Instrument Security Factor (ISF) of Current Transformer

Instrument security Factor (ISF) is used for specifying current limit, by which current will be delivered by a current transfprmer (CT) during fault conditions. 

ISF is used for metering class CTs. For protection class CTs, there is similar term called Accuracy Limit Factor (ALF).

Metering instruments like energy meters, transducers etc. are meant for measuring electrical quantities with high accuracy under normal operation. Measurement under fault condition is not required and insignificant due to short duration of faults. Metering instruments are designed to carry current upto a certain level. 

For example, an energy meter used with CT of 1000/1A ratio carries maximum 1A under full load. Therefore its input current carrying paths needs to be designed for current upto 150 to 200% of rated current (say 2A continuous) . For fault conditions, which will prevail for short duration, it should withstand upto 5A or 10A. But in case of severe fault of 40kA or 50kA on primary side, secondary current may go upto 40A or 50A. If meter is to be designed for such heavy currents its size and connections will be bulky. Its cost will go up and accuracy will go down. Therefore, CT core is designed to saturate at such heavy currents and maximum secondary current is specified as ISF.

ISF is ratio of rated instrument limit primary current to rated primary current. If ISF is specified as 5 for a 1000/1A CT, then CT metering core will saturate at 5000A primary and current in secondary will not go above 5A. Sometimes, auxiliary reactors are used to achieve ISF limit.

Equivalent circuit of CT:

I_P = Primary current

I_S = Secondary current

V_S = Secondary voltage

Z_E = Exciting impedance

I_E = Exciting current

R_S = Secondary resistance

X_L = Leakage reactance

Z_B = Burden impedance 

Effect of burden on ISF: ISF is specified for a certain burden. If actual burden is different from rated burden, ISF will change as below:

ISF at actual burden = ISF at rated burden x (Rated burden / Actual burden)

* Internal secondary winding resistance of CT to be added for calculating burden.

For example, for a CT with rated burden 20VA and ISF 5, if actual burden is 10VA, ISF will be =5 x (20/10) =10. I.e. maximum secondary current will be 10A.

Resistance, Reactance and Impedance

Resistance: When electric current flows through a material there will always some opposition to this flow. Resistance (R) is measure of opposition to flow of electrical current. Resistance is measured in ohm (Ω). The resistance depends on resistivity (ρ), length (l) and area (a) of material:

1. Length of material: Resistance is directly proportional to length of material. That’s why poor voltage conditions are observed in remote locations.
2. Area of material: Resistance is inversely proportional to the area of material. That’s why we use thicker wires for heavy duty appliances.
3. Resistivity of material: Resistivity is fundamental property of material by which it opposes flow of electric current. Resistivity of some of common materials at 20⁰C is given below:

Resistance will cause energy loss, which is equal to I²R. For transfer of electrical power from source to appliance, low losses are expected, therefore materials with lower resistivity are used as conductors like copper, aluminum etc. In some cases, these losses may be intentional. Like in heaters, I²R losses should be higher. Therefore, we use material with high resistivity like Nichrome etc. Rubber, Air and PVC are used for insulation material due to higher resistivity.

Resistance of a material is practically considered constant over a short working range of temperature. However, it varies with temperature due to change in resistivity.  For example resistivity of copper at 21⁰C will be 1.68 x 10^-8 x (1 + 0.00404) Ωm.

Resistance will be same for DC as well as AC currents.   

Reactance (X): In AC systems, voltage and current are continuously varying due to sine wave form. Due to this, in addition to resistance, one more component opposes the flow of current called reactance (X). Reactance have two components, Capacitive reactance (Xc) and inductive reactance (XL).  

Xc will be due to capacitance (C) in the circuit and XL will be due to inductance (L) in the circuit. 

For Parallel RLC circuit, same voltage will be applied across three elements:

Current in Capacitance (IXc) leads voltage by 90⁰ and current in Inductance (IXL) lags voltage by 90⁰, Threrefore, IXc and IXL are 180⁰ apart. Total current due to reactance (IX) will be difference of IXc and IXL. Being parallel connected circuit, total reactance (X) of the circuit will be 

For series RLC circuit same current will flow through all the three elements. Voltage across three elements will be different in magnitude and angle. In this case magnitude of voltage across these components can be higher than source voltage magnitude (VAC).

Reactance of the circuit will be

Impedance (Z): Impedance is combined effect of resistance (R) and total reactance (X). Currents flowing through resistance and reactance are 90⁰ apart. 

For parallel RLC circuit Impedance will be 

For Series RLC circuit impedance will be

For example: If in above circuit of Parallel connected RLC:

V=100V

R = 10 Ω

C = 100 μF

L = 10 mH

f = 50 Hz

Current through resistor

IR = 100/10 = 10A @0⁰

Xc = 1/ (2 x 3.1428 x 50 x 100 x 10^-6)

Xc = 31.82 Ω

Current through capacitor

IXc = 100/31.82 = 3.142A @90⁰

XL = 2 x 3.1428 x 50 x 10 x 10^-3

XL = 3.14 Ω

Current through inductor

IXL = 100/3.14 = 31.85A @-90⁰

Total reactance X will be

X = 1/(1/3.14 - 1/31.82) = 3.48 Ω

IX = 100/3.48 = 28.7A @-90⁰

Or 

IX = IXL - IXc = 31.85 - 3.142 = 28.7A

Z = 1/[√{(1/R)² + (1/X)²}]

Z = 1/[√{(1/10)² + (1/3.48)²)}] = 3.29 Ω

Or calculation of impedance by caculating current first:

IZ = √(IR² + IX²)

IZ = √(10² + 28.7²) = 30.39 A

Z = V/IZ = 100/30.39 = 3.29 Ω

From the above it can be seen that current through inductor is 31.85A which is higher than current supplied by source.

Now, if same components are connected in series, the impedance of circuit will be as below:

V=100V

R = 10 Ω

C = 100 μF

L = 10 mH

f = 50 Hz

Xc = 1/ (2 x 3.1428 x 50 x 100 x 10^-6)

Xc = 31.82 Ω

XL = 2 x 3.1428 x 50 x 10 x 10^-3

XL = 3.14 Ω

Total reactance X will be

X = Xc - XL = 31.82 - 3.14 = 28.68 Ω

Z = √(R² + X²) 

Z = √(10² + 28.68²) = 30.37 Ω

Current through circuit = V/Z = 100/30.37 = 3.29 A

Voltage across resistor = IR = 3.29 x 10 = 32.9V @0⁰

Voltage across capacitor = IXc = 3.29 x 31.82 = 104.7V @-90⁰

Voltage across inductor = IXL = 3.29 x 3.14 = 10.3V @90⁰

From the above it can be seen that voltage across capacitor is 104.7V wich is higher than source voltage of 100V. Therefore, while working with RLC circuits, precuations must be taken to avoid electric shock from higher voltages in the circuit.