Which would achieve a 120/69V secondary voltage?

[Robert0]

13. On a given 480/277V system, which of the following would be used to achieve a 120/69V secondary voltage?
Your Answer: (3) 4:1 PTs connected Delta-Delta
Correct Answer: (3) 4:1 PTs connected Wye-Wye

[nwilliams]

13. On a given 480/277V system, which of the following would be used to achieve a 120/69V secondary voltage?
Your Answer: (3) 4:1 PTs connected Delta-Delta
Correct Answer: (3) 4:1 PTs connected Wye-Wye

3 PTs connected Delta-Delta with a Turns Ratio of 4:1 will get you close to the result of 120/69V but not close enough. The problem with this configuration is that the Line-to-Line voltage and Winding voltage will both be 120V. We need the Winding voltage to be 69V. You need a Wye-configuration on the secondary to achieve this. When you use 3 PTs connected Wye-Wye with a Turns Ratio of 4:1, you get a Line-to-Line voltage of 120V and a winding voltage (can be called Line-to-Neutral in a wye configuration) of 69V.


If you see a system described in terms of 2 voltages like 120/69V, it's likely a a wye configuration. You can check this by dividing the smaller voltage into the bigger voltage. The result will be 1.732...

If only one voltage is given, odds are it's a delta.

[Kalbi_Rob]

13. On a given 480/277V system, which of the following would be used to achieve a 120/69V secondary voltage?
Your Answer: (3) 4:1 PTs connected Delta-Delta
Correct Answer: (3) 4:1 PTs connected Wye-Wye

Acronym WiDe works really well here. For Wye configuration current (I) is equal between phase and line. For Delta configuration voltage (E) (Electromotive force, which is measured in Volts (V)) is equal between phase and line. (phase to ground and line to line)
For Wye configuration, Line Voltage = sqrt (3) * Phase Voltage (ex: Line Voltage = 277 * 1.732 = 480)
For Delta configuration, Line Current = sqrt (3) * Phase Current

sqrt(3) ~ 1.732