An ideal transformer is one which has no losses (no iron loss and no copper loss) and no leakage flux i.e. all the flux produced by the primary winding is linking with the secondary winding.
 
In actual practice, it is impossible to make such a transformerbut to understand the concepts of transformer it is better to start with an ideal transformer and then extend to a practical transformer. In this article I am discussing the ideal transformer on load.
 
What is ideal transformer?
 
Consider an ideal transformer whose secondary is open as shown in the figure. When it is connected to AC supply voltage V1 a current Im flows through its primary winding.
 
Since the resistance of the primary coil is zero (i.e. it is purely inductive) the current Im lags behind the applied voltage V1 by 90o. The losses are zero (as assumed) and no load is applied to the transformer, therefore, the magnitude of current Im will be very small.
 
This current is called magnetizing current Im as it produces flux φ in the core. Because flux is produced by Im hence it is in phase with Im.
 
This alternating flux links with both secondary and primary windings. When this flux links with secondary winding it produces mutually induced EMF E2 in the secondary winding. When this flux links with primary winding it produces self-induced EMF E1 in the primary winding.
 
Both the EMFs (E1 and E2) oppose the cause (applied voltage V1) producing it (as per Lenz’s law). Hence these are shown in opposite direction in the phasor diagram.
 
V1 = primary applied voltage
E1 = self-induced EMF in the primary winding
E1 = V1 (because it is an ideal transformer with no resistance and reactance on primary side)
 
E2 = mutually induce EMF on the secondary side
V2 = secondary terminal voltage
V2 = E2 (because at no load I2 is zero)
 
I1 = Im, magnetizing current on primary side
I2 = 0 (no load)
 

Ideal Transformer on Load

 
When a load is applied to the secondary side of an ideal transformer, a finite value of secondary current (I2) starts flowing. The magnitude and phase of secondary current, I2 w.r.t. secondary terminal voltage, V2 depends upon the nature of the load.
 
If the load is resistive, I2 is in phase with V2.
If the load is inductive (R-L) type, I2 lags V2 by some angle φ2.
If the load is capacitive (R-C) type, I2 leads V2 by some angle φ2.
 
When I2 flows, it sets up its own MMF (N2I2) and hence creates its own flux φ2.
 
However, φ2 opposes the main flux φm setup by magnetizing current (as per Lenz’s law).
 
This momentarily weakens the main flux φm and therefore primary back EMF E1 tends to decrease. Due to the reduction in E1, the difference between V1 and E1 increases and the additional primary current I2’ starts flowing. The I2’ component of primary current is called load component of the primary current.

Voltage regulation, efficiency, phasor diagrams of ideal transformer on load and no load.

The current component I2’, sets up an MMF N1I2’ to counter the effect of secondary produced MMF N2I2
 
i.e N1I2’ = N2I2
or I2’ = N2I2 / N1 = KI2
 
and it is 180o out of phase with I2. The net primary current I1 is phasor sum of I2’ and Im (because Iw is zero in an ideal transformer).
 
Thus due to load on the secondary side, the primary current of the transformer increases to supply the additional power to the load.
 
Since, winding resistance of an ideal transformer is zero, therefore, its voltage regulation will be zero and its efficiency will be 100%.

 

Transformer 2 | Objective Type Question Answers

 

#1 Full load copper loss in a transformer is 1600 watts. At half load the losses will be

400 W

#2 Bucholzes relay is generally not provided on transformer below

500 kVA

#3 Operating time of Buchholz relay is in the order of

0.1 second

#4 For a transformer, operating at a constant load current, maximum efficiency will occur at

unity power factor

#5 “A” class insulation can withstand

maximum temperature rise of 105 degree Celsius

#6 If full load copper loss of a transformer is 800 W, its copper loss at 75% load will be

450 W

#7 The secondary of a current transformer is always short circuited under operating conditions because

it avoids core saturation and high voltage induction

#8 The main advantage of an autotransformer over a two winding transformer is that

only one winding is used as a result there is substantial saving in material

#9 If a break occurs in a 200 : 100 volt autotransformer in the winding which is common to high voltage as well as low voltage side, then the output voltage on the low voltage side will be

200

#10 In an autotransformer, effective saving on copper and copper loss will occur when the transformation ratio is

nearly equal to 1

#11 Which of the following conditions should be fullfilled for the parallel operation of two transformers?

all of the above.

#12 If the transformer core is made of copper

eddy current losses will be more

#13 If the secondary of a 5:1 step-down transformer is connected to the primary of 10:1 step-down transformer, the total step-down ratio of transformation will be

50:1

#14 An engineer uses an ohmmeter to measure the DC resistance of a transformer winding. The engineer notices that, when one of the ohmmeter leads is disconnected from the transformer, a spark results. What accounts for this phenomenon?

Faraday’s law

#15 The efficiency of two identical transformers under load conditions can be determined by

back to back test

#16 Oil impregnated paper condenser bushing generally used on transformers operating at

132 kV

#17 If a transformer core is made of copper and coils are made up of steel wires, then

copper losses in the winding will be more

#18 As a general rule, the kVA rating ratio of transformers operating in parallel should be within

3:1

#19 If the two transformers not having the same percentage impedances are connected in parallel for sharing a load then

load sharing of the transformers will not be proportional to kVA ratings

#20 Harmonic currents in a transformer cause

all of the above.

#21 Which of the following is the most likely source of harmonics in a transformer?

core saturation

#22 A tap changer is used on transformer for

adjustments in secondary voltage

#23 A 30/10 P 15 current transformer will have output of

30 VA

#24 A voltage transformer of which class of accuracy can be used for precision testing in standard laboratories?

0.1 class

#25 The reactance of a transformer depends on

leakage flux

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