**Instrument Transformer and Power Management (P1) Course**

**Chapter (6) : Current Transformers****6.7 Current transformer burden**

__Introduction :__

Current transformer burdens consists of the combined impedances of the secondary leads, the meters , relays and auxiliary or summating transformers used in the current circuits.

The secondary leads are generally kept short , their impedance are usually neglected.

Meter burdens, relays normally specified in volt amperes per element are published by each meter manufacturer.

However, since transformer accuracy class ratings are given for specific impedances, this volt ampere burdens must be converted to impedance values before they can be used.

To do this, the meter burdens expressed in volt amperes is divided by the rated current squared.

The burden expressed in ohms is: V A / (Rated current )

^{2}

Note :

The rated secondary current for most current transformers is 5 amperes.

To be mathematically correct, burdens should be added vectors. However, since arithmetic additions are easier, and produce burden calculation higher than actual values (vectors), they are usually close enough to be used for quick checking.

__Summary :__

The circuit connected to the secondary winding is termed as the " burden” of the current transformer.

Burden is expressed in terms of impedance of the circuit connected to the secondary ( resistance and reactance).

The British method is to specify the burden on C.T in volt amperes at rated secondary current, Thus we may express the burden in the following two forms: “e.g. 0.5 ohm impedance or 12.5 volt ampere at 5 amperes,

Let the rated burden be VA at rated secondary current Is, then the ohmic impedance of burden Z

_{b}can calculated as follows:(Zb = VA / I

_{s}

^{2}) ohms.

But we must know that, for current transformer we want the current be Constant, so the burdens are connected in series.

The ideal case is when IsEc is very high then the burden must be low as possible.

__Example :__

For a current transformer of 500V A and turns ratio of .100015 , if the primary current is 500 A -it is required to connect some relays which have the following coil resistances :

10 Ω - 15 Ω - 25 Ω - 20 Ω - 7 Ω - 15 Ω - 8 Ω

Show if that is possible or not ? If not what can you do to use them.

__Solution :__

Since Ip / Is = N2/ N1

Z burden = VA/ IS2 = 500 / (5)2 = 20 Ω

Z load = .10 + 15 + 25 + 20 + 7 + 15 + 8 = 100 Ω

Then, that not possible to connect all relays because Z load must be equal or less than Z burden

Then, We must to neglect the relay of 25 Ω, 15 Ω, 15 Ω, 10 Ω, 8 Ω & 7 Ω or the relays of 15 Ω, 15 Ω, 25 Ω, 7 Ω, 20 Ω & 8 Ω and so on

However, we must have a total load less than or equal to 20 Ω

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