Transformer MCQ Quiz - Objective Question with Answer for Transformer - Download Free PDF

The correct answer is option 3.

Leakage flux

• The flux created by the load current in the primary winding, which links only with the primary winding, represents leakage flux.
• This flux doesn't pass through the core to link with the secondary winding and instead, flows through the winding insulation and insulating oil. This leakage flux contributes to the primary leakage reactance. 
• This flux does not contribute to energy transfer between windings and is considered a form of energy loss.

Magnetic flux

• ​Magnetic flux is defined as the number of magnetic field lines passing through a given closed surface. It provides the measurement of the total magnetic field that passes through a given surface area.

Mutual flux

• Mutual flux is the magnetic flux that links both the primary and secondary windings of a transformer, contributing to its operation. 

Induced flux

• Induced flux refers to the change in magnetic flux through a loop of wire, which results in the generation of an electromotive force (EMF) or voltage. This process is a fundamental principle in electromagnetic induction.

The correct answer is option 3.

Function of the explosion vent in the distribution transformer:

• When an internal fault occurs within the transformer, such as a short circuit or insulation breakdown, it can generate high-pressure gases.
• These gases can rapidly increase the pressure inside the transformer tank. 
• The explosion vent is designed to release this excess pressure when it reaches a predetermined level, typically around 5 psi. 
• By releasing the pressure, the explosion vent prevents the transformer tank from bursting or being severely damaged. 
• It also helps to prevent potential hazards, such as fire or injury, associated with a transformer explosion. 

Equivalent circuit of a transformer

In the above diagram, R_o represents the core losses. Neglecting the core loss component, the circuit is modified as:

• In an equivalent circuit where core losses are neglected, the magnetising branch consists solely of a magnetising reactance (X_m).
• This reactance accounts for the inductive component required to establish the magnetic field within the core.

Concept

The same polarity condition for the parallel operation of a transformer is necessary because:

• Circulating Current: When transformers have opposite polarities, their secondary EMFs act in opposition, creating a closed loop that allows a significant current to flow even without a load. This circulating current can damage the transformers and the system. 
• Load Sharing: With the same polarity, the EMFs of the transformers add constructively, allowing them to share the load effectively. Opposite polarity would result in a destructive interaction, preventing proper load sharing and potentially overloading one transformer. 
• Short Circuit Prevention: Incorrect polarity can lead to a short circuit in the parallel connection, causing severe damage to the equipment. 

Concept:

The Buchholz relay is a gas-actuated protection device used in oil-immersed transformers. It is designed to detect faults occurring inside the transformer, such as insulation failure or short-circuits. The relay responds to gas accumulation or oil flow caused by internal faults.

Working Principle:

It is installed in the pipe connecting the transformer oil tank and the conservator tank. When a fault occurs, gases are generated due to oil decomposition and are detected by the relay, which then triggers an alarm or trips the transformer.

• Short circuit test performs on the high-voltage (HV) side of the transformer where the low-voltage (LV) side or the secondary is short-circuited.
• A wattmeter is connected to the high voltage side. An ammeter is connected in series with the high voltage side

No load losses of the electrical machine:

When the load is not connected to the electric machine, the machine draws a low value of current to keep the machine active or running

• The current which is taken by the machine is called no-load current
• Current is drawn by the core of the machine 
• The equivalent circuit is represented by a high value of resistance in parallel.
• The power factor of the machine will be very low.

Magnetizing inrush current:

• Magnetizing inrush current in transformer is the current which is drawn by a transformer at the time of energizing it.
• To have minimum inrush current in the transformer the switch-on instant should be at maximum input voltage
• This inrush current is transient in nature and exists for few milliseconds.
• The inrush current may be up to 10 times higher than the normal rated current of the transformer.
• Although the magnitude of inrush current is high it generally does not create any permanent fault in the transformer as it exists for a very small time.
• But still, inrush current in a power transformer is a problem, because it interferes with the operation of circuits as they have been designed to function.
• Some effects of high inrush current include nuisance fuse or breaker interruptions, as well as arcing and failure of primary circuit components, such as switches. Another side effect of high inrush is the injection of noise.

The differences between core type and shell-type transformers are given below.

Confusion PointsCore type transformers have a shorter mean coil length. This is because the low voltage windings are wrapped around the core, closest to it, in a low-high configuration. The low voltage section carries more current and uses more material, so placing it closer to the core which reduces the average winding length and the amount of material needed.

Concept:

The magnitude of net emf of an ideal transformer is given by the formula:

E = 4.44 × f × N × ϕ_m 

Where E = RMS value of applied voltage.

f = frequency of the transformer.

N = number of turns.

ϕm =  the magnitude of maximum flux in the core.

Calculation:

E = 111 V, f = 50 Hz, N = 200

111 = 4.44 × 50 × 200 × ϕm

ϕm = 111 / (4.44 × 50 × 200)

= 2.5 mWb

Concept:

When dc is applied inductor behaves as a short circuit.

Total loss = copper loss + iron loss

Copper loss = I²R

Calculation:

Given dc supply = 15 volt

DC current = 1.5 amp

Resistance \(R = \frac{{15}}{{1.5}} = 10\;{\Omega }\)

Total loss = 60 watt

Current drawn = 2 amp

Copper loss = 2² × 10 = 40 watt

The correct answer is option 4): 41.38%

Concept:

The maximum efficiency occurs at the proportion of load x

\(x=\sqrt\frac{P_i}{P_{cufl}}\)

Where,

x = Fraction of load

S = Apparent power in kVA

Pi = Iron losses

Pcu= Copper losses

Maximum efficiency = \(\eta_{max}=\frac{x\ S\ \cosϕ}{x\ S\ \cosϕ\ + \ \ 2 P_i\ }\)

Calculation

Given

S = 5 kVA

No-load loss (Pi) = 2.5kW

Copper loss(Pcu) =  5 Kw 

cos ϕ = 1

x =\(\sqrt{ Pi \over pcu} \)

= \(\sqrt {2.5\over 5}\)

= 0.707

\(\eta_{max}=\frac{x\ S\ \cosϕ}{x\ S\ \cosϕ\ + \ \ 2 P_i\ }\)

= \(0.707 \times 5 \over (0.707 \times 5 ) + 5\)

= 0.41 38

\(\eta_{max} %\)  = 41.38 %

Voltage regulation is the change in secondary terminal voltage from no load to full load at a specific power factor of load and the change is expressed in percentage.

 V0 = no-load secondary voltage

V = full load secondary voltage

Voltage regulation for the transformer is given by the ratio of change in secondary terminal voltage from no load to full load to no load secondary voltage.

\(Volatge\;regulation = \frac{{{V_0} - {V}}}{{{V_0}}}\;\)  

Additional InformationVoltage regulation can be expressed as a fraction or unit-change of the no-load terminal voltage and can be defined in one of two ways,

1. Voltage regulation-down, (Regdown) and
2. Voltage regulation-up, (Regup)
    

• When the load is connected to the second output terminal, the terminal voltage goes down, or when the load is removed, the secondary terminal voltage goes up.
• Hence, the regulation of the transformer will depend on which voltage value is used as the reference voltage, load or non-load value.
   

Transformer Voltage Regulation as a Percentage Change:

Transformer when connected to load:

%Reg = \(\frac{V_{NL}-V_{FL}}{V_{NL}}\)

Transformer under no load:

%Reg = \(\frac{V_{NL}-V_{FL}}{V_{FL}}\)

Eddy Current:​

• Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday’s law of induction.
• Eddy currents flow in closed loops within conductors, in-plane perpendicular to the magnetic field.
• By Lenz law, the current swirls in such a way as to create a magnetic field opposing the change; for this to occur in a conductor, electrons swirl in a plane perpendicular to the magnetic field.
• Because of the tendency of eddy currents to oppose, eddy currents cause a loss of energy.
• Eddy currents transform more useful forms of energy, such as kinetic energy, into heat, which isn’t generally useful.
• Thus eddy currents are a cause of energy loss in alternating current (AC) inductors, transformers, electric motors and generators, and other AC machinery, requiring special construction such as laminated magnetic cores or ferrite cores to minimize them.
• Eddy currents are also used to heat objects in induction heating furnaces and equipment, and to detect cracks and flaws in metal parts using eddy-current testing instruments.

• The magnitude of the current in a given loop is proportional to the strength of the magnetic field, the area of the loop, and the rate of change of flux, and inversely proportional to the resistivity of the material.

Concept:

Voltage regulation:

Voltage regulation is the change in secondary terminal voltage from no load to full load at a specific power factor of load and the change is expressed in percentage no-load voltage.

Let's consider,

E2 = no-load secondary voltage

V2 = full load secondary voltage

Voltage regulation \(= \frac{{{E_2} - {V_2}}}{{{E_2}}}\)

It can also be expressed as,

Regulation \(= x.\frac{{{I_2}{R_{02}}\cos {ϕ _2} \pm {I_2}{X_{02}}\sin {ϕ _2}}}{{{E_2}}}\)

⇒p.u regulation = x.[(p.u R).cosϕ₂ + (p.u X).sinϕ₂]

⇒ Regulation = x.(%R cos ϕ ± %X sin ϕ)

Where

x = fraction of load

%R = ohmic drop

%X = reactance drop

And, 

'+' sign is used for lagging loads

'-' sign is used for leading loads

Calculation:

Given that

Per unit impedance Z = (R + j0.4) p.u

Case 1:

Regulation at full load = 32% = 0.32 p.u

Power factor = cosϕ = 0.8 lag

⇒ sinϕ = 0.6

p.u regulation = (p.u R).cosϕ2 + (p.u X).sinϕ2

⇒ 0.32 =  (p.u R) . 0.8 + 0.4 × 0.6

⇒ R(p.u) = (0.32 - 0.24) / 0.8 = 0.1 p.u

Case 2:

Powe factor = cosϕ = 0.8 lead

Fraction of load x = 0.5

⇒ V.R (p.u) = 0.5 (0.1 × 0.8 - 0.4 × 0.6) = - 0.08

⇒ % Voltage regulation = V.R(p.u) × 100 = - 8%