# Ratios of Instrument Transformers MCQ || Instrument Transformer ratio Questions and Answers

1. Transformation ratio of an instrument is defined as  ________

1. Ratio of primary to secondary phasor
2. Ratio of secondary to primary phasor
3. Reciprocal of the primary phasor
4. Reciprocal of the secondary phasor

Answer.1. Ratio of primary to secondary phasor

Explanation:

The transformation ratio of the instrument transformer is defined as the ratio of the magnitude of the actual primary phasor to the magnitude of the secondary phasor.

2. The 1000/5 A 50 Hz ring cored current converter has a bar primary. The secondary resistor (Burden) is a net resistance of 1 Ω and carries a turbulent current of 1 A at a power factor of 0.5. What will be its ratio error?

1. 0.05%
2. – 0.05%
3. 0.5%
4. – 0.5%

Explanation:

The secondary burden is purely resistive.

∴ cos ϕ = 1,

ϕ = cos-1(1) = 0°

so, δ = 0°

The p.f of the exciting current is 0.5.

cos (90 – α ) =  cos-1 (0.5)

90 – α = 60

α = 30

Exciting current I0 = 1 A

Nominal Ratio  K = 1000/5 = 200

Since there is no turn compensation, the turn ratio is equal to the nominal ratio or n = Kn = 200.

When the primary winding carries a rated current of 1000 A.

Rated secondary Is = 5A

n Is = 200× 5 = 1000 A

Actual transformation ratio,

$\begin{array}{l} R = n + \frac{{{I_0}}}{{{I_s}}}sin(\delta + \alpha )\\ \\ = 200{\rm{\;}} + {\rm{\;}}\frac{1}{5}sin(0 + 30) \end{array}$

Ratio error  = (Kn − R)/R

(200 − 200.1)/200.1

= −0.5%

3. For a C.T. the transformation ratio is given by which of the following relation?

1. R = Is/Ip
2. R = Ip/Is
3. R = 1/Is
4. R = 1/Ip

Explanation:

In the current transformer, the transformation ratio of the instrument transformer is defined as the ratio of the magnitude of the actual primary phasor to the magnitude of the secondary phasor.

R = Rated Primary winding current/Rated secondary winding current

R = Ip/Is

where

R is the transformation ratio
Ip is the actual primary winding current
Is is the actual secondary winding current.

4. The ratio of transformation in the case of potential transformers

1. Increases with increases in power factor of secondary burden
2. Remains constant irrespective of the power of secondary burden
3. Decreases with increases in power factor of Secondary burden
4. None of the above

Answer.3. Decreases with increases in power factor of Secondary burden

Explanation:

The burden across the secondary of an instrument transformer is specified as V2/I2

The ratio of transformation in the case of potential transformers decreases with increases in power factor of Secondary burden.

The nominal ratio of an instrument transformer does not remain constant as the load on the secondary charges. It changes because of effect of secondary current and this causes errors in the measurement. The specific loading at rated secondary winding voltage is specified such that the errors do not exceed the limits. Such a permissible load is called the burden of an instrument transformer.

5. For a P.T. the transformation ratio is given by which of the following relation?

1. R = Vs/Vp
2. R = Vp/Vs
3. R = 1/Vs
4. R = 1/Vp

Explanation:

In the Potential transformer, the transformation ratio of the instrument transformer is defined as the ratio of the magnitude of the actual primary phasor to the magnitude of the secondary phasor.

R = Rated Primary winding voltage/Rated secondary winding voltage

R = Vp/Vs

where

R is the transformation ratio
Vp is the actual primary winding voltage
Vs is the actual secondary winding voltage.

6. In a current transformer, the ratio of rated primary current to rated secondary current is called:

1. nominal ratio
2. ratio error
3. phase angle error
4. actual transformation ratio

Explanation:

Nominal Ratio:

In instrument transformers, the nominal ratio is defined as the ratio of rated primary winding quantity to the rated secondary winding quantity.

In current transformers, nominal ratio = rated primary current / rated secondary current

In potential transformers, nominal ratio = rated primary voltage / rated secondary voltage

Note:

• Ratio error = (nominal ratio – actual ratio) / Actual ratio × 100
• Actual transformation ratio = (Actual primary current) / (Actual secondary current)

7. Nominal ratio of an instrument transformer is defined as the __________

1. Reciprocal of the rated primary value
2. Ratio of rated secondary value to primary value
3. Reciprocal of the rated secondary value
4. Ratio of rated primary value to secondary value

Answer.4. Ratio of rated primary value to secondary value

Explanation:

Nominal Ratio:

In instrument transformers, the nominal ratio is defined as the ratio of rated primary winding quantity to the rated secondary winding quantity.

In current transformers, nominal ratio = rated primary current / rated secondary current

In potential transformers, nominal ratio = rated primary voltage / rated secondary voltage

Note:

• Ratio error = (nominal ratio – actual ratio) / Actual ratio × 100
• Actual transformation ratio = (Actual primary current) / (Actual secondary current)

8. A 500 A/5A, 50 Hz current transformer has a bar primary. The secondary burden is a pure resistance of 1 Ω and it draws a current if 5 A If the magnetic core requires 250 AT for magnetization the percentage ratio error is

1. 10.56
2. -10.56
3. 11.80
4. -11.80

Explanation:

Nominal Ratio = 500/5 = 100

Magnetization required = 250 A

Primary turns = 1

Io = 250A

K = I1/I2 = 100

I2 = KI1 = 500 A

${{I_P} = \sqrt {{{500}^2} + {{250}^2}} = 559.069}$

Ratio = Ip/Is = 555.069/5 = 111.8

Ratio error = $= \frac{{100 – 111.8}}{{111.8}} \times 100$

9. For a C.T. the nominal ratio is given by which of the following relation?

1. Kn = Vs(rated)/Vp(rated)
2. Kn = Vp(rated)/Vn(rated)
3. Kn = 1/Vp(rated)
4. Kn = 1/Vs(rated)

Explanation:

Nominal Ratio:

In instrument transformers, the nominal ratio is defined as the ratio of rated primary winding quantity to the rated secondary winding quantity.

In potential transformers, nominal ratio = rated primary voltage / rated secondary voltage

Kn = Vp(rated)/Vn(rated)

where

R is the transformation ratio
Vp(rated) is the rated primary winding voltage
Vs(rated) is the rated secondary winding voltage.

10. Instrument transformers are known to introduce magnitude and phase errors in measurements. These are primarily due to

1. Improper connections on the primary side
2. Measurement errors inherent in the meter connected to the transfer secondary
3. Open and short circuit parameters of the instrument transformers
4. None

Answer.3. Open and short circuit parameters of the instrument transformers

Explanation:

Instrument transformers are known to introduce magnitude and phase errors in measurements. These are primarily due to the Open and short circuit parameters of the instrument transformers.

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.

11. Ratio correction factor of instrument transformer is defined as _________

1. Reciprocal of nominal ratio
2. Ratio of nominal ratio to transformation ratio
3. Ratio of transformation ratio to nominal ratio
4. Reciprocal of transformation ratio

Answer.3. Ratio of transformation ratio to nominal ratio

Explanation:

Ratio correction factor: The ratio correction factor is the transformation (actual) ratio divided by nominal ratio.

R.C.F = Kactual /Knominal

It is a factor by which the turns ratio (nominal ratio) of the current transformers must be multiplied to get the true ratio of the current transformers.

12. The primary winding exacting current of a current transformer with a bar primary, nominal ratio 100/1, operating on an external burden of 1.6 Ω non inductive, the secondary winding resistance being 0.2 Ω, is 1.9 A, lagging 40.6° to the secondary voltage reversed there being 100 secondary turns. Voltage reversed there being 100 secondary turns with 1 A flowing in the secondary winding. Calculate actual ratio

1. 72.37
2. 101.44
3. 127.34
4. 93.42

Explanation:

Primary winding turns Np­ = 1

Secondary winding turns Ns = 100

Turns ratio, n = Ns/Np = 100

Total secondary circuit resistance = 1.6 + 0.2 = 1.8 Ω

Secondary circuit resistance = 0

The secondary P.f. is unity and its phase angle

δ = 0

α = angle between Io and ϕ

= 90° – (angle between Io and Es reversed)

= 90° – 40.6° = 49.4°

cosα = 0.65, sinα = 0.759

Actual ratio

$R = n + \frac{{{I_o}}}{{{I_s}}}\sin \left( {\delta + \alpha } \right)$

$= 100 + \frac{{1.9}}{{1.0}}\sin \left( {{0^ \circ } + {{49.4}^ \circ }} \right)$

=100+1.91.0sin⁡(0∘+49.4∘)

= 101.44

13. For a current transformer the turns ratio is defined as the _________

1. n = Np ⁄ Ns
2. n = 1 ⁄ Np
3. n = Ns
4. n = Ns ⁄ Np

Answer.4. N = Ns ⁄ Np

Explanation:

The current transformer turn ratio is defined by the ratio of the number of turns on the primary winding to the number of turns in the secondary winding.

N = Ns ⁄ Np

where

n is the turns ratio
Ns is the secondary turns
Np is the primary turn.

14. The magnetizing and loss component of exciting current of a current transformer rated 1000/5 A, are 15 A and 9 A respectively. The phase angle between secondary winding induced voltage and current is 40 degree. The phase angle error of the transformer is

1. 4.56
2. 3.27
3. 2.38
4. 7.24

Explanation:

Im = 15 A

Ic = 9 A

δ = 40°

Is = 5 A

Phase angle error,

$\theta = \frac{{180}}{\pi }\left[ {\frac{{{I_m}\cos \delta – {I_c}\sin \delta }}{{n{I_s}}}} \right]$

n = 100/5 = 20

$\theta = \frac{{180}}{\pi }\left[ {\frac{{15\cos 40 – 9\sin 40}}{{20 \times 5}}} \right] = 3.27$

15.  For a potential transformer the turns ratio is defined as the _________

1. n = Np ⁄ Ns
2. n = 1 ⁄ Np
3. n = Ns
4. n = Ns ⁄ Np

Answer.1. N = Np ⁄ Ns

Explanation:

The potential transformer turn ratio is defined by the ratio of the number of turns on the primary winding to the number of turns in the secondary winding.

N = Ns ⁄ Np

where

n is the turns ratio
Ns is the secondary turns
Np is the primary turn.

16. A 100/5 A, 50 Hz current transformer has a bar primary and rated secondary burden of 10 VA the secondary winding has 186 turn and leakage inductance of 0.96 mH. With purely resistive burden at rated full load, the magnetization MMF is 14 A and loss excitation 9 A. Find the phase angle error.

1. 0.14°
2. 0.10°
3. 0.05°
4. 0.28°

Explanation:

Secondary burden = 10 VA

Secondary current Is = 5A

Secondary circuit impedance = 10/(5)2 = 0.4Ω

Secondary circuit reactance = 2π 50 × 0.96 × 10-3

= 0.30144

Phase angle of secondary circuit

δ = sin−1 0.30144/0.4 = 48.9°

sinδ = 0.7536

cosδ = 0.65

Turn ratio = 186

Magnetizing current Im = 14

Loss component Ie = 9

Phase angle

$\theta = \frac{{180}}{\pi }\left[ {\frac{{{I_m}\cos \delta – {I_e}\sin \delta }}{{n{I_s}}}} \right]$

$= \frac{{180}}{\pi }\left[ {\frac{{14 \times 0.65 – 9 \times 0.7536}}{{186 \times 5}}} \right]$

=180π[14×0.65−9×0.7536186×5]

θ = 0.1428°

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