Answer:
The value is [tex]V_s = 1.01 + j0.05 \ pu[/tex]
Explanation:
From the question we are told that
The series impedance is [tex]R = 0.01 + j0.05 \ pu[/tex]
The active power is [tex]P = 1.0 \ pu[/tex]
The voltage at the load terminal is [tex]V_l = 1.0 \ pu[/tex]
Generally the voltage at the source side of the transformer at the source side of the transformer is mathematically represented as
[tex]V_s = V_l + I * R[/tex]
Here I is the current through the load which is mathematically represented as
[tex]I = \frac{P< \theta}{V_l}[/tex]
Here [tex]\theta[/tex] is the power factor which is mathematically represented as
[tex]\theta = cos^{-1}(P)[/tex]
[tex]\theta = cos^{-1}(1)[/tex]
[tex]\theta = 0^o[/tex]
So
[tex]I = \frac{ 1 \angle 0^o}{1}[/tex]
[tex]I = 1 \angle 0^o[/tex]
So
[tex]V_s = 1 + (0.01 + j0.05)(1)[/tex]
[tex]V_s = 1.01 + j0.05 \ pu[/tex]
