Understanding the 3-Phase Transformer CT Ratio Calculation

This problem involves calculating the current transformer (CT) ratio on the high voltage (HV) side of a 3-phase transformer, given its rating, connection type, and the CT ratio on the low voltage (LV) side.

Given Information:

• Transformer Rating: 33 kV (HV) / 6.6 kV (LV)
• Connection: HV side is Star (Y), LV side is Delta ($\Delta$)
• LV CT Ratio: 500 : 5 (Primary Current : Secondary Current)

Step 1: Calculate Voltage and Current Ratios

First, let's find the voltage transformation ratio:

$$ V_{ratio} = \frac{V_{HV}}{V_{LV}} = \frac{33 \text{ kV}}{6.6 \text{ kV}} = 5 $$

For a 3-phase transformer, the ideal relationship between line currents ($I_{LL}$) and line voltages ($V_{LL}$) depends on the connection. In a Star-Delta connection:

• HV Side (Star - Y): Line current equals phase current ($I_{LL\_HV} = I_{ph\_HV}$).
• LV Side (Delta - $\Delta$): Line current is $\sqrt{3}$ times the phase current ($I_{LL\_LV} = \sqrt{3} \times I_{ph\_LV}$).

The fundamental transformer relationship holds for phase quantities: $$ \frac{I_{ph\_LV}}{I_{ph\_HV}} = \frac{V_{HV\_phase}}{V_{LV\_phase}} $$ Where: $$ V_{HV\_phase} = \frac{V_{LL\_HV}}{\sqrt{3}} = \frac{33}{\sqrt{3}} \text{ kV} $$ $$ V_{LV\_phase} = V_{LL\_LV} = 6.6 \text{ kV} $$ Substituting these values:

$$ \frac{I_{ph\_LV}}{I_{ph\_HV}} = \frac{33/\sqrt{3}}{6.6} = \frac{33}{6.6 \times \sqrt{3}} = \frac{5}{\sqrt{3}} $$

Now, let's relate this to line currents:

Simplifying this, we get the relationship between the line currents:

$$ \frac{I_{LL\_LV}}{I_{LL\_HV}} = 5 $$

This means the line current on the LV side is 5 times the line current on the HV side.

Step 2: Determine the LV Line Current

The CT on the LV side has a ratio of 500:5. This ratio represents the current the CT monitors (primary) to the current it outputs (secondary).

$$ N_{LV} = \frac{\text{Primary Current}}{\text{Secondary Current}} = \frac{I_{LL\_LV}}{I_{Sec\_LV}} = \frac{500}{5} = 100 $$

Assuming the standard secondary current of 5 A, we can find the LV line current:

$$ I_{LL\_LV} = N_{LV} \times I_{Sec\_LV} = 100 \times 5 \text{ A} = 500 \text{ A} $$

Step 3: Calculate the HV Line Current

Using the relationship derived in Step 1 ($I_{LL\_LV} = 5 \times I_{LL\_HV}$):

$$ 500 \text{ A} = 5 \times I_{LL\_HV} $$

Solving for the HV line current:

$$ I_{LL\_HV} = \frac{500 \text{ A}}{5} = 100 \text{ A} $$

Step 4: Determine the HV CT Ratio

The CT on the HV side needs to measure this line current, $I_{LL\_HV}$. Therefore, the primary rating of the HV CT should be 100 A.

The question asks for the HV CT ratio ($I_{HV\_primary} : I_{HV\_secondary}$). Based on the options provided and the derived HV primary current (100 A), the correct option is $100 : \frac{5}{\sqrt{3}}$.

This means:

• HV CT Primary Rating = $I_{LL\_HV} = 100$ A
• HV CT Secondary Rating = $\frac{5}{\sqrt{3}}$ A

The ratio value for the HV CT is:

$$ N_{HV} = \frac{100 \text{ A}}{5/\sqrt{3} \text{ A}} = \frac{100 \sqrt{3}}{5} = 20\sqrt{3} \approx 34.64 $$

Note: While standard CT secondary ratings are typically 1 A or 5 A, the value $\frac{5}{\sqrt{3}}$ A in the correct answer suggests a specific design or calculation method related to the Y-$\Delta$ connection characteristics, ensuring the ratio of CT ratio values ($N_{LV} / N_{HV}$) matches the phase current transformation ratio ($I_{ph\_LV} / I_{ph\_HV}$).

Conclusion:

The ratio of the current transformer on the HV side is determined to be $100 : \frac{5}{\sqrt{3}}$.