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

In one line

In a balanced delta (Δ) connection, line voltage equals phase voltage, while line current is \( \sqrt3 \) times phase current and lags it by 30°.

Explained simply

In a delta, the three phases form a closed triangle, and the lines are taken from the corners. Each phase is connected directly between two lines, so it gets the full line voltage. But each line corner is shared by two phases, so the line current is the phasor difference of two phase currents that are 120° apart, giving \( \sqrt3 \) times the phase current.

Technically, by KCL at a corner: \( \bar I_R = \bar I_{RY} - \bar I_{BR} \). Its magnitude is \( 2I_{ph}\cos30^\circ = \sqrt3I_{ph} \).

How it works

1. Draw phase currents \( I_{RY}, I_{YB}, I_{BR} \) at 120° to each other.
2. At line R: \( I_R = I_{RY} - I_{BR} \). Reverse \( I_{BR} \) and add it to \( I_{RY} \); the angle between them is 60°.
3. Magnitude \( = \sqrt3I_{ph} \), lagging \( I_{RY} \) by 30°.
4. Voltage: \( V_L = V_{ph} \) because each phase sits between two lines.
5. Power: \( P = 3V_{ph}I_{ph}\cos\phi = \sqrt3V_LI_L\cos\phi \) (same formula as star).

Delta has no neutral. It is common in motors and transformer windings.

Key formula

\( V_L = V_{ph}, \quad I_L = \sqrt3I_{ph}, \quad P = \sqrt3V_LI_L\cos\phi \)

\( V_L, V_{ph} \) = line and phase voltage (V), \( I_L, I_{ph} \) = line and phase current (A), \( \phi \) = angle between phase voltage and phase current, \( P \) = total active power (W).

Quick example

Three 10 \( \Omega \) resistors in delta on 415 V: \( I_{ph} = 41.5 \) A, \( I_L = \sqrt3\times41.5 = 71.9 \) A, \( P = 3\times415\times41.5 = 51.7 \) kW. The same resistors in star took 17.2 kW, so delta takes three times the power. This is the idea behind star-delta starters, which start a motor in star to reduce starting current, then switch to delta.

Common mistakes

• Writing \( V_L = \sqrt3V_{ph} \) for delta.
• Forgetting that the 30° shift is a lag of line current behind phase current.
• Using \( P = 3V_LI_L\cos\phi \) (it is either \( 3V_{ph}I_{ph}\cos\phi \) or \( \sqrt3V_LI_L\cos\phi \)).

How MU asks it

Asked as "Derive the relationship between line and phase currents in a balanced delta connection" (5–8 marks) and as numericals (e.g. "Three identical coils each of R and X are connected in delta across 415 V; find line current and power"). For full marks: draw the delta with phase current directions, write KCL at a node, draw the phasor diagram, derive \( \sqrt3 \), state the 30° lag, and give the power formula.