In a balanced three-phase wye-connected system, what is the relationship between line voltage V_L and phase (phase-neutral) voltage V_Phase?

In a balanced three-phase wye-connected system, what is the relationship between line voltage V_L and phase (phase-neutral) voltage V_Phase?

• A. V_Phase = V_L / √3
• B. V_L = V_Phase
• C. V_L = 3 × V_Phase
• D. V_L = √3 V_Phase

Answer: (D)

This question tests how line-to-line voltage relates to phase-to-neutral voltage in a balanced three-phase wye. In a wye system, each phase-to-neutral voltage has the same magnitude, V_Phase, and the phase angles are 120 degrees apart. The line-to-line voltage is the voltage difference between two phases, which is a vector (phasor) difference of two equal-magnitude voltages separated by 120 degrees. Using the vector difference, the difference between two phase voltages has magnitude: sqrt( V_Phase^2 + V_Phase^2 − 2 V_Phase^2 cos(120°) ) = sqrt( 2 V_Phase^2 − 2 V_Phase^2(−1/2) ) = sqrt( 3 V_Phase^2 ) = √3 × V_Phase. So the line voltage equals √3 times the phase-to-neutral voltage. Equivalently, the phase voltage is the line voltage divided by √3.

This question tests how line-to-line voltage relates to phase-to-neutral voltage in a balanced three-phase wye. In a wye system, each phase-to-neutral voltage has the same magnitude, V_Phase, and the phase angles are 120 degrees apart. The line-to-line voltage is the voltage difference between two phases, which is a vector (phasor) difference of two equal-magnitude voltages separated by 120 degrees.

Using the vector difference, the difference between two phase voltages has magnitude:

sqrt( V_Phase^2 + V_Phase^2 − 2 V_Phase^2 cos(120°) )

= sqrt( 2 V_Phase^2 − 2 V_Phase^2(−1/2) )

= sqrt( 3 V_Phase^2 )

= √3 × V_Phase.

So the line voltage equals √3 times the phase-to-neutral voltage. Equivalently, the phase voltage is the line voltage divided by √3.