How to Calculate CT Burden: Screening Secondary Circuit Load for Accurate Protection and Metering
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Electrical Engineering April 26, 2026 11 min read

How to Calculate CT Burden: Screening Secondary Circuit Load for Accurate Protection and Metering

Problem Framing

A current transformer (CT) with a correctly selected ratio can still cause relay misoperation or metering errors if the secondary burden exceeds the CT's rated capacity. Neglecting lead resistance is a documented source of metering inaccuracy: a 600:5 CT rated 0.3B0.1 on a 200 ft run of #12 AWG copper can accumulate enough I²R burden to shift ratio error beyond the 0.3% class limit, producing a revenue metering error that persists undetected for months. The CT burden calculation (total VA load on the secondary from wiring and connected devices) is the screening step that prevents this.

When burden exceeds the CT's rated class, the core saturates earlier during faults, reducing relay pickup and potentially delaying fault clearing. For metering, high burden increases ratio and phase-angle errors. This calculator uses a fixed steady-state model to estimate total burden from secondary current, loop resistance, and device burden. It is the first check before confirming CT application against nameplate data and standards like IEEE C57.13-2016 or IEC 61869-2:2012. For related screening methods, see How to Calculate Cable Ampacity: Applying Correction and Adjustment Factors for Conductor Screening: conductor sizing directly affects lead resistance and thus CT burden.

Exact Formula / Method

The total CT burden is the sum of two components: burden from secondary wiring resistance (I²R) and burden from connected devices (relays, meters, transducers). The wiring resistance must account for the total loop: both outgoing and return conductors.

R_loop = R_direct                   (if direct resistance entered)
R_loop = 2 × L × R_per_unit        (if lead length entered)
B_resist = I² × R_loop
B_total = B_resist + B_device

Variables
- I: CT secondary current (A). Typical values: 1 A or 5 A (standard CT secondaries). 5 A is common in North America; 1 A in IEC systems for long runs.
- R_loop: Total secondary loop resistance (Ω). For a 5 A circuit, typical range 0.05–0.2 Ω; for 1 A, up to 0.5–1.0 Ω may be acceptable.
- L: One-way lead length (m or ft). Realistic: 10–500 m (33–1640 ft).
- R_per_unit: Conductor resistance per unit length (Ω/m or Ω/ft). For uncoated copper at 20°C reference per NEC 2023 Chapter 9 Table 8 (Conductor Properties): #12 AWG = 0.001588 Ω/ft (0.00521 Ω/m); #10 AWG = 0.000999 Ω/ft (0.00328 Ω/m). For operating temperature 75°C, multiply by approximately 1.21 per NEC Chapter 9 Table 9 temperature correction. CT secondary circuits typically operate near ambient temperature, so 20°C values are appropriate for screening calculations.
- B_resist: Burden from resistance (VA). Grows with I²; doubling I quadruples B_resist.
- B_device: Connected device burden (VA). Typical relay burden: 0.1–0.5 VA (modern numerical relays) or 1–5 VA (electromechanical). Metering: 0.2–2 VA.
- B_total: Total CT burden (VA). Compared against CT rated burden (e.g., 15 VA for a C100 relay class, or 2.5 VA for a 0.3B0.1 metering class).

The I²R term captures the physical fact that secondary current flowing through wire resistance dissipates power as heat, which the CT must supply. Longer runs and smaller gauge wire increase this burden. The device burden is additive because each connected device draws its own VA from the CT.

Inputs Explained

The two critical inputs are the total secondary loop resistance and the connected device burden. Engineers commonly underestimate both.

Loop resistance can be entered directly (measured with a low-resistance ohmmeter or calculated from wire tables) or derived from lead length and conductor resistance per unit length. When using lead length mode, remember that the calculator automatically doubles the one-way length to account for the full loop. For a 5 A circuit, a 0.1 Ω loop adds 2.5 VA, enough to push a lightly rated metering CT over its class. On a 1 A circuit, the same resistance adds only 0.1 VA, making 1 A secondaries advantageous for long runs.

Connected device burden must be obtained from manufacturer data sheets. Modern microprocessor relays often quote burden as 0.1–0.3 VA per phase, but older electromechanical relays can be 2–5 VA. If multiple devices are connected in series (e.g., relay + meter + transducer), sum their burdens. Omitting the device burden is a common mistake that can understate total burden by 50% or more.

Worked Example

Scenario: A 600:5 CT feeds a protection relay (0.3 VA burden) located 200 ft (61 m) from the CT. The secondary wiring is #12 AWG copper (0.00159 Ω/ft). Determine total CT burden.

Metric calculation:

I = 5 A
L = 61 m
R_per_unit = 0.00521 Ω/m (from wire table)
R_loop = 2 × 61 × 0.00521 = 0.6356 Ω
B_resist = 5² × 0.6356 = 25 × 0.6356 = 15.89 VA
B_device = 0.3 VA
B_total = 15.89 + 0.3 = 16.19 VA

Imperial calculation:

I = 5 A
L = 200 ft
R_per_unit = 0.00159 Ω/ft
R_loop = 2 × 200 × 0.00159 = 0.636 Ω
B_resist = 25 × 0.636 = 15.9 VA
B_device = 0.3 VA
B_total = 15.9 + 0.3 = 16.2 VA

Interpretation: B_total = 16.2 VA. Compare against CT class ratings per IEEE C57.13-2016:

Protection class C100:
- C-rating: 100 V at 20× rated secondary current (100 A for 5 A CT) without exceeding 10% ratio error
- Burden impedance at rated current: V/I = 100 V / 100 A = 1.0 Ω
- Burden VA at rated current: I² × Z = 5² × 1.0 = 25 VA
- Verdict: 16.2 VA < 25 VA — CT meets C100 class for protection.

Metering class 0.3B0.1:
- Standard burden 0.1 Ω at 5 A → 2.5 VA rated burden per IEEE C57.13-2016 Table 6
- Verdict: 16.2 VA >> 2.5 VA — CT does not meet 0.3B0.1 class for revenue metering. Ratio and phase-angle errors will exceed 0.3% accuracy specification.

Decision matrix:

(1) Upgrade conductor #12 AWG → #10 AWG: R_per_unit drops from 0.00159 Ω/ft to 0.001 Ω/ft. R_loop = 2 × 200 × 0.001 = 0.4 Ω. B_resist = 25 × 0.4 = 10 VA. B_total = 10.3 VA. Still exceeds 2.5 VA for 0.3B0.1 metering.

(2) Switch to 1 A secondary CT: B_resist scales with I², so 1 A vs 5 A reduces I²R term by factor 25. R_loop unchanged at 0.636 Ω. B_resist = 1² × 0.636 = 0.636 VA. B_total = 0.636 + 0.3 = 0.94 VA. Meets 0.3B0.1 class with margin.

(3) Relocate meter closer to CT (reduce L from 200 ft to 50 ft): R_loop = 2 × 50 × 0.00159 = 0.159 Ω. B_resist = 25 × 0.159 = 3.98 VA. B_total = 4.28 VA. Still exceeds 0.3B0.1 (2.5 VA), needs further reduction.

(4) Combination: 1 A secondary + #10 AWG conductor. R_loop = 2 × 200 × 0.001 = 0.4 Ω. B_resist = 1² × 0.4 = 0.4 VA. B_total = 0.7 VA. Comfortably meets 0.3B0.1 class.

For protection function alone (C100), no changes needed. For revenue-grade metering combined with protection, option (4) provides cleanest solution per IEEE C57.13-2016 best practice.

What the Result Means

Engineering interpretation by B_total magnitude (compared against IEEE C57.13-2016 standard burden classes):

Below 2.5 VA: meets all standard metering classes per IEEE C57.13-2016 Table 6: B-0.1 (0.1 Ω, 2.5 VA at 5 A), B-0.2 (0.2 Ω, 5 VA), B-0.5 (0.5 Ω, 12.5 VA), B-0.9 (0.9 Ω, 22.5 VA), B-1.8 (1.8 Ω, 45 VA). Suitable for revenue-grade metering applications with accuracy class 0.3 (within 0.3% ratio error).

2.5–12.5 VA: meets B-0.2 through B-0.5 metering classes. Acceptable for general metering and standard relay protection (B-1, B-2 relay classes per IEEE C57.13-2016 Table 7).

12.5–25 VA: exceeds typical metering classes. Verify against B-1 (1 Ω, 25 VA) and B-2 (2 Ω, 50 VA) standard relay burdens. Acceptable for C100 protection class CTs (25 VA equivalent at 5 A rated).

Above 25 VA: exceeds C100 protection rating. Verify against C200 (50 VA equivalent), C400 (100 VA), C800 (200 VA) higher protection classes per IEEE C57.13-2016 Section 6.4. May require CT redesign or burden reduction.

Verification path: if B_total exceeds CT rated burden (from nameplate), CT may not meet accuracy class during steady state or saturate during faults. For protection applications, perform CT saturation analysis using knee-point voltage per IEEE C57.13-2016 Section 7 (Saturation Voltage), which requires separate excitation curve test data. For CT secondary feeder verification, see Cable Ampacity Calculator: NEC 310.15 ampacity verification for CT secondary feeder conductors. For terminal voltage analysis at metering devices, see Voltage Drop Calculator: terminal voltage analysis for CT secondary circuits at rated current.

Common Mistakes

Ignoring the square-law effect of secondary current: A 5 A secondary produces 25× the I²R burden of a 1 A secondary for the same resistance. Engineers moving from 5 A to 1 A designs often forget that lead burden drops dramatically. Conversely, using 5 A on a long run without checking burden is a frequent cause of saturation.

Using one-way lead length instead of total loop: The calculator automatically doubles the one-way length, but when manually estimating, many engineers use only the one-way distance. This halves the calculated resistance burden, leading to an optimistic result. In the example above, using 200 ft instead of 400 ft of conductor would give 7.95 VA instead of 15.9 VA.

Omitting device burden or using outdated data: A relay data sheet might list burden as "< 0.5 VA" but that is per phase at rated current. If the relay has multiple CT inputs (e.g., for differential protection), each input adds burden. Also, older electromechanical relays have higher burden than modern numerical ones; using generic values can understate burden by 1–2 VA.

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When This Method Is Not Enough

This calculator estimates steady-state burden at rated secondary current. It does not model CT saturation during faults, where high fault current (up to 20× rated) drives the core into nonlinear operation. The burden impedance seen by the CT during saturation changes because the relay's burden may be nonlinear (e.g., microprocessor relays have a constant VA characteristic, not constant impedance). For protection applications, the engineer must verify that the CT's knee-point voltage exceeds the required voltage for the maximum fault current and total burden, per IEEE C57.13-2016 or IEC 61869-2:2012.

Additionally, this method assumes all connected devices present a constant VA load. In reality, some relays have a burden that varies with current (e.g., electromechanical induction discs). The fixed model also does not account for the resistance of terminals, fuses, or test switches, which add small but cumulative ohms. For critical protection schemes, use the exact lead resistance measured in the field and confirm with a CT saturation curve or excitation test.

FAQ

How do I measure total secondary loop resistance for a CT circuit?

Use a low-resistance ohmmeter (milliohmmeter) at the CT terminals with all secondary devices disconnected. Measure resistance between the two secondary leads. This gives the total loop resistance including wiring, terminals, and connections. Typical values for a 5 A circuit range from 0.05 Ω (short run, #10 AWG) to 0.5 Ω (500 ft, #12 AWG).

What is the difference between CT burden and CT accuracy class?

CT burden is the load (VA) connected to the secondary. Accuracy class (e.g., 0.3B0.1) specifies the maximum allowable error at a given burden. A CT can have a low burden but still fail accuracy if the burden exceeds its class rating. For example, a 0.3B0.1 CT is accurate within 0.3% at 0.1 Ω burden (2.5 VA at 5 A). If actual burden is 5 VA, accuracy degrades.

Can I connect multiple relays in series on the same CT secondary?

Yes, but the total device burden is the sum of each relay's burden. Ensure the total burden does not exceed the CT's rated burden for the required accuracy class. Also, check that the CT can drive the total impedance during faults; each relay's burden may be in series, increasing the loop impedance.

Why does a 1 A secondary reduce burden compared to 5 A?

Because I²R burden scales with the square of current. For the same loop resistance, a 1 A secondary produces 1/25th the I²R burden of a 5 A secondary. This allows longer lead runs or smaller conductors without exceeding CT burden ratings. Many IEC installations use 1 A for this reason.

What should I do if the calculated burden exceeds the CT nameplate rating?

First, verify the CT nameplate rating; it may be listed as a voltage class (e.g., C100) rather than VA at rated current. Convert: VA = (voltage class) / (20 × I_rated). For C100 at 5 A, VA = 100 / 100 = 1 Ω → 25 VA. If burden still exceeds rating, options include: use larger conductor (reduce R_loop), reduce lead length, switch to 1 A secondary, or select a CT with higher burden rating (e.g., C200 or C400).

What secondary current rating should I choose for a new CT installation: 1 A or 5 A?

For North American installations per IEEE C57.13-2016, 5 A is the standard and is suitable for short runs within switchgear or motor control centers. For runs exceeding 50 m (160 ft), 1 A is preferred because I²R burden scales with the square of current: the same loop resistance produces 25 times less burden at 1 A than at 5 A. IEC installations per IEC 61869-2:2012 commonly specify 1 A for this reason. If retrofitting into an existing 5 A protection scheme with long runs, calculate the actual burden first; if it exceeds the CT's rated class, upgrading conductor gauge or switching to a 1 A secondary CT are the primary corrective measures.

How does CT saturation during faults affect protection coordination?

CT saturation occurs when fault current drives the magnetic core into nonlinear region, distorting secondary current waveform and delaying or preventing relay operation. Per IEEE C57.13-2016 Section 7 (Saturation Voltage): saturation onset typically occurs when secondary terminal voltage exceeds knee-point voltage V_kp, defined as the point where 10% increase in voltage causes 50% increase in excitation current per IEC 61869-2:2012 Annex 2A. Required CT performance for protection: V_kp,required = I_fault × (R_CT_secondary + R_loop + Z_burden), where I_fault is maximum symmetric fault current referred to secondary, R_CT_secondary is the CT's internal secondary winding resistance from nameplate, and Z_burden is total connected device impedance. For C-class CTs per IEEE C57.13-2016 Section 6.4, the CT delivers at least the C-rating voltage at 20× rated current (typically 100 A for 5 A CTs). For fault currents below 20× rated, the CT operates linearly within its accuracy class. Effects of saturation on protection coordination include overcurrent relay delayed pickup due to RMS current underestimation, differential relay false trip due to mismatched CT saturation between zone CTs, and distance relay impedance measurement error. Mitigation: select C-class CT with V_kp exceeding V_kp,required by at least 1.5× margin per IEEE C37.110-2007 (IEEE Guide for the Application of Current Transformers Used for Protective Relaying Purposes). Alternative: use TPS or TPY class CTs per IEC 61869-2:2012 designed specifically for transient performance.

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