
Wyedelta transformer secondary line current
A wyedelta transformer has a turns ratio of 2:1. It has a secondary line voltage of 480 V. The secondary is connected to a wyeconnected threephase resistive load of 50 ohms in each phase. How much is the secondary line current?
My answer is 9.6 Amps, I used 480V since it wont matter in the delta side (V line = V phase) and divide 50 ohms from it resulting 9.6 amps, however I got the question wrong. I know I line = 1.73 * I phase. I cant remember how to solve for it.

NETA Level 2 Practice Exam
a watthour meter has a watthour constant, kh, of 1.2 it is used with 1200/5 ratio current transformers and 14,400/120 potential transformers. what is the primary watthour constant kh?
is it as simple as multiplying 1.2 by the CT Ratio and PT Ratio?

Originally Posted by
RGarcia
A wyedelta transformer has a turns ratio of 2:1. It has a secondary line voltage of 480 V. The secondary is connected to a wyeconnected threephase resistive load of 50 ohms in each phase. How much is the secondary line current?
My answer is 9.6 Amps, I used 480V since it wont matter in the delta side (V line = V phase) and divide 50 ohms from it resulting 9.6 amps, however I got the question wrong. I know I line = 1.73 * I phase. I cant remember how to solve for it.
You have to keep in mind the connection of the load. The load is connected Wye. Take the Delta secondary voltage and convert the phase voltage into line to ground. 480V/1.732 = 277V. 277V /50 ohms = 5.54A.

Originally Posted by
RGarcia
a watthour meter has a watthour constant, kh, of 1.2 it is used with 1200/5 ratio current transformers and 14,400/120 potential transformers. what is the primary watthour constant kh?
is it as simple as multiplying 1.2 by the CT Ratio and PT Ratio?
yes

Originally Posted by
RGarcia
A wyedelta transformer has a turns ratio of 2:1. It has a secondary line voltage of 480 V. The secondary is connected to a wyeconnected threephase resistive load of 50 ohms in each phase. How much is the secondary line current?
My answer is 9.6 Amps, I used 480V since it wont matter in the delta side (V line = V phase) and divide 50 ohms from it resulting 9.6 amps, however I got the question wrong. I know I line = 1.73 * I phase. I cant remember how to solve for it.
480/1.732=277
277/50=5.54A
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