What are line and phase voltage and current relations in a balanced star connection?

In one line

In a balanced star (Y) connection, line voltage is \( \sqrt{3} \) times phase voltage and leads it by 30°, while line current equals phase current.

Explained simply

In a star connection, one end of each of the three phase windings is joined at a common point (neutral), and the other three ends go out as lines. Each line wire is in series with one phase, so the same current flows through both. But the voltage between two lines is across two phases at once, and since those phase voltages are 120° apart, the result is \( \sqrt3 \) times bigger, not 2 times.

Technically, line voltage \( \bar V_{RY} = \bar V_R - \bar V_Y \). Subtracting two equal phasors 120° apart gives a phasor of magnitude \( 2V_{ph}\cos30^\circ = \sqrt3V_{ph} \).

How it works

1. Draw \( V_R, V_Y, V_B \) at 120° to each other.
2. Reverse \( V_Y \) and add it to \( V_R \) to get \( V_{RY} \). The angle between \( V_R \) and \( -V_Y \) is 60°.
3. \( V_{RY} = 2V_{ph}\cos30^\circ = \sqrt3V_{ph} \), leading \( V_R \) by 30°.
4. Similarly for \( V_{YB} \) and \( V_{BR} \); the line voltages are also 120° apart.
5. Current: \( I_L = I_{ph} \). Total power \( P = 3V_{ph}I_{ph}\cos\phi = \sqrt3V_LI_L\cos\phi \), where \( \phi \) is the angle between phase voltage and phase current.

Star allows a neutral wire, so both 415 V (line) and 230–240 V (phase) are available; used in distribution.

Key formula

\( V_L = \sqrt3V_{ph}, \quad I_L = I_{ph}, \quad P = \sqrt3V_LI_L\cos\phi \)

\( V_L \) = line voltage (V), \( V_{ph} \) = phase voltage (V), \( I_L, I_{ph} \) = line and phase currents (A), \( \phi \) = phase angle of the load impedance, \( P \) = total power (W).

Quick example

Three 10 \( \Omega \) resistors in star on a 415 V supply. \( V_{ph} = 415/\sqrt3 = 239.6 \) V, \( I_{ph} = I_L = 23.96 \) A, \( P = \sqrt3\times415\times23.96\times1 = 17.2 \) kW.

Common mistakes

• Taking \( \phi \) as the angle between line voltage and line current; it is between phase voltage and phase current.
• Writing \( I_L = \sqrt3I_{ph} \) for star (that is delta).
• Using \( V_L = 2V_{ph} \) by adding magnitudes.

How MU asks it

Asked as "Derive the relation between line and phase quantities in a balanced star connection with phasor diagram" (5–8 marks) and as numericals on balanced loads (find phase current, line current, power). For full marks: draw the star circuit with neutral, the full phasor diagram showing \( V_{RY} = V_R - V_Y \), derive \( \sqrt3 \) using \( \cos30^\circ \), state the 30° lead, and derive the power formula.