Question

An auto-transformer is most efficient when the transformation ratio ($K = V_2/V_1$) is:

• A. Near zero
• B. Near unity (1.0)
• C. Exactly 0.5
• D. Greater than 10

Hint:

The weight of copper in an autotransformer is \(W_{\text{auto}} = (1 - K) \cdot W_{\text{tw}}\).
As \(K \to 1\), the required copper weight approaches zero, which makes the autotransformer extremely lightweight, cheap, and highly efficient compared to a two-winding transformer.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to identify the value of the transformation ratio (\(K\)) at which an autotransformer achieves its highest operating efficiency.

Step 2: Key Formula or Approach:
In an autotransformer, power is transferred from the primary to the secondary side through two mechanisms:
1. Inductive transfer (through magnetic coupling, as in a two-winding transformer):
\[ P_{\text{inductive}} = (1 - K) \cdot P_{\text{input}} \] 2. Conductive transfer (directly through the electrical connection):
\[ P_{\text{conductive}} = K \cdot P_{\text{input}} \] where \(K = \frac{V_2}{V_1}\) (with \(V_2 < V_1\)).

Step 3: Detailed Explanation:

• Conductive power transfer does not involve magnetic core losses (hysteresis and eddy current) or winding leakage flux. Consequently, conductive transfer is highly efficient.

• As the transformation ratio \(K\) approaches unity (1.0), the fraction of power transferred conductively (\(K \cdot P_{\text{input}}\)) increases, while the inductively transferred power decreases toward zero.

• This means that for \(K \approx 1.0\), only a very small portion of the power needs to be transferred magnetically.

• Consequently, the physical core and winding size required are minimized, reducing both core losses and copper losses (\(I^2 R\)).

• This combination of minimized losses results in maximum efficiency when the transformation ratio is near unity (1.0).

Step 4: Final Answer:
An autotransformer is most efficient when the transformation ratio is near unity (1.0).