Question

A Three-Phase transformer has a total series impedance of 0.01+j0.05 pu. The total impedance of the magnetization part of the transformer is very big and can be ignored in the modeling. The transformer delivers a load with 1.0 pu active power at unity power factor. If the voltage at the load terminals is 1.0 pu, what will be the voltage at the source side of the transformer?

Answer 1

The value is
`V_s  = 1.01 + j0.05 \  pu`

Step-by-step explanation:

From the question we are told that

The series impedance is
`R  =  0.01 + j0.05 \ pu`

The active power is
`P  =  1.0 \  pu`

The voltage at the load terminal is
`V_l   = 1.0 \ pu`

Generally the voltage at the source side of the transformer at the source side of the transformer is mathematically represented as

`V_s  = V_l +   I  * R`

Here I is the current through the load which is mathematically represented as

`I  =  (P< \theta)/(V_l)`

Here
`\theta` is the power factor which is mathematically represented as

`\theta =  cos^(-1)(P)`

`\theta =  cos^(-1)(1)`

`\theta =  0^o`

So

`I  =  ( 1 \angle 0^o)/(1)`

`I  =  1 \angle 0^o`

So

`V_s  = 1  +   (0.01 + j0.05)(1)`

`V_s  = 1.01 + j0.05 \  pu`