company.

Factors Affecting the Electricity Tariffs:

1.) Type of load

  • The load is mainly classified into three types, i.e., domestic, commercial, or industrial.
  • Industrial consumers use more energy for a longer time than domestic consumers, and hence the tariff for industrial consumers is more than the domestic consumers.

2.) Maximum demand

  • The cost of the electrical energy supplied by a generating station depends on the installed capacity of the plant and the kWh generated.
  • Increased maximum capacity increased the installed capacity of the generating station.

3.) The power factor of the load

  • The power factor plays a major role in plant economics.
  • The low power factor increases the load current which increases the losses in the system.
  • Thus, the regulation becomes poor. For improving the power factor, the power factor correction equipment is installed at the generating station.
  • Thus, the cost of the generation increases.

4.) The time at which load is required

  • The time at which the maximum load is required is also essential for the electricity tariff.
  • If the maximum demand coincides with the maximum demand of the consumer, then an additional plant is required.
  • And if the maximum demand of the consumers occurs during off-peak hours, the load factor is improved, and no extra plant capacity is needed.
  • Thus, the overall cost per kWh generated is reduced.

Question 177

UPPCL Technician 2021• Domain Knowledge

"The net emf around a closed-circuit loop is equal to the sum of the potential drops around the loop" - This statement is related to which of them?

  1. AOption (a)

    Kirchhoff's Law of current 

  2. BOption (b)

    Kirchhoff's law of voltage 

  3. COption (c)

    Kirchhoff's Faraday's Law

  4. DOption (d)

    Lange's law

View answer & explanation

Correct answer

Answer: (b)B.

Kirchhoff's law of voltage 

Why this answer is correct

The correct answer is option 2

Concept:

Kirchhoff’s Voltage Law (KVL): "The net emf around a closed-circuit loop is equal to the sum of the potential drops around the loop"

Mathematically, KVL implies that

\(\mathop \sum \limits_{m = 1}^M {v_m} = 0 \)

Where M is the number of voltages in a loop or number of branches in a loop

And, vm is mth voltage.

Consider a circuit shown below in which R1 and R2 are two resistance, v1, v2 are two voltage source which causes current (I) flow in the loop.

The sign of voltage drop across the passive element is such a way that the current entering from the positive terminal.

Applying KVL on this circuit

V1 + V4 – IR1 – IR2 = 0

Or, V1 + V4 = IR1 + IR2

Hence, the Sum of Voltage drops = Sum of Voltage rises

Note: KVL deals with the conservation of energy.

Kirchhoff’s Current Law (KCL): It states that the algebraic sum of currents entering a node or closed boundary is zero.

Mathematically, KCL implies that

\(\mathop \sum \limits_{n = 1}^N {i_n} = 0 \)

Where N is the number of branches connected to the node

And in is the nth current entering or leaving the node.

By this law current entering a node may be regarded as positive and current leaving a node may be regarded as negative or vice-versa.

Considered the node in the figure shown below, in which i1, i3, and i5 are incoming current and i2, i4, and i6 are outgoing current.

Applying KCL gives,

i1 + i3 + i5 = i2 + i4 + i6

Hence, the sum of currents entering a node is equal to the sum of the currents leaving the node.

Note: KCL is based on the conservation of charge.

Question 178

UPPCL Technician 2021• Domain Knowledge

Fluorescent lamp should operate on _______.

  1. AOption (a)

    Grid supply frequency for better performance

  2. BOption (b)

    Less than grid supply frequency for better performance

  3. COption (c)

    Low frequency signals for better performance

  4. DOption (d)

    High frequency signals for better performance

View answer & explanation

Correct answer

Answer: (d)D.

High frequency signals for better performance

Why this answer is correct

Fluorescent lamp

  • A fluorescent lamp is a low-pressure mercury-vapor gas-discharge lamp that uses fluorescence to produce visible light.
  • An electric current in the gas excites mercury vapor, which produces short-wave ultraviolet light that then causes a phosphor coating on the inside of the lamp to glow.
  • Traditionally, fluorescent lamps have been operated at the same 50Hz or 60Hz frequency as the periodicity of the mains electricity supply. Operating at high-frequency mode results in low power loss at the cathodes which means more power is available in the arc.
  • A fluorescent lamp converts electrical energy into useful light much more efficiently than an incandescent lamp.
  • The typical luminous efficacy of fluorescent lighting systems is 50–100 lumens per watt.

Question 179

UPPCL Technician 2021• Domain Knowledge

What is the correct relation between the number of parallel paths (A) and the number of poles (P) in a lap-wound DC machine?

  1. AOption (a)

    A = P

  2. BOption (b)

    A > P

  3. COption (c)

    A + P = 0

  4. DOption (d)

    A < P

View answer & explanation

Correct answer

Answer: (a)A.

A = P

Why this answer is correct

Types of winding in DC machine

1) Lap winding

​In lap winding, the conductors are joined in such a way that their parallel paths and poles are equal in number.

The end of each armature coil is connected to the adjacent segment on the commutator. 

A = P

A = No. of parallel paths

P = No. of poles

2.) Wave winding

In this winding, the conductors are connected to two parallel paths irrespective of the number of poles of the machine.

The finishing end of one armature coil is connected to the starting end of the other armature coil commutator segment at some distance apart.

Question 180

UPPCL Technician 2021• Domain Knowledge

What are major components the of the battery?

  1. AOption (a)

    UPS

  2. BOption (b)

    thermal power plant

  3. COption (c)

    hydroelectric plant

  4. DOption (d)

    transformer

View answer & explanation

Correct answer

Answer: (a)A.

UPS

Why this answer is correct

The battery is a major component of the UPS.

Components of UPS

There are four main components in any online double conversion uninterruptible power supply (UPS) system: Rectifier; UPS Batteries; Inverter; and Static Bypass Switch.  

1.) UPS Rectifier

  • The UPS rectifier carries out several key functions. The first is to convert the input power from AC (Alternating Current) to DC (Direct Current).
  • Its second main role is to recharge the batteries, while the DC power routes to the inverter too.

2.) Batteries

  • The batteries in a UPS system provide emergency power when the main supply fails.
  • Either the rectifier or a separate charger ensures that the batteries are always charged.

3.) Inverter 

  • This component fulfills the second half of the double conversion by switching the DC voltage from the rectifier or battery back to an AC output that powers the critical load.

4.) Static Bypass Switch

  • This component is a safeguard in case there’s a failure within the UPS system.
  • In the event of a UPS fault, the static bypass switch automatically connects the load to the mains supply, bypassing the rectifier, batteries, and inverter.