# 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

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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%

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Correct Answer - Option 2 : - 0.05%

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 iron loss component of no-load current or core excitation current.
• The error in the current transformer introduced due to this difference is called the 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.

Calculation:

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,

$R = n+\frac{I_0}{I_s}sin(δ+ \alpha)=200\ +\ \frac{1}{5}sin(0+30)$

R = 200.01

Ratio error = $\frac{{K_n}-R}{R}\times100$

= $\frac{200-200.1}{200.1}\times100$

= - 0.05%