Transformers
Losses in Transformer
Learning Objectives:
- Understand the differences between two types of losses in a transformer.
- Calculate efficiency of a transformer by evaluating iron and copper losees.
No load losses or core losses
No load losses remains the same irrespective of the load connected to the transformer.
It is the power consumed to sustain the magnetic field in the transformer’s core. It is of two types - hysteresis loss and eddy current loss
Hysteresis loss is the energy lost by reversing the magnetizing field in the core as the AC changes direction in every cycle.
Eddy current loss is a result of induced currents circulating in the core.
Hysteresis loss is minimized by using steel of high silicon content for the core Eddy current loss is minimized by using very thin laminations polished with varnish.
Load losses or copper losses
They are associated with load current flow in the transformer windings
Copper loss is power lost in the primary and secondary windings of a transformer due to the Ohmic resistance of the windings.
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
Principle and Construction of Transformer
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
Construction:
Laminated Core: Cores are designed not to have current in it. But still there is a leakages current Or the eddy current present in it. To minimize these current the cores are laminated.
Winding: Made up of copper or aluminum coated with very thin layer of insulation.
Tank: The main function of conservator tank of transformer is to provide adequate space for expansion of oil inside the transformer. It is usually Cylindrical or cubical.
Isolation: It is provided by using Synthetic Oil between Core and Tank.
Breather: When the temperature changes occur in Transformer insulating oil, the oil expands or contracts and there an exchange of air also occurs when transformer is fully loaded. When transformer gets cooled, the oil level goes down and air gets absorbed within.
\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: 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
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.
\begin{align*} \dfrac{V_1}{V_2} &= \dfrac{N_1}{N_2} = \mathrm{Turns\ ratio}\\ \dfrac{V_1}{V_2} &= \dfrac{3}{1}\\ \dfrac{60}{V_2} &= 3 \\ V_2 &= 20\ V \end{align*}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
A 12 V, 60 W lamp is connected to the secondary of a step-down transformer, whose primary is connected to AC mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?
A. 0.27 A
B. 2.7 A
C. 3.7 A
D. 0.37 A
\begin{align*} (VI)_{\mathrm{primary}} &= (VI)_{\mathrm{secondary}}\\ 220I &= 60\\ I &= \dfrac{60}{220}\\ &= 0.27\ A \end{align*}
Correct Answer: A
Single-phase and Three-phase
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
Reflected Impedance
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:
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