Instrument transformers formulas

Instrument transformers. 15 formulas for GATE EE, IN Electrical and Electronic Measurements, each with when to use it. Each is explained in the topic notes.

All 15 Instrument transformers formulas

NOMINAL RATIO

K_{n} = \frac{rated primary current (or voltage)}{rated secondary current (or voltage)}

Read off the rating plate and fixed for the transformer. For example, 1000/5 A gives $K_{n} = 200$ and 11000/110 V gives $K_{n} = 100$ . Readings are normally multiplied by $K_{n}$ .

ACTUAL RATIO (TRANSFORMATION RATIO)

R = \frac{I _{p}}{I _{s}} (CT), R = \frac{V _{p}}{V _{s}} (PT)

The ratio of the actual primary magnitude to the actual secondary magnitude at the operating point. It is not constant: it changes with the primary current or voltage, the burden and the burden power factor.

TURNS RATIO

n = \frac{N _{s}}{N _{p}} (CT), n = \frac{N _{p}}{N _{s}} (PT)

Both are defined so that $n \approx K_{n}$ . A bar (single-turn) primary CT has $N_{p} = 1$ , so $n = N_{s}$ . With turns compensation, $n$ is made slightly smaller than $K_{n}$ .

RATIO CORRECTION FACTOR

RCF = \frac{R}{K _{n}}, true primary value = R \times (secondary reading) = RCF \times K_{n} \times (secondary reading)

An RCF above 1 means the nominal-ratio figure is too low. For an ideal transformer RCF = 1. In terms of RCF, ratio error $= (\frac{1}{RCF} - 1) \times 100%$ .

RATIO ERROR (CURRENT ERROR OR VOLTAGE ERROR)

ratio error = \frac{K _{n} - R}{R} \times 100% = \frac{K _{n} I _{s} - I _{p}}{I _{p}} \times 100%

The error in the value obtained with the nominal ratio, as a percentage of the true value. Positive means $K_{n} \times$ reading is high and negative means it is low. For a PT, replace $I$ with $V$ . If the ratio error is $e$ (as a fraction), then $R = K_{n} / (1 + e)$ .

CT SECONDARY EMF AND CORE FLUX

E_{s} = I_{s} Z_{total} = 4.44 f N_{s} Φ_{m} = 4.44 f N_{s} B_{m} A_{c}

$Z_{total}$ is the impedance magnitude of the whole secondary circuit (winding, leads and burden); $Φ_{m}$ is the peak core flux, $B_{m}$ the peak flux density and $A_{c}$ the core cross-section, for sinusoidal flux at frequency $f$ . The flux the CT must carry, and so its exciting current, is set by $I_{s}$ and the burden impedance, which is why the core flux follows the line current.

CT ACTUAL RATIO

R \approx n + \frac{I _{e} cos δ + I _{m} sin δ}{I _{s}} = n + \frac{I _{0} sin ( δ + α )}{I _{s}}

Valid when $I_{0} ≪ n I_{s}$ . Since $0 < δ + α < 18 0^{\circ}$ , $sin (δ + α) > 0$ : the correction term is always positive, so $R > n$ . An uncompensated CT ( $n = K_{n}$ ) therefore has $R > K_{n}$ and a negative ratio error, approximately $- \frac{I _{0} sin ( δ + α )}{K _{n} I _{s}} \times 100%$ , which means it reads low.

CT PHASE ANGLE

θ_{C T} \approx \frac{I _{m} cos δ - I _{e} sin δ}{n I _{s}} = \frac{I _{0} cos ( δ + α )}{n I _{s}} rad

Positive (the reversed secondary current leads $I_{p}$ ) while $δ + α < 9 0^{\circ}$ , zero at $δ + α = 9 0^{\circ}$ and negative beyond. Here $n I_{s}$ is the secondary current referred to the primary.

CT WITH A PURELY RESISTIVE SECONDARY CIRCUIT

δ = 0 : R \approx n + \frac{I _{e}}{I _{s}}, θ_{C T} \approx \frac{I _{m}}{n I _{s}} rad

Use this when the secondary circuit, winding included, is purely resistive. The loss component then sets the ratio error and the magnetising component sets the phase angle.

BURDEN IN OHMS

Z_{b} = \frac{VA _{rated}}{I _{s, rated}^{2}} (CT), Z_{b} = \frac{V _{s, rated}^{2}}{VA _{rated}} (PT)

The burden includes the leads. On a CT the instruments are in series, so their impedances add. On a PT they are in parallel, so their admittances add. Examples: 15 VA on a 5 A CT is 0.6 Ω, 15 VA on a 1 A CT is 15 Ω, and 100 VA on a 110 V PT is 121 Ω.

PT ACTUAL RATIO

R \approx n + \frac{\frac{I _{s}}{n} ( R _{P} cos Δ + X _{P} sin Δ ) + I _{e} r _{p} + I _{m} x _{p}}{V _{s}}

Valid when the winding drops are small compared with $V_{p}$ (the normal case), so the exciting-current components can be taken in phase with and in quadrature to the reversed secondary voltage. At no load ( $I_{s} = 0$ ) this becomes $R_{0} \approx n + (I_{e} r_{p} + I_{m} x_{p}) / V_{s}$ . The load term is proportional to $I_{s}$ , so for a burden of fixed power factor the ratio, and hence the ratio error, changes almost linearly with the burden VA.

PT PHASE ANGLE

θ_{P T} \approx \frac{I _{m} r _{p} - I _{e} x _{p} - \frac{I _{s}}{n} ( X _{P} cos Δ - R _{P} sin Δ )}{n V _{s}} rad

Sign as on this page: positive when the reversed secondary voltage leads $V_{p}$ . At no load it is $(I_{m} r_{p} - I_{e} x_{p}) / (n V_{s})$ . The load term is negative whenever $tan Δ < X_{P} / R_{P}$ , and in that case adding burden swings the reversed secondary voltage towards lagging $V_{p}$ . Setting $θ_{P T} = 0$ gives the burden at which the phase angle vanishes; for a unity-power-factor burden ( $Δ = 0$ ) that is $I_{s} = n (I_{m} r_{p} - I_{e} x_{p}) / X_{P}$ .

ANGLE SEEN BY THE WATTMETER

θ_{m} = φ - θ_{C T} + θ_{P T}

$φ$ is positive for a lagging load. $θ_{C T}$ and $θ_{P T}$ are positive when the reversed secondary phasor leads its primary. A leading CT phase angle makes a lagging load look closer to unity power factor.

TRUE POWER FROM THE WATTMETER READING

P = R_{v} R_{i} W \frac{cos φ}{cos θ _{m}}, correction factor = \frac{cos φ}{cos θ _{m}}

$R_{v}$ and $R_{i}$ are the actual PT and CT ratios at the operating point. The same factors multiply the energy registered by an energy meter. If ratio errors are neglected, replace $R_{v} R_{i}$ with $K_{v} K_{i}$ .

ERROR OF THE NOMINAL-RATIO POWER (SMALL ERRORS)

ε_{P} \approx e_{v} + e_{i} + 100 (θ_{C T} - θ_{P T}) tan φ (%)

$ε_{P} = (K_{v} K_{i} W - P) / P \times 100$ . Here $e_{v}$ and $e_{i}$ are the PT and CT ratio errors in percent, and the phase angles are in radians. Valid while the ratio errors are small and $(θ_{C T} - θ_{P T}) tan φ ≪ 1$ ; for a leading load $φ$ , and so $tan φ$ , is negative. Because of the $tan φ$ factor, phase-angle errors are negligible near unity power factor and large at low power factor.

Where this was asked: 2020 (1), 2024 (2). See the 3 previous-year questions on Instrument transformers. This shows what was asked, not what will be asked.

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Instrument transformers: the notes and 3 solved previous-year questions