Question

A current transformer has a rated current ratio of 600:5 A, Z2 = 0.1 ± j0.5 Ω and ZB = 6.8 ± j1.5 Ω. The core area is 2.8X10-3 m3 and the magnetization curve at 60 Hz is given in the figure Q4. The CT must operate at maximum primary current of 1500A.

Answer 1

I didn't know the answer....

sorry

Answer 2

Answer:

The CT must operate at maximum primary current of 1500A is $20.83$

Explanation:

Given: A current transformer has a rated current ratio of 600:5 A, Z2 = 0.1 ± j0.5 Ω and ZB = 6.8 ± j1.5 Ω. The core area is 2.8X10-3 m3 and the magnetization curve at 60 Hz

To find: CT

Concept:

Current Transformer:

• A current transformer (CT) is an instrument transformer in which the secondary current is substantially proportional to the primary current and differs in phase from it by ideally zero degrees
• Instrument current transformer is used in electrical power system for stepping down currents for metering and protection purposes.

Solution:

Transformation Ratio of CT,

$\mathrm{n}=\frac{\mid \text { Primary Current } \mid}{\mid \text { Secondary Current } \mid}=\frac{\left|I_{p}\right|}{\left|I_{s}\right|}$

Given,

$\begin{aligned}&n=\frac{100}{5}=20 \\&\bar{I}_{s}=5 \angle 0^{\circ} A . \\&\left|\bar{I}_{s}\right|=5 A . \\&\bar{I}_{p}=n \bar{I}_{s}+\bar{I}_{o} \\&\bar{I}_{p}=n \bar{I}_{s}+\left(I_{w}+j I_{m}\right) \\&\bar{I}_{p}=20 \times 5 \angle 0^{\circ}+(4+j 6) \\&\bar{I}_{p}=104+j 6 \\&\left|\bar{I}_{p}\right|=104.172\end{aligned}$

So, the Transformation Ratio of CT

$\mathbf{n}=\frac{104.172}{5}=\mathbf{2 0 . 8 3}$