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:
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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.
\( \left| \frac{I_A}{I_{ad}} \right| = \frac{1}{2\sqrt{3}} \) and \( I_{ad} \) lags \( I_A \) by \( 30^\circ \).
\( \left| \frac{I_A}{I_{ad}} \right| = \frac{1}{2\sqrt{3}} \) and \( I_{ad} \) leads \( I_A \) by \( 30^\circ \).
\( \left| \frac{I_A}{I_{ad}} \right| = 2\sqrt{3} \) and \( I_{ad} \) lags \( I_A \) by \( 30^\circ \).
\( \left| \frac{I_A}{I_{ad}} \right| = 2\sqrt{3} \) and \( I_{ad} \) leads \( I_A \) by \( 30^\circ \).
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The Correct Option isA
Solution and Explanation
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 \( 30^\circ \).
