GATE 2020 IN Q.52: solution and answer

GATE 2020 · INQ.52 · 2 marks · NAT

A 1000/1 A, 5 VA, UPF bar-primary measuring current transformer has 1000 secondary turns. The current transformer exhibits a ratio error of $- 0.1 %$ and a phase error of 3.438 minutes when the primary current is 1000 A. At this operating condition, the rms value of the magnetization current of the current transformer (in amperes, rounded off to two decimal places) is ____

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Answer and solution

ANSWER

The rms magnetizing current is $I_{m} \approx 1.00 A$ .

Official answer: 0.95 to 1.05Source: Official answer key, GATE 2020 IN (IIT Delhi)

Bar primary means $N_{p} = 1$ , so the turns ratio is $n = N_{s} / N_{p} = 1000$ . The nominal ratio is $K_{n} = 1000/1 = 1000$ .
Ratio error $= \frac{K _{n} - R}{R} = - 0.1 % = - 0.001$ . This gives the actual ratio $R = \frac{1000}{0.999} = 1001.001$ . The secondary current is therefore $I_{s} = \frac{I _{p}}{R} = \frac{1000 \times 0.999}{1000} = 0.999 A$ . Referred to the primary, this is $n I_{s} = 999 A$ .
The burden is at UPF, and the secondary winding impedance is taken as resistive because no leakage reactance is given. So the secondary circuit angle is $δ = 0$ , and $I_{s}$ is in phase with $E_{s}$ . In the phasor diagram the flux is at $9 0^{\circ}$ to the induced emf. The reflected secondary current $n I_{s}$ and the core-loss component $I_{e}$ of the exciting current therefore lie along the same line. The magnetizing component $I_{m}$ , which is in phase with the flux, is at $9 0^{\circ}$ to them.
Take $n I_{s}$ as the reference direction. The primary current is the phasor sum $I_{p} = (n I_{s} + I_{e}) + j I_{m}$ . The phase error $θ$ is the angle between $I_{p}$ and $n I_{s}$ , so $I_{p} sin θ = I_{m}$ and $I_{p} cos θ = n I_{s} + I_{e}$ .
Convert the phase error to radians: $θ = 3.438 min = \frac{3.438}{60} \times \frac{π}{180} = 1.000 \times 1 0^{- 3} rad$ .
The quadrature component gives $I_{m} = I_{p} sin θ = 1000 \times sin (1.000 \times 1 0^{- 3}) = 1.00 A$ . The standard approximation with $δ = 0$ , $θ \approx \frac{I _{m}}{n I _{s}}$ , gives the same value to two decimals: $I_{m} = 1 0^{- 3} \times 999 = 0.999 \approx 1.00 A$ .
Consistency check using the in-phase component: $I_{e} = I_{p} cos θ - n I_{s} = 999.9995 - 999 = 0.9995 A$ . The approximate formula $R \approx n + \frac{I _{e}}{I _{s}}$ agrees: $I_{e} = (1001.001 - 1000) (0.999) = 1.00 A$ . This is the loss component, not the magnetizing component, so it does not change the answer.
Therefore the rms magnetizing current is $I_{m} \approx 1.00 A$ .

KEY IDEA

A UPF burden with a resistive secondary gives $δ = 0$ . The reflected secondary current $n I_{s}$ then lies along the core-loss component $I_{e}$ , and only the magnetizing component $I_{m}$ sits in quadrature. So the phase error fixes $I_{m}$ directly: $I_{m} = I_{p} sin θ \approx n I_{s} θ$ . The ratio error fixes $I_{e}$ . Remember that $3.438$ minutes is exactly $1 mrad$ .

WATCH OUT FOR

The usual slips are computing $I_{e}$ from the ratio error and reporting it as the magnetizing current, and plugging $3.438$ minutes (or degrees) straight into the formula without converting to radians. With $δ = 0$ , the ratio error comes from $I_{e}$ and the phase error comes from $I_{m}$ . The two values are nearly equal here (about $1 A$ each), which hides the mix-up but does not justify it.

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GATE 2020 IN Q.52: solution and answer

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