Engine: Electricity

Question 1,118 of 1,386

A 440/110 V single phase transformer supplies a load of 5 kW at 0.8 power factor load. What will be its secondary current ignoring transformer power losses?

0 ✓ | 0 ✕ 10 free left today

Tip: press A-D to answer

Mariner discussion

Comments from mariners who studied this question.

gi gigem_das05 · Jun 1, 2017

.8PF X 5kW = 4KVA
I=4KVA/110 = 36.36 Amps

Can someone tell me what I am doing wrong?

Wb Wbbartsch · Apr 9, 2018

5kva is the load, not the supply.

TB TBird999 · Dec 21, 2018

IDk if this is right but this is how I did it.
5,000W / 0.8 = 6250W

6250W /110V = 56.82A

AJ AJGoan · Jun 21, 2019

P.F = 0.8 is equal to both sides of the xfrmr

110(.8) = secondary voltage

P=IE

5000w = 88I

56.82= I

Hubert page 184-185

Re Reynz22 · Sep 26, 2019

AC Power calculation P=EI*PF

P=load power
E=secondary voltage where load is connected
PF= power factor

solve for I, I=P/(E*PF)= 5000/(110*0.8)= 56.818A

Engine: Electricity

A 440/110 V single phase transformer supplies a load of 5 kW at 0.8 power factor load. What will be its secondary current ignoring transformer power losses?

More Electricity questions

Mariner discussion