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9. Understand the principle of parallel operation of transformer.

9.6. Describe the parallel operation of transformers with equal voltage ratio.

Why do transformers connected in parallel have the same voltage ratio?
Explanation: The transformers must have the same voltage-ratio to avoid no-load circulating current when transformers are in parallel on both primary and secondary sides. Since the leakage impedance is low, even a small voltage difference can give rise to considerable no-load circulating current and extra I2R loss.
Condition for Parallel Operation of Transformer

For parallel connection of transformers, primary windings of the Transformers are connected to source bus-bars and secondary windings are connected to the load bus-bars.

Various conditions that must be fulfilled for the successful parallel operation of transformers:

  1. Same voltage and Turns Ratio (both primary and secondary voltage rating is same)
  2. Same Percentage Impedance and X/R ratio
  3. Identical Position of Tap changer
  4. Same KVA ratings
  5. Same Phase angle shift (vector group are same)
  6. Same Frequency rating
  7. Same Polarity
  8. Same Phase sequence

Some of these conditions are convenient and some are mandatory.

The convenient conditions are: Same voltage Ratio and Turns Ratio, Same Percentage Impedance, Same KVA Rating, Same Position of Tap changer.

The mandatory conditions conditions are: Same Phase Angle Shift, Same Polarity, Same Phase Sequence and Same Frequency. When the convenient conditions are not met paralleled operation is possible but not optimal.

Conditions for Parallel Operation of Transformers

When two or more transformers run in parallel, they must satisfy the following conditions for satisfactory performance. These are the conditions for parallel operation of transformers.

  1. Same voltage ratio of transformer.
  2. Same percentage impedance.
  3. Same polarity.
  4. Same phase sequence.

Example Connecting two 2000 kVA, 5.75% impedance transformers in parallel, each with the same turn ratios to a 4000 kVA load.

  • Loading on the transformers-1 =KVA1=[( KVA1 / %Z) / ((KVA1 / %Z1)+ (KVA2 / %Z2))]X KVAl
  • kVA1 = 348 / (348 + 348) x 4000 kVA = 2000 kVA.
  • Loading on the transformers-2 =KVA1=[( KVA2 / %Z) / ((KVA1 / %Z1)+ (KVA2 / %Z2))]X KVAl
  • kVA2 = 348 / (348 + 348) x 4000 kVA = 2000 kVA
  • Hence KVA1=KVA2=2000KVA

  • Example Connecting 3000 kVA and 1000 kVA transformers in parallel, each with 5.75% impedance, each with the same turn ratios, connected to a common 4000 kVA load.

    • Loading on Transformer-1=kVA1 = 522 / (522 + 174) x 4000 = 3000 kVA
    • Loading on Transformer-1=kVA2 = 174 / (522 + 174) x 4000 = 1000 kVA

    From above calculation it is seen that different kVA ratings on transformers connected to one common load, that current division causes each transformer to only be loaded to its kVA rating. The key here is that the percent impedance are the same.


  • Example Two 2000 kVA transformers in parallel, one with 5.75% impedance and the other with 4% impedance, each with the same turn ratios, connected to a common 3500 kVA load.

    • Loading on Transformer-1=kVA1 = 348 / (348 + 500) x 3500 = 1436 kVA
    • Loading on Transformer-2=kVA2 = 500 / (348 + 500) x 3500 = 2064 kVA

    It can be seen that because transformer percent impedances do not match, they cannot be loaded to their combined kVA rating. Load division between the transformers is not equal. At below combined rated kVA loading, the 4% impedance transformer is overloaded by 3.2%, while the 5.75% impedance transformer is loaded by 72%.

    xample Two transformers in parallel with one 3000 kVA (kVA1) with 5.75% impedance, and the other a 1000 kVA (kVA2) with 4% impedance, each with the same turn ratios, connected to a common 3500 kVA load.

    • Loading on Transformer-1=kVA1 = 522 / (522 + 250) x 3500 = 2366 kVA
    • Loading on Transformer-2=kVA2 = 250 / (522 + 250) x 3500 = 1134 kVA

    Because the percent impedance is less in the 1000 kVA transformer, it is overloaded with a less than combined rated load.


  • Example Two 2000 kVA transformers connected in parallel, each with 5.75% impedance, same X/R ratio (8), transformer 1 with tap adjusted 2.5% from nominal and transformer 2 tapped at nominal. What is the percent circulating current (%IC)

    • %Z1 = 5.75, So %R’ = %Z1 / √[(X/R)2 + 1)] = 5.75 / √((8)2 + 1)=0.713
    • %R1 = %R2 = 0.713
    • %X1 = %R x (X/R)=%X1= %X2= 0.713 x 8 = 5.7
    • Let %e = difference in voltage ratio expressed in percentage of normal and k = kVA1/ kVA2
    • Circulating current %IC = %eX100 / √ (%R1+k%R2)2 + (%Z1+k%Z2)2.
    • %IC = 2.5X100 / √ (0.713 + (2000/2000)X0.713)2 + (5.7 + (2000/2000)X5.7)2
    • %IC = 250 / 11.7 = 21.7

    The circulating current is 21.7% of the full load current.


  • Example Two transformers connected in parallel, 2000 kVA1 with 5.75% impedance, X/R ratio of 8, 1000 kVA2 with 4% impedance, X/R ratio of 5, 2000 kVA1 with tap adjusted 2.5% from nominal and 1000 kVA2 tapped at nominal.

    • %Z1 = 5.75, So %R’ = %Z1 / √[(X/R)2 + 1)] = 5.75 / √((8)2 + 1)=0.713
    • %X1= %R x (X/R)=0.713 x 8 = 5.7
    • %Z2= 4, So %R2 = %Z2 /√ [(X/R)2 + 1)]= 4 / √((5)2 + 1) =0.784
    • %X2 = %R x (X/R)=0.784 x 5 = 3.92
    • Let %e = difference in voltage ratio expressed in percentage of normal and k = kVA1/ kVA2
    • Circulating current %IC = %eX100 / √ (%R1+k%R2)2 + (%Z1+k%Z2)2.
    • %IC = 2.5X100 / √ (0.713 + (2000/2000)X0.713)2 + (5.7 + (2000/2000)X5.7)2
    • %IC = 250 / 13.73 = 18.21.

    The circulating current is 18.21% of the full load current.


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