The **copper loss in the **transformer is equal to the I^{2}R loss. The copper loss is very important for the calculation of the transformer efficiency. The efficiency of the transformer can be improved by minimizing the **copper loss and core loss.**

In the transformer, the copper loss in the primary winding is Ip^{2}Rp, and the copper loss in the secondary winding is Is^{2}Rs loss, where I_{p} and I_{s} are primary and secondary current of transformer and R_{p} and R_{s} are resistances of primary and secondary winding respectively.

The copper loss is the wastage of power due to I^{2}R loss in the transformer winding, and the energy is wasted as heat. As both primary and secondary currents depend upon the percentage loading of the transformer, the current increases as the load on the transformer is increased. With an increase in the primary and the secondary current, the copper loss in the transformer gets increased with the load.

The copper loss is the wastage of power due to I^{2}R loss in the transformer winding, and the energy is wasted as heat. As both primary and secondary currents depend upon the percentage loading of the transformer, the current increases as the load on the transformer are increased. With an increase in the primary and the secondary current, the **copper loss in the **transformer gets increased with the load.

The copper loss value is not constant, and it varies as the load on the transformer is increased. That is why the copper loss is also called a variable loss.

The total copper loss (I^{2}R) loss in the transformer is load dependent. The copper losses are proportional to the square of the RMS current flowing in the winding and also proportional to the resistance of the winding. The resistance of the conductor varies with the rise in temperature.

The copper loss in the transformer is proportional to the square of the current flowing through the winding. When the load on the transformer is increased the copper loss varies because of the increased current and increased resistance caused by temperature rise.

The resistance value of the copper or aluminum must be corrected for the maximum permissible rise of the transformer winding at the rated capacity of the transformer. For example, the resistance of the winding measured at 30° C must be corrected for 75° C for an oil-cooled transformer.

**For copper winding,** the increased value of resistance with temperature can be calculated using the following formula.

R_{L}=R_{0} [(TL+235)/(To+235)] |

**Where,**

R_{L} = Resistance at T_{L} temperature

R_{0} = Resistance at ambient temperature

R_{L} = Operating temperature

T_{0} = Ambient temperature

**For aluminum winding,** the increased value of resistance with temperature can be calculated using the following formula.

R_{L}=R_{0}[(T_{L}+225)/(T_{o}+225)] |

Read More : No Load Losses in Transformer |

**Example on increase of resistance with temperature**

Let the resistance of three HV winding of the transformer is 0.967 Ω,0.968 Ω, and 0.967 Ωat 23.8° C. The average HV winding resistance per phase at 23.8°C is;

=(0.967+0.968+0.967)/3 =0.967 Ω

The average HV winding resistance per phase at 75°C is;

R_{L }= R_{0} * [(TL+235)/(To+235)]

R_{75 }= 0.967 * [(75+235)/(23.8+235)]

= 0.967 * [(75+235)/(23.8+235)

R_{75 }= 1.159 Ω

**Percentage increase in resistance**

=(1.159-0.967)/0.967 *100

= 19.85%

From the above calculation, it is clear that the resistance increases with an increase in temperature caused by the flow of current in a conductor.

Read More : Effect of Temperature on Resistance |

**Copper Loss with Transformer Loading**

**Example on Copper Loss**

The full load copper loss of a transformer is 1000 watts, the copper loss at half load(50% Load) will be;

P_{cu }= ( % Load/100)^{2} x P

= ( 50/100)^{2} x 1000

= ( 1/2)^{2} x 1000

= 1/4 x 1000**P _{cu }= 250 Watts**

Thus the copper loss of the transformer at 50 % load is equal to the 1/4 th of the full load copper loss.

**The copper loss at 3/4 th load is equal to**

P_{cu} = (3/4)^{2 } x 1000 = 562.5 Watts

**The copper loss at 1/4 th load is equal to**

P** _{cu}** = (1/4)

^{2 }x 1000 = 62.5 Watts

For a given transformer,the manufacturer can supply values for no-load loss, P_{no-load}, and load loss, P_{LOAD}.

The total transformer loss, P_{total}, at any load level can then be calculated from;

P_{total }= P_{no-load}+ (% load/100)^{2 }x P_{Load} |

**Where,**

_{P no- load = No- load loss ( constant loss)}% Load = load on transformer

Pload = Copper loss at rated full load

**How to Minimize Copper Loss in a Transformer**

The copper loss can be reduced by use of copper wire of proper gauge for transformer winding. The thicker wire has less resistance and the lower copper loss. The large cross-sectional area conductor winding should be used for minimization of the copper loss.

The wire used for transformer winding is chosen of high conductivity. The copper wire has less resistance hence it has less copper loss as compared to the aluminum conductor.

The copper loss can be minimized by vacuum pressure impregnation of the transformer. The transformer winding is kept in vacuum and high pressure varnish is passed sp that air gaps if any gets filled.

The transformer which is used for high frequency application needs to be designed in such a way to have lower skin and proximity effect. The conductor area reduces due to skin and proximity effect and thus, the resistance of the conductor gets increased and it offers higher copper loss.

_{Related Posts on Transformer:}

**No Load Losses in Transformer****Difference between Copper Loss and Iron Loss****Eddy Current Loss Formula Derivation****Eddy Current Loss**

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