How Does Transformer Impedance Impact Primary Voltage and Current Calculations?

[cs23]

Homework Statement

A Single phase transformer has primary to secondary ratio of 2, its primary R1=0.05 ohms,X1=0.25ohms and the secondary R2=0.02ohms and X2=0.03ohms. The paramet of the magentizing branch referred to the primary side are Rc1=2kohms and Xm1=500 ohms. The secondary load current I2=85 A at V2=220V and 0.9 power factor lagging. Calculate the primary voltage V1 and Current I1.

Homework Equations

kvl,Req and Xeq

The Attempt at a Solution

I took KVl of the right side loop and got V1+Z-aV2. Then i found z using Req + JXeq. But I'm getting confused on how to use the power lagging

 

[KataKoniK]

factor to find the values of V1 and I1. Can someone help me out with this?

Dear fellow scientist,

Thank you for bringing this problem to our attention. After examining the given parameters, I have come up with the following solution:

First, we can calculate the equivalent impedance of the secondary load using the given values of R2 and X2. This gives us Z2 = 0.0375 + j0.03 ohms. Now, using the transformer turns ratio of 2, we can calculate the equivalent impedance referred to the primary side as Z1 = (Z2/2)^2 = 0.009375 + j0.0075 ohms.

Next, we can calculate the equivalent impedance of the magnetizing branch referred to the primary side using the given values of Rc1 and Xm1. This gives us Zm1 = 1.959 + j0.49 ohms.

Now, using KVL in the right side loop, we can write the following equation:

V1 - I1(Z1 + Zm1) = V2

Substituting the calculated values of Z1 and Zm1, we get:

V1 - I1(0.009375 + j0.0075 + 1.959 + j0.49) = 220

Simplifying this equation, we get:

V1 - I1(1.968375 + j0.4975) = 220

Since we know that the secondary load current is 85 A at a power factor of 0.9 lagging, we can write the following equation:

V2 = I2Z2*cos(θ)

Where θ is the angle between the voltage and current in the secondary circuit, which in this case is the power factor angle of 0.9 lagging. Substituting the given values, we get:

220 = 85*(0.0375 + j0.03)*cos(0.9)

Simplifying this equation, we get:

220 = 3.1875 + j2.565*cos(0.9)

Now, we can equate the real and imaginary parts of the two equations we have derived to get:

V1 - 1.968375I1 = 3.1875

-0.4975I1 = 2.565*cos(0.9)

Solving these equations, we get:

V1 =