Engine : Electricity - 1118/1386
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| « Previous 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 |
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| Comments |
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| Reynz22 - 2019-09-26 23:48:53 Expired Member (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 |
| AJGoan - 2019-06-21 09:06:51 Member (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 |
| TBird999 - 2018-12-21 11:35:31 Member (4) |
| IDk if this is right but this is how I did it. 5,000W / 0.8 = 6250W 6250W /110V = 56.82A |
| Wbbartsch - 2018-04-09 10:46:15 Member (1) |
| 5kva is the load, not the supply. |
| gigem_das05 - 2017-06-01 20:31:23 Member (4) |
| .8PF X 5kW = 4KVA I=4KVA/110 = 36.36 Amps Can someone tell me what I am doing wrong? |