| Question #1118 | |
|---|---|
| Link to Question
|
|
| 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? | |
| A) 14.2 A | |
| B) 56.82 A | |
| C) 45.45 A | |
| D) 11.36 A | |
| Posted: 01 Jun 2017 20:31 UTC | Post #1 |
|---|---|
| gigem_das05 Deck & Engine |
Member Total Posts: 4 |
| .8PF X 5kW = 4KVA I=4KVA/110 = 36.36 Amps Can someone tell me what I am doing wrong? |
|
| Posted: 09 Apr 2018 10:46 UTC | Post #2 |
| Wbbartsch Engine |
Member Total Posts: 1 |
| 5kva is the load, not the supply. | |
| Posted: 21 Dec 2018 11:35 UTC | Post #3 |
| TBird999 Engine |
Member Total Posts: 4 |
| IDk if this is right but this is how I did it. 5,000W / 0.8 = 6250W 6250W /110V = 56.82A |
|
| Posted: 21 Jun 2019 09:06 UTC | Post #4 |
| AJGoan Engine |
Member Total Posts: 2 |
| 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 |
|
| Posted: 26 Sep 2019 23:48 UTC | Post #5 |
| Reynz22 Engine |
Expired Member Total Posts: 2 |
| 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 |
|
1 / 1