Learning Objectives:

• Understand the differences between two types of losses in a transformer.
• Calculate efficiency of a transformer by evaluating iron and copper losees.

Solved Example: 9978-01

Hysteresis and eddy current losses of single-phase transformer working on 200 V, 50 Hz is $P_h$ and $P_c$ respectively. The percentage decrease in $P_h$ and $P_c$, when the transformer operates on 160 V, 40 Hz supply, will respectively be:

A. 32, 36

B. 25, 50

C. 20, 36

D. 40, 80

Correct Answer: C

Solved Example: 9978-02

A single phase transformer when supplied with 220 V, 50 Hz has eddy current loss of 50 W. If the transformer is connected to a voltage of 330 V, 50 Hz the eddy current loss will be:

A. 168.75 W

B. 112.5 W

C. 75 W

D. 50 W

New eddy current loss = \[\left(\dfrac{330}{220}\right)^2 \times 50 = 112.5\ W\]

Correct Answer: B

Solved Example: 9978-03

In a single-phase transformer, the total iron loss is 2500 W at nominal voltage of 440 V and frequency 50 Hz. The total iron loss is 850 W at 220 V and 25 Hz. Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are:

A. 1600 W and 900 W

B. 900 W and 1600 W

C. 600 W and 250 W

D. 250 W and 600 W

Correct Answer: B

Solved Example: 9978-04

When are eddy-current losses in a transformer reduced?

A. If laminations are thick

B. If number of turns in primary winding is reduced

C. If the number of turns in secondary winding is reduced

D. If laminations are thin

Correct Answer: D

Solved Example: 9978-05

Generally the no-load losses of an electrical machine is represented in its equivalent circuit by a:

A. Parallel resistance with a low value

B. Series resistance with a low value

C. Parallel resistance with a high value

D. Series resistance with a high value

Correct Answer: C

Solved Example: 9978-06

While conducting short-circuit test on a transformer the following side is short circuited:

A. High voltage side

B. Low voltage side

C. Primary side

D. Secondary side

Correct Answer: B

Solved Example: 9978-07

If the supply frequency to the transformer is increased, the iron loss will:

A. Not change

B. Decrease

C. Increase

D. Any of the above

Correct Answer: C

Solved Example: 9978-08

The efficiency of a transformer will be maximum when:

A. Copper losses = hysteresis losses

B. Hysteresis losses = eddy current losses

C. Eddy current losses = copper losses

D. Copper losses = iron losses

Correct Answer: D

Learning Objectives:

• Describe principle of working of an ideal transformer.
• Analyze operation of such a transformer.

Transformer is a static device. It transfer electrical energy from one part of the electrical or electronic circuit to other part of circuit without changing the frequency. It works on the Michael Faraday’s law of Electromagnetic Mutual Induction.

Principle:

When the current is provided to the primary winding it behaves as electromagnet due to this the EMF is induced in the secondary winding as it comes in the area having magnetic field lines due to primary Winding. \[V_s = N_s \dfrac{d\phi}{dt}\] where,
\(V_s\) = Secondary Winding Voltage
\(N_s\) = Secondary Winding Turns

BillC at the English-language Wikipedia, CC BY-SA 3.0, via Wikimedia Commons

\begin{align*} \mathrm{Turns\ Ratio} &= \dfrac{\mathrm{No.\ windings\ in\ Secondary}}{\mathrm{No.\ Windings\ in\ Primary}} \\ &= \dfrac{\mathrm{Voltage\ in\ Secondary}}{\mathrm{Voltage\ in\ Primary}} \\ &= \dfrac{V_s}{V_p}\\ &= \dfrac{N_s}{N_p} \end{align*}

Solved Example: 345345

What is the typical efficiency range of a modern transformer?

A. 50-60%

B. 70-80%

C. 90-98%

D. 99-100%

Correct Answer: C

Solved Example: 45645

In a step-up transformer, the voltage on the secondary side is:

A. Lower than the primary side

B. Equal to the primary voltage

C. Higher than the primary side

D. Unpredictable

Correct Answer: C

Solved Example: 546456

Which principle is primarily responsible for the operation of a transformer?

A. Electrolysis

B. Electromagnetic Induction

C. Capacitance

D. Resistance

Correct Answer: B

Solved Example: 9167-01

A single-phase 111-V, 50-Hz supply is connected to a coil with 200 turns of a coil-core assembly. Find the magnitude of maximum flux in the core.

A. 10 mWb

B. 2.5 mWb

C. 1 mWb

D. 25 mWb

\begin{align*} E &= 4.44 f N \phi_m\\ 111 &= 4.44 \times (50) \times (200) \phi_m\\ \phi_m &= 2.5\ mWb \end{align*}

Correct Answer: B

Solved Example: 9167-02

A transformer cannot raise or lower the voltage of a D.C. supply because:

A. There is no need to change the D.C. voltage

B. A D.C. circuit has more losses

C. Faraday's laws of electromagnetic induction are not valid since the rate of change of flux is zero

D. None of the above

Correct Answer: C

Solved Example: 9167-03

The primary coil of a transformer is connected to a 60 V AC source. The secondary coil is connected to a 330 $\Omega$ load. The turns ratio is 3:1. What is the secondary voltage?

A. 2 V

B. 20 V

C. 180 V

D. 18 V

Secondary voltage is independent of the load and depends only upon the turns ratio.

Correct Answer: B

Solved Example: 9167-04

In a transformer the resistance between its primary and secondary should be:

A. Zero

B. 10 $\Omega$

C. 1000 $\Omega$

D. $\infty$

Correct Answer: D

Solved Example: 9167-05

Primary winding of a transformer:

A. Is always a low voltage winding

B. Is always a high voltage winding

C. Could either be a low voltage or high voltage winding

D. None of the above

Correct Answer: C

Solved Example: 9167-06

The transformer ratings are usually expressed in terms of:

A. Volts

B. Amperes

C. kW

D. kVA

Correct Answer: D

Solved Example: 9167-07

The number of turns in primary and the secondary coils in a transformer are 800 and 24000, respectively, If the current in the secondary is 6.0 A, the current in the primary is:

A. 90 A

B. 120 A

C. 150 A

D. 180 A

\begin{align*} \dfrac{N_P}{N_S} &= \dfrac{I_S}{I_P}\\ \dfrac{800}{24000} &= \dfrac{6}{I_P}\\ I_P &= 180\ \mathrm{A} \end{align*}

Correct Answer: D

Solved Example: 9167-08

An ideal transformer has 100 turns in primary and N turns in secondary. If the input is 220 V A.C., and we want 11 V output, what is the value of N ?

A. 1

B. 5

C. 100

D. 500

\begin{align*} \dfrac{V_1}{V_2} &= \dfrac{N_1}{N_2}\\ \dfrac{220}{11} &= \dfrac{100}{N_2}\\ N_2 &= 5 \end{align*}

Correct Answer: B

Three phase system has the following advantages as compare to single phase system

• Power to weight ratio of 3-$\phi$ alternator is high as compared to 1-$\phi$ alternator.
• That means for generation for same amount of Electric Power, the size of 3-$\phi$ alternator is small as compare to 1-$\phi$ alternator.
• Hence, the overall cost of alternator is reduced for generation of same amount of power.
• Moreover, due to reduction in weight, transportation and installation of alternator become convenient and less space is required to accommodate the alternator in power houses.
• For electric power transmission and distribution of same amount of power, the requirement of conductor material is less in 3-$\phi$ system as compare to 1-$\phi$ system.

Solved Example: 9254-02

Polyphase generators produce simultaneous multiple sinusoidal voltages that are separated by:

A. Certain constant phase angles

B. Certain constant frequencies

C. Certain constant voltages

D. Certain constant currents

Correct Answer: A

Solved Example: 9254-03

In a Y-connected source feeding a ∆-connected load,

A. Each phase of the load has one-third of the full line voltage across it

B. Each phase of the load has two-thirds of the full line voltage across it

C. Each phase of the load has the full line voltage across it

D. Each phase of the load has a voltage across it equal to $\sqrt{3}$

Correct Answer: C

Solved Example: 9254-04

A two-phase generator is connected to two 90 $\Omega$ load resistors. Each coil generates 120 V AC. A common neutral line exists. How much current flows through the common neutral line?

A. 1.33 A

B. 1.88 A

C. 2.66 A

D. 1.77 A

Correct Answer: D

Solved Example: 9254-05

In a $\Delta$-connected source feeding a Y-connected load:

A. Each phase voltage equals the difference of the corresponding load voltages

B. Each phase voltage equals the corresponding load voltage

C. Each phase voltage is one-third the corresponding load voltage

D. Each phase voltage is 60$^\circ$ out of phase with the corresponding load voltage

Correct Answer: A

Learning Objectives:

• Determine the reflected impedance and draw an equivalent circuit based on it.

• In a transformer, the load 'seen' from the voltage source is different from the actual load.
• The effective resistance as seen by the primary source will depend upon the turns ratio of the primary and secondary sides of the transformer.
• The effective impedance as seen by the voltage source is called reflected impedance.
• Once the reflected impedance is evaluated, the ciruit can be considered like a simple resistive series circuit.
\[R_p = R \left(\dfrac{N_p}{N_s}\right)^2\]
• The impedance seen at the input is:
$Z_P = a^2Z_S$       Page 366

Solved Example: 9169-01

Which is to be short circuited on performing short circuit test on a transformer?

A. Low Voltage Side

B. High Voltage Side

C. Primary side

D. Secondary side

Correct Answer: A

Solved Example: 9169-02

For a transformer, the load connected to the secondary has an impedance of 8 $\Omega$. Its reflected impedance on primary is observed to be 648 $\Omega$. The turns ratio of this transformer is:

A. 6:1

B. 10:1

C. 9:1

D. 8:1

\[R_p = R \left(\dfrac{N_p}{N_s}\right)^2\] \[648 = 8 N^2\] \[N = \sqrt{\dfrac{648}{8}} = 9\]

Correct Answer: C

Solved Example: 9169-03

When a 200 $\Omega$ load resistor is connected across the secondary winding of a transformer with a turns ratio of 4, the source "sees" a reflective load of:

A. 50 $\Omega$

B. 12.5 $\Omega$

C. 800 $\Omega$

D. 3200 $\Omega$

The impedance seen at the input is \begin{align*} Z_P &= a^2Z_S\\ &= 4^2(200)\\ &= 3200\ \Omega \end{align*}

Correct Answer: D