Question

The load current of a step down 250/200 V auto-transformer is 100 A. The conductive and inductive powers transformed are

• A. \( 4\text{ kVA and } 20\text{ kVA} \)
• B. \( 4\text{ kVA and } 16\text{ kVA} \)
• C. \( 16\text{ kVA and } 4\text{ kVA} \)
• D. \( 20\text{ kVA and } 4\text{ kVA} \)

Hint:

In an auto-transformer, the closer the value of $K$ is to $1$, the higher the portion of power transferred directly via conduction. This is why auto-transformers are highly efficient and compact when the primary and secondary voltage values are close to each other.

Answer 1

The Correct Option is C

Concept: An auto-transformer consists of a single continuous winding shared by both the primary and secondary sides. Unlike a standard two-winding transformer where power is transferred completely via electromagnetic induction (inductively), an auto-transformer transfers power through two distinct mechanisms simultaneously:
• Conduction: Power transferred directly through the electrical connectivity between the input and output circuits.
• Induction: Power transferred magnetically across the shared winding core common to both sections. Let $V_H$ be the higher voltage, $V_L$ be the lower voltage, and $K$ be the transformation ratio defined as: $$K = \frac{V_L}{V_H} \quad (\text{where } K < 1)$$ The formulas governing the division of power transformation are: $$\text{Total Power Rating } (S_{\text{total}}) = V_L \cdot I_L \quad \text{or} \quad V_H \cdot I_H$$ $$\text{Conductively Transformed Power } (S_{\text{cond}}) = K \cdot S_{\text{total}}$$ $$\text{Inductively Transformed Power } (S_{\text{ind}}) = (1 - K) \cdot S_{\text{total}}$$

Step 1: Extract parameters and calculate the transformation ratio \( K \).
From the question text, the given values are:
• Higher voltage side, $V_H = 250\text{ V}$
• Lower voltage side, $V_L = 200\text{ V}$
• Secondary load current, $I_L = 100\text{ A}$ Let us calculate the auto-transformer turns/voltage ratio $K$: $$K = \frac{V_L}{V_H} = \frac{200}{250} = \frac{4}{5} = 0.8$$

Step 2: Calculate the total power output capacity of the auto-transformer.
The total power output delivered to the load (apparent power $S_{\text{total}}$) is given by the product of the output voltage and the output current: $$S_{\text{total}} = V_L \cdot I_L$$ $$S_{\text{total}} = 200\text{ V} \cdot 100\text{ A} = 20000\text{ VA} = 20\text{ kVA}$$

Step 3: Calculate the conductive power component.
Using the standard relational ratio for directly conducted apparent power: $$S_{\text{cond}} = K \cdot S_{\text{total}}$$ $$S_{\text{cond}} = 0.8 \cdot 20\text{ kVA} = 16\text{ kVA}$$

Step 4: Calculate the inductive power component.
Using the remaining fractional ratio for inductively transferred power: $$S_{\text{ind}} = (1 - K) \cdot S_{\text{total}}$$ $$S_{\text{ind}} = (1 - 0.8) \cdot 20\text{ kVA} = 0.2 \cdot 20\text{ kVA} = 4\text{ kVA}$$ Alternatively, it can be verified via basic subtraction: $$S_{\text{ind}} = S_{\text{total}} - S_{\text{cond}} = 20\text{ kVA} - 16\text{ kVA} = 4\text{ kVA}$$ Hence, the conductive power is $16\text{ kVA}$ and the inductive power is $4\text{ kVA}$, which matches option (3).