Chapter 10, Problem 10.1P

To determine

The value of the CT secondary current (I’) and VT secondary voltage (V’).

Answer to Problem 10.1P

  $\begin{array}{l}V\text{'}=115V\\ I\text{'}=4.184A\end{array}$

Explanation of Solution

Given Information:

  $\begin{array}{l}{V}_{LL}=345KV\\ S_{3\varphi }=600MVA\\ \text{pf}=0.95\text{lagging}\\ {\text{n}}_1\left(\text{CT}\text{Ratio}\right)=1200:5\\ \text{n}\left(\text{VT}\text{Ratio}\right)=3000:1\end{array}$ .

Calculation:

The secondary voltage of the voltage transformer (V’) is,

  $V\text{'}=\left(\frac{1}{n}\right)V_{LL}$

Substitute the values in the above equation,

  $\begin{array}{l}V\text{'}=\left(\frac{1}{3000}\right)\times \left(345\times {10}^3\right)\\ V\text{'}=115V\end{array}$

Now, the current (I) entering the primary of the current transformer is,

  $I=\frac{S_{3\varphi }}{\sqrt{3}{V}_{LL}}$

On substituting the values in the above equation,

And get,

  $\begin{array}{l}I=\frac{600\times {10}^6}{\sqrt{3}\times (345\times {10}^3)}\\ I=\frac{600\times {10}^6}{597.55\times {10}^3}\\ I=1004A\end{array}$

Here, the CT equivalent circuit is,

  

Fig: CT equivalent circuit diagram

Now, the secondary current of CT is,

  $I_2=\left(\frac{1}{n_1}\right)\times I$

Substituting the values in the above equation,

  $\begin{array}{l}{I}_2=\left(\frac{5}{1200}\right)\times 1004\\ I_2=4.184A\end{array}$

From the above equivalent circuit diagram, the secondary current consists of two components, there are $I_{e}andI\text{'}$ . Therefore the expression of the secondary current of CT is,

  $I_2=I\text{'}+I_e$

Whereas, I’ is the relay current and I_e is the exciting current.

CT error is Zero in the given question.

So that the exciting current is Zero, (I_e=0), therefore,

  $\begin{array}{l}{I}_2=I\text{'}+I_e\\ 4.184A=I\text{'}+0\\ I\text{'}=4.184A\end{array}$