**Introduction**

Transformer voltage drop is important to know since it is one of the factors that affect the performance of an electrical system where it is installed. Obviously high voltage drop in transformer could lead into low voltage at the load side of the system.

**Formula**

Single Phase Transformer: Vd = I (R cos theta + X sin theta)

Three Phase Transformer: Vd = sqrt(3) x I (R cos theta + X theta)

where:

Vd = voltage drop

R = Resistance

X = Reactance

theta = power factor angle

Read: What is the Importance of X/R Ratio?

Find the voltage drop of the single phase transformer supplying a 50 HP motor with a power factor of 0.70. The transformer has manufacturer rating given below.

In this case the resistance and reactance of the transformer is given in its percentage form, therefore we need to determine the actual value of these quantities. In determinin23g the actual values we need to use the following formula,

R actual =

KVA transformer

R actual =

100 KVA

X actual =

100 KVA

X actual =

100 KVA

Determine the value of the current

P = 50 HP x

HP

I = P / VL * pf

I = 37, 300 Watts / 230 V * 0.7

I = 231 Amperes

cos theta = 0.7

sin theta = 0.7

Vd = I (R cos theta + X sin theta)

Vd = 231 A x [ (0.01185)( 0.7) + (0.0182) (0.7) ]

Find the voltage drop of the three phase transformer supplying a 100 KVA load with a power factor of 0.80. The transformer has manufacturer rating given below.

The process is still the same above but we differ only in the final calculation of the current since the value that we can get is to be multiplied with sqrt(3) or 1.73.

R actual =

KVA transformer

R actual =

100 KVA

X actual =

100 KVA

X actual =

100 KVA

Determine the value of the current

I = S / (1.73 x VL)

I = 100,000 VA / (1.73 x 230)

I = 251 Amperes

cos theta = 0.8

sin theta = 0. 6

Vd = 1.73 x I (R cos theta + X sin theta)

Vd = 1.73 x 251 A x [ (0.0031)( 0.8) + (0.0047) (0.6) ]

**Example 1 (Single Phase Transformer)**Find the voltage drop of the single phase transformer supplying a 50 HP motor with a power factor of 0.70. The transformer has manufacturer rating given below.

- voltage rating = 12.7KV / 230V
- KVA rating = 100 KVA
- % R = 2.24 %
- % X = 3.34 %

__Solution:__In this case the resistance and reactance of the transformer is given in its percentage form, therefore we need to determine the actual value of these quantities. In determinin23g the actual values we need to use the following formula,

R actual =

__10 (%R) (KV secondary) ^2__KVA transformer

R actual =

__10 (2.24%) (0.23 KV) ^ 2__>> use 230 V secondary as base voltage100 KVA

**R actual = 0.01185 ohms >> value of the actual resistance**X actual =

__10 (%X) (KV secondary) ^ 2__100 KVA

X actual =

__10 (3.44%) (0.23 KV) ^2__>> use 230 V secondary as base voltage100 KVA

**X actual = 0.0182 ohms >> value of the actual reactance**Determine the value of the current

P = 50 HP x

__746 W__= 37,300 wattsHP

I = P / VL * pf

I = 37, 300 Watts / 230 V * 0.7

I = 231 Amperes

cos theta = 0.7

sin theta = 0.7

Vd = I (R cos theta + X sin theta)

Vd = 231 A x [ (0.01185)( 0.7) + (0.0182) (0.7) ]

**Vd = 4.85 Volts or****%Vd =**__(4.85 V) x 100__= 2.11 %**230 Volts Base****Example 2 (Three Phase Transformer)**Find the voltage drop of the three phase transformer supplying a 100 KVA load with a power factor of 0.80. The transformer has manufacturer rating given below.

- voltage rating = 12.7KV / 230V
- KVA rating = 150 KVA
- % R = 1.08 %
- % X = 1.63 %

__Solution:__The process is still the same above but we differ only in the final calculation of the current since the value that we can get is to be multiplied with sqrt(3) or 1.73.

R actual =

__10 (%R) (KV secondary) ^2__KVA transformer

R actual =

__10 (1.08%) (0.23 KV) ^ 2__>> use 230 V secondary as base voltage100 KVA

**R actual = 0.0031 ohms >> value of the actual resistance**X actual =

__10 (%X) (KV secondary) ^ 2__100 KVA

X actual =

__10 (1.63%) (0.23 KV) ^2__>> use 230 V secondary as base voltage100 KVA

**X actual = 0.0047 ohms >> value of the actual reactance**Determine the value of the current

I = S / (1.73 x VL)

I = 100,000 VA / (1.73 x 230)

I = 251 Amperes

cos theta = 0.8

sin theta = 0. 6

Vd = 1.73 x I (R cos theta + X sin theta)

Vd = 1.73 x 251 A x [ (0.0031)( 0.8) + (0.0047) (0.6) ]

**Vd = 2.30 Volts or****%Vd =**__(2.30 V) x 100__= 1.0 %**230 Volts Base**