In a three-phase four-wire system, the line-to-line voltage equals the line-to-neutral voltage multiplied by which factor?

In a three-phase four-wire system, the line-to-line voltage equals the line-to-neutral voltage multiplied by which factor?

• A. 1.0
• B. 1.732
• C. 2.0
• D. 0.577

Answer: (B)

The line-to-line voltage is higher than the line-to-neutral voltage by a factor of √3. In a balanced three-phase system with a neutral, each phase voltage has magnitude V_LN and the line-to-line voltage is the difference between two phase voltages that are 120 degrees apart. Using phasor math, |V_line_line| = |V_phase − V_phase∠−120°| = √(V^2 + V^2 − 2V^2 cos 120°) = √(3) V = √3 · V_LN. So the line-to-line voltage equals the line-to-neutral voltage multiplied by √3, which is about 1.732.

The line-to-line voltage is higher than the line-to-neutral voltage by a factor of √3. In a balanced three-phase system with a neutral, each phase voltage has magnitude V_LN and the line-to-line voltage is the difference between two phase voltages that are 120 degrees apart. Using phasor math, |V_line_line| = |V_phase − V_phase∠−120°| = √(V^2 + V^2 − 2V^2 cos 120°) = √(3) V = √3 · V_LN. So the line-to-line voltage equals the line-to-neutral voltage multiplied by √3, which is about 1.732.