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

Single phasing in 3ϕ windings

• Single phasing occurs when one of the three phases supplying voltage to a three-phase motor is lost. This can happen due to a blown fuse, broken wire, worn contact, or mechanical failure.
• When single phasing occurs in a three-phase motor, the windings in the remaining two phases are likely to burn out.

Explanation

In this figure, the phases likely to burn out due to single phasing depend on the configuration and load balance. Single phasing occurs when one phase of a three-phase system is lost or open, and the motor continues to operate on the remaining phases. This causes excessive current to flow in the two remaining phases, leading to potential burnout in the affected coils.

• For the Delta Connection: In a delta configuration, if one phase is lost (e.g., phase R opens), the other two phases (e.g., B and Y) will have to carry more current. This can cause overheating and eventually burning out of these two phases' windings. So, in the delta connection, the phases that are likely to burn out are the two remaining connected phases.
   
• For the Star Connection: In a star connection, if one phase is lost, the motor may still run on the remaining phases, but with an unbalanced load. Typically, the phase directly across from the open phase will experience an increased load, causing potential overheating. The phases likely to burn out in a star configuration are the two remaining phases or the phase directly opposing the open circuit.

• In a 3-phase induction motor, the stator is laminated to reduce eddy current losses.
• Laminating the stator core minimizes the paths for eddy currents, thus reducing the circulating currents induced by the alternating magnetic field in the stator. This helps in lowering the energy lost as heat due to eddy currents.
• Although lamination also slightly reduces hysteresis losses, the primary purpose is to prevent eddy current losses.
   

Concept:

• The stator is a stationary part of the induction motor. A stator winding is placed in the stator of the induction motor and the three-phase supply is given to it.
• It carries a 3-phase winding and is fed from a 3-phase supply
• The stator of the induction motor consists of three parts

1. Stator Frame
2. Stator core
3. Stator winding
    

Stator Frame:

• It is the outer part of the three-phase induction motor.
• Its main function is to support the stator core and the field winding.
• It acts as a covering, and it provides protection and mechanical strength to all the inner parts of the induction motor.
• The frame is either made up of die-cast or fabricated steel.
• The frame of three phase induction motor should be strong and rigid as the air gap length of three phase induction motor is very small.
• Otherwise, the rotor will not remain concentric with the stator, which will give rise to an unbalanced magnetic pull.
   

Stator core:

• The main function of the stator core is to carry the alternating flux. In order to reduce the eddy current loss, the stator core is laminated.
• These laminated types of structures are made up of stamping which is about 0.4 to 0.5 mm thick.
• All the stamping is stamped together to form a stator core, which is then housed in a stator frame.
• The stamping is made up of silicon steel, which helps to reduce the hysteresis loss occurring in the motor.
   

Stator winding

• The slots on the periphery of the stator core of the three-phase induction motor carry three-phase windings
•  The three phases of the winding are connected either in star or delta depending upon which type of starting method we use. We start the squirrel cage motor mostly with star-delta starter and hence the stator of the squirrel cage motor is delta connected.
• It is supplied with 3 phase AC supply

Additional Information

• A commutator is a rotary electrical switch in certain types of electric motors and electrical generators that periodically reverses the current direction between the rotor and the external circuit. It is used to convert AC to DC
• The air gap of a motor is the gap between the stator teeth or core and the rotor magnets. This gap is a key component in the motor design and affects the overall strength of the magnetic circuit and motor efficiency.

• The magnetic frame (core) of a large transformer is typically built from cold rolled grain-oriented (CRGO) silicon steel.
• This material is specifically used because it has low hysteresis and eddy current losses, which improves the transformer's efficiency.
• The grain orientation in CRGO steel aligns the magnetic domains, reducing core losses significantly in the direction of the magnetic flux, making it ideal for transformer cores.

Construction of transformer

• The transformer consists of a central core on which primary and secondary winding is placed.
• The core of a shell-type transformer is made up of a ferromagnetic material such as soft iron.
• In most types of transformer construction, the central iron core is constructed from a highly permeable material commonly made from thin silicon steel laminations. 
• These thin laminations are assembled together to provide the required magnetic path with the minimum magnetic losses.
• The resistivity of the steel sheet itself is high, thus reducing any eddy current loss by making the laminations very thin.

Concept

The turn ratio for a 3ϕ transformer is given by:

\(Turn\space ratio={V_{ph1}\over V_{ph2}}\)

Here, V_ph1 and V_ph2 represent the phase voltages on the primary and secondary sides of the transformer.

Calculation

Given, Turn ratio = 44

V_L1  = 11 kV

\(V_{ph1}={11\over \sqrt{3}}\space kV\)

\(44={11\over \sqrt{3}V_{ph2}}kV\)

\(V_{ph2}={1\over 4\sqrt{3}}kV\)

\(V_{L2}=\sqrt{3}V_{ph2}\)

\(V_{L2}=\sqrt{3}\times {1\over 4\sqrt{3}}kV\)

\(V_{L2}=250\space V\)

Concept

The power on the primary and secondary sides of the transformer are equal.

\(V_pI_p=V_sI_s\)

Here, V_p and V_s are voltages on the primary and secondary sides respectively.

I_p and I_s are currents on the primary and secondary sides respectively.

Calculation

Given, Vp = 120 V

Vs = 20 V

Is = 4 A

\(I_p={20\times 4\over 120}=0.6667\space A\)

The primary copper loss is given by:

\(P_p=(0.6667)^2\times 5=2.2225\space W\)

The secondary copper loss is given by:

\(P_s=(4)^2\times 0.3=4.8\space W\)

The total copper loss is:

\(P_T=2.2225+4.8\)

P_T = 7.02 W

• 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.

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:

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

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.

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

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 = \(\frac{V_{NL}-V_{FL}}{V_{FL}}\)

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 %

Concept:

Open delta or V-V connection of 3 ϕ transformer:

V-V connection has only two windings instead of three windings in both primary and secondary.

It is used when one winding of transformer goes in maintenance.

Power transfer capacity reduces to 57.7%(or 1/√ 3) times of Δ-Δ connection or closed delta connection.

The capacity of the transformer is measured in VA = √3 V_LI_L

For closed delta, transformer output VA = 3 V_LI_ph

For open delta, transformer output VA =√ 3 VLIph

Calculation:
% increase in capacity = \(\frac{(VA)_{\Delta-\Delta}-(VA)_{V-V}}{(VA)_{V-V}}\times100\)

=\(\frac{3-√3}{√3}\times100\)

= 73.2 %

Concept:

In a transformer,

Turns ratio (n) \(=\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{{{V}_{1}}}{{{V}_{2}}}\) 

\(\therefore {{N}_{1}}\propto {{V}_{1}}~\And ~{{N}_{2}}\propto {{V}_{2}}\)

Where,

N₁ = Number of turns in primary side

N₂ = Number of turns in secondary side

V₁ = Primary side voltage

V₂ = Secondary side voltage

Calculation:

V₁ = 250 V = LV side voltage,

V₂ = 3000 V = HV side voltage,

EMF or voltage per turn = 8 V

Number of LV turns of the transformer is

\({{N}_{1}}={{N}_{LV}}=\frac{250}{8}\approx 32\)

Number of HV turns of the transformer is