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

Concept:

Current transformer:

• A current transformer is a device that is used to measure high alternating current in a conductor.
• In the figure shown below the conductor act as the primary winding of a single turn that passed through the circular laminated iron core.
• The secondary winding consists of a large number of turns of fine wire wrapped around the core.
• Due to transformer action, the secondary current transforms into a lower value.

​

Let, Np is a number of turn in the primary winding

Ns is the number of turns in the secondary winding.

Ip and Is are primary and secondary turns respectively.

Therefore, the secondary current is given by,

$I_s=I_p\times \frac{N_p}{N_s}$

• Therefore, the current transformer changes the current into a lower value that can easily be measured by the measuring instrument.
• Along with a Potential transformer (PT), a current transformer (CT) can measure Power and energy also.
• Hence, CT used with Ammeter, Wattmeter, and Watt-hour meter.
   

​Calculation:

Given, N_p = 1, N_s = 8, I_p = 50 A

$I_s=I_p\times \frac{N_p}{N_s}$

I_s = 50 × (1 / 8)

I_s = 6.25 A

Ratio error in current transformer:

• In the current transformer, the primary current Ip should be exactly equal to the secondary current multiplied by turns ratio, i.e. KTIs.
• But there is a difference between primary current Ip should be exactly equal to the secondary current multiplied by the turns ratio.
• This difference is contributed by the core excitation or magnetizing component of no-load current.
• The error in the current transformer introduced due to this difference is called current error or ratio error.

The actual ratio of transformation varies with operating conditions and the error in secondary voltage is defined as

Percentage ratio error $=\frac{K_n-R}{R}\times 100$

Kn is the nominal ratio

R is the actual ratio

It can be reduced by secondary turns compensation i.e. slightly decreasing the secondary turns.

Phase angle error:

In an ideal voltage transformer, there should not be any phase difference between the primary voltage and the secondary voltage reversed. However, in an actual transformer, there exists a phase difference between Vo­ and V­s reversed.

The phase angle is taken as +ve when secondary voltage reversed leads the primary voltage.

The angle is -ve when the secondary voltage reversed lags the primary voltage.

It can be reduced by keeping the primary and secondary windings are wound as closely as possible.

Current Transformer:

The secondary side of the current transformer is always kept short-circuited in order to avoid core saturation and high voltage induction so that the current transformer can be used to measure high values of currents.

• The current transformer works on the principle of shorted secondary.
• It means that the burden on the system Zb is equal to 0.
• Thus, the current transformer produces a current in its secondary which is proportional to the current in its primary.

Important Points:

• The most important precaution in the use of a CT is that in no case should it be open-circuited (even accidentally).
• As the primary current is independent of the secondary current, all of it acts as a magnetizing current when the secondary is opened.
• This results in deep saturation of the core which cannot be returned to the normal state and so the CT is no longer usable.
• Again, due to large flux in the core, the flux linkage of secondary winding will be large which in turn will produce a large voltage across the secondary terminals of the CT.
• This large voltage across the secondary terminals will be very dangerous and will lead to insulation failure and there is a good chance that the person who is opening the CT secondary while primary is energized will get a fatal shock.

Phase angle error:

• Phase angle (δ) of the current transformer is the angle between the primary current Ip and secondary current Is when reversed.
• The primary and secondary phasor difference is to be 180° in the case of an ideal current transformer.
• But in practice, this angle having a difference of θ is called phase angle error.
• So, for practical current transformer θ + δ = 180°.
• The total primary current is not actually transformed in the current transformer. One part of the primary current is consumed for core excitation and the remaining is actually transformers with a turns ratio of the current transformer.
• Both ratio error and phase angle error are due to the excitation current Ie.
• Especially, the ratio error is due to the iron loss component Iw and phase angle error is due to magnetizing component I_m.

Calculation:

Given that,

Phase angle error θ = 5°

Phase angle between the primary current IP and secondary current IS  is given by 

 δ = 180° - θ

= 180° - 5°

= 175° 

Given that, impedance of secondary winding = 0.01 Ω

Rated current = 5 A

fault current = 20 times of rated current

= 20 × 5 = 100 A

The correct answer is option 1):$\theta =\frac{\mathrm{I}\mathrm{m}\mathrm{C}\mathrm{o}\mathrm{s}\delta -\mathrm{I}\mathrm{c}\mathrm{S}\mathrm{i}\mathrm{n}\delta }{\mathrm{n}\mathrm{I}\mathrm{s}}\mathrm{d}\mathrm{e}\mathrm{g}\mathrm{r}\mathrm{e}\mathrm{e}$

Concept:

• The phase angle between the primary and secondary of the current transformer must be 180 degrees.
• The primary and secondary current must be out of phase.
• The deviation in the phase angle of the primary and secondary current is called the phase angle error.

​​θ=Im Cosδ−Ic SinδnIsdegree" id="MathJax-Element-81-Frame" role="presentation" style="position: relative;" tabindex="0">θ=Im Cosδ−Ic SinδnIsdegreeθ=Im Cosδ−Ic SinδnIsdegree" id="MathJax-Element-30-Frame" role="presentation" style="position: relative;" tabindex="0">θ=Im Cosδ−Ic SinδnIsdegree$\theta =\frac{\mathrm{I}\mathrm{m}\mathrm{C}\mathrm{o}\mathrm{s}\delta -\mathrm{I}\mathrm{c}\mathrm{S}\mathrm{i}\mathrm{n}\delta }{\mathrm{n}\mathrm{I}\mathrm{s}}\mathrm{d}\mathrm{e}\mathrm{g}\mathrm{r}\mathrm{e}\mathrm{e}$ θ=ImCosδ−IcSinδnIsdegree $\rm \theta=\frac{Im \ Cos \delta-Ic \ Sin \delta}{nIs} degree$

m = Magnetizing component of exciting current,

δ = Angle between secondary induced voltage and secondary current

Ic = Core loss component of exciting current

n = turns ratio

Is = Secondary winding current  

Ip= Primary winding current 

The correct answer is option 2):(Current transformer )

Concept:

• A current transformer's primary coil is always connected in series with the main conductor giving rise to it also being referred to as a series transformer. The nominal secondary current is rated at 1A or 5A for ease of measurement
• A current transformer is also called  ‘series transformer’
• A current transformer (CT) is an instrument transformer in which the secondary current is substantially proportional to the primary current and differs in phase from it by ideally zero degrees.
• The primary current of the CT does not depend upon whether the load or burden is connected to the secondary & also not on the impedance value of the burden.

Power transformer

• A power transformer is a static machine used for transforming power from one circuit to another without changing the frequency.
• A power transformer can increase or decrease the voltage and current, hence can be used as either a step-up or step-down transformer.

Current transformer

• A current transformer is a type of transformer that is mainly designed to reduce the high currents to a lower value for measurement and protection purposes.
• Basically, a current transformer is a voltage step-up and current step-down transformer which means when it reduces the current, it increases the voltage in the same ratio.
• A power transformer and a current transformer are always connected in opposite phase connection.

If a power transformer has a star-connected primary and a delta-connected secondary, the CT connections on its primary and secondary sides should be connected in delta and star respectively.

In order to nullify the phase shift of the Power transformer (PT), The Current transformers (CT) are connected in a reverse configuration.

Concept:

Phase angle:

The angle by which the secondary current phasor, when reversed, differs in phase from the primary current, is known as the phase angle of the transformer.

This angle is taken to be +ve if the secondary current reversed leads to the primary current. The angle is taken as -ve if the secondary current reversed lags behind the primary current.

The angle between Is reversed and Ip is θ. Therefore, the phase angle is θ.

From the phasor diagram,

$\tan \theta =\frac{bc}{\partial b}=\frac{bc}{0a+ab}=\frac{I_0\cos \left(\delta +\alpha \right)}{n{I}_s+I_0\sin \left(\delta +\alpha \right)}$

As θ is very small,

$\theta =\frac{I_0\cos \left(\delta +\alpha \right)}{n{I}_s+I_0\sin \left(\delta +\alpha \right)}rad$

Now, I0 is very small as compared to n Is and therefore, we can neglect the term I0 sin (δ + α)

$\theta \approx \frac{I_0\cos \left(\delta +\alpha \right)}{n{I}_s}rad$

$\theta =\frac{180}{\pi }\left(\frac{I_m\cos \delta -I_c\sin \delta }{n{I}_s}\right)$ degree

This phase angle is involved in the measurement of power and energy, the phase angle error is necessary to reduce in these measurements.

Calculation:

N₁ = 1, N₂ = 500, I_s = 5A, R₂ = 1Ω

Magnetizing ampere turns = 200 AT

$I_0=\frac{\mathrm{M}\mathrm{a}\mathrm{g}\mathrm{n}\mathrm{e}\mathrm{t}\mathrm{i}\mathrm{z}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}\mathrm{s}}{\mathrm{P}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}\mathrm{s}}=200A$

Transformation ratio

$\begin{array}{l}n=\frac{N_2}{N_1}=\frac{500}{1}=500\\ \theta =\frac{180}{\pi }\left(\frac{I_o\cos \left(\alpha +\delta \right)}{n{I}_s}\right)\\ =\frac{180}{\pi }\left[\frac{200\times cos{0}^{\circ }}{500\times 5}\right]={4.6}^{\circ }\end{array}$

Instrument transformers are of two types.

1. Potential transformer

2. Current transformer

These are used to extend the range of induction type of instruments.

Important Points The advantages of instrument transformers over ammeter shunts and voltage multiplies are the metering circuit is electrically isolated from the power circuit thereby providing safety in use to both the instruments and the operator and small power losses in comparison to those in ammeter shunt and voltmeter multiplier.