Instrument transformers: CT and PT

Why it matters

You cannot connect a 5 A ammeter to a 2000 A busbar or a 110 V voltmeter to an 11 kV line. Current transformers (CTs) and potential (voltage) transformers (PTs) scale these down and isolate the instruments and relays from the high-voltage circuit. Their ratio and phase-angle errors decide whether a tariff meter bills correctly and whether a relay trips when it should.

Key ideas

Purpose. Instrument transformers extend AC meter ranges, isolate low-voltage instruments from the power circuit, and allow standard instruments (5 A or 1 A, 110 V) to be used everywhere. Measurements multiply the instrument reading by the transformer ratio.

Current transformer.

• The primary is in series with the line: a few turns, or a single bar (bar-primary CT). The secondary has many turns and feeds ammeters, wattmeter current coils and relays — a very low-impedance burden, so the CT works almost short-circuited.
• The primary current is set by the external circuit, not by the CT burden. The core flux is small because the secondary ampere-turns nearly cancel the primary ampere-turns.
• Never open-circuit a CT secondary while the primary carries current. With no secondary ampere-turns to oppose it, the whole primary MMF magnetises the core: flux rises to saturation, a dangerously high peak voltage appears across the open secondary, the core overheats, and it may be left with residual magnetism that changes its ratio. Short the secondary with the shorting link before removing an instrument.

Potential transformer.

• The primary is connected in parallel (across the line), the secondary feeds voltmeters, wattmeter pressure coils and relays — a high-impedance burden, so it works like a lightly loaded power transformer.
• A PT secondary must never be short-circuited (fuses are fitted). One end of the secondary is earthed for safety.

Errors. Because some primary current is needed to magnetise the core and supply core loss, and because of winding resistance and leakage reactance, the actual ratio is not exactly the turns ratio and the secondary quantity is not exactly 180° (reversed) from the primary.

• Ratio error: difference between the nominal ratio Kn (marked on the nameplate, e.g. 1000/5) and the actual ratio R = Ip/Is (or Vp/Vs).
• Phase-angle error θ: the angle between the primary phasor and the reversed secondary phasor; positive when the reversed secondary leads. It does not matter for ammeters or voltmeters but causes errors in wattmeters, energy meters and directional relays, especially at low power factor.
• CT errors grow with burden (larger secondary voltage needs more flux and more exciting current) and at low primary current. Reduced by low-loss, high-permeability cores (CRGO, nickel-iron), short magnetic paths, and turns compensation (one or two secondary turns fewer than nominal).
• PT errors grow with burden (more voltage drop in the windings). Reduced by low winding resistance and reactance and a well-designed core.
• Burden is the load on the secondary, in VA at rated secondary current or voltage, with its power factor. Accuracy class (e.g. 0.2, 0.5, 1 for metering; 5P, 10P for protection) defines permissible ratio and phase errors — take the limits from the relevant standard.

Metering versus protection CTs. Metering CTs must be accurate near rated current and should saturate during faults to protect the instruments; protection CTs must stay reasonably accurate up to many times rated current (accuracy-limit factor).

Formulas

Kn = rated primary / rated secondary (e.g. 1000/5 = 200) R = Ip / Is (CT) ; R = Vp / Vs (PT) — actual ratio Ratio error (%) = (Kn − R) / R × 100 Ip = R·Is ; Vp = R·Vs ; corrected primary from the reading.

R ≈ n + (Im·sin δ + Iw·cos δ) / Is (CT) θ ≈ (Im·cos δ − Iw·sin δ) / (n·Is) rad (CT; × 180/π for degrees) n = turns ratio Ns/Np; Is = secondary current (A); Im, Iw = magnetising and core-loss components of the exciting current referred to the primary (A); δ = phase angle of the secondary burden (lagging positive).

Primary power = wattmeter reading × CT ratio × PT ratio (neglecting phase-angle errors)

Worked examples

Example 1 — CT errors. A bar-primary CT has 200 secondary turns (nominal ratio 1000/5 = 200). At rated current the secondary burden has a power factor of 0.8 lagging. Referred to the primary, the magnetising component is 2 A and the core-loss component 1 A. Find the actual ratio, ratio error and phase-angle error.

1. n = 200/1 = 200; cos δ = 0.8, sin δ = 0.6.
2. R = 200 + (2 × 0.6 + 1 × 0.8)/5 = 200 + 2.0/5 = 200.4.
3. Ratio error = (200 − 200.4)/200.4 × 100 = −0.20 %.
4. θ = (2 × 0.8 − 1 × 0.6)/(200 × 5) = 1.0/1000 = 0.001 rad = 0.0573° = 3.4 minutes.

Answer: R = 200.4, ratio error −0.20 % (the meter reads low), phase-angle error ≈ 3.4′.

Example 2 — turns compensation (GATE level). The CT of Example 1 has its secondary reduced to 199 turns, with the same excitation components and burden. Find the new ratio error.

1. n = 199; R = 199 + 0.4 = 199.4.
2. Ratio error = (200 − 199.4)/199.4 × 100 = +0.30 %.

Answer: +0.30 %. One turn over-corrects at this burden; in practice the compensation is chosen so that the error stays smallest over the working range of current and burden.

Example 3 — power through instrument transformers. A wattmeter connected through a 100/5 A CT and an 11 000/110 V PT reads 400 W. Neglecting errors, find the primary power.

1. CT ratio = 20; PT ratio = 100.
2. P = 400 × 20 × 100 = 800 000 W.

Answer: 800 kW.

Common mistakes

• Treating the CT ratio 1000/5 as a turns ratio Np:Ns of 1000:5 — the turns ratio is the inverse of the current ratio (Ns/Np ≈ 200).
• Opening the CT secondary to remove an ammeter; always short it first.
• Short-circuiting a PT secondary.
• Defining ratio error with the actual ratio in the numerator (sign comes out wrong).
• Ignoring phase-angle error in power measurement at low power factor — a few minutes of arc matter when cos φ is small.
• Assuming CT error is constant; it rises with burden and at low current.

For GATE IN

Expect NAT questions on actual ratio, ratio error and phase-angle error from excitation components and burden power factor, turns compensation, and primary power from a reading through CT and PT. MCQs test why a CT secondary must not be open-circuited, burden and accuracy class, and metering versus protection CTs. Practise the CT phasor diagram until the formulas are obvious.

Quick check

1. A 200/5 A CT feeds an ammeter reading 3.5 A. Primary current?
2. What happens if a loaded CT's secondary is opened?
3. Nominal ratio 400, actual ratio 402. Ratio error?
4. Is a CT's burden a high or a low impedance? Answers: 1. 140 A; 2. the core saturates, a dangerous high voltage appears across the secondary and the core overheats; 3. (400 − 402)/402 = −0.50 %; 4. low impedance (the CT works almost short-circuited).