Question

The transformer connection given in the figure is part of a balanced 3-phase circuit where the phase sequence is “abc”. The primary to secondary turns ratio is 2:1. If \( I_a + I_b + I_c = 0 \), then the relationship between \( I_A \) and \( I_{ad} \) will be: 

• A. \( \left| \frac{I_A}{I_{ad}} \right| = \frac{1}{2\sqrt{3}} \) and \( I_{ad} \) lags \( I_A \) by \( 30^\circ \).
• B. \( \left| \frac{I_A}{I_{ad}} \right| = \frac{1}{2\sqrt{3}} \) and \( I_{ad} \) leads \( I_A \) by \( 30^\circ \).
• C. \( \left| \frac{I_A}{I_{ad}} \right| = 2\sqrt{3} \) and \( I_{ad} \) lags \( I_A \) by \( 30^\circ \).
• D. \( \left| \frac{I_A}{I_{ad}} \right| = 2\sqrt{3} \) and \( I_{ad} \) leads \( I_A \) by \( 30^\circ \).

Hint:

In a Delta-Star transformer, line current transformation involves both magnitude change and a \( 30^\circ \) phase shift. Always apply vector phasor relationships when analyzing such systems.

Answer 1

The Correct Option is A

The transformer has a Delta secondary and Star (Y) primary configuration with a turns ratio of 2:1 (primary:secondary).
For such a configuration: 
- There is a \( 30^\circ \) phase shift between line currents. 
- The magnitude scaling from line current on delta side to line current on star side is: \[ \left| \frac{I_Y}{I_\Delta} \right| = \frac{1}{\sqrt{3}} \times \frac{1}{n} = \frac{1}{\sqrt{3} \cdot 2} \] where \( n = 2 \) is the turns ratio from primary to secondary. Hence: - \( \left| \frac{I_A}{I_{ad}} \right| = \frac{1}{2\sqrt{3}} \) - And for a delta-star transformer, the delta side current lags the star side current by \(30^\circ\) So, \( I_{ad} \) lags \( I_A \) by \( 30^\circ \).