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A transformer having efficiency 90% is working on 100 V and at 2.0 kW power. If the current in the secondary coil is 5A, calculate (i) the current in the primary coil and (ii) voltage across the secondary coil. |
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Answer» Solution :Here `eta=90%=(9)/(10),I_(s)=5A,E_(p)=100V`, (i) `E_(p)I_(p)=2kW=2000W` `I_(p)=(2000)/(E_(p)) or I_(p)=(2000)/(100)=20A` (ii) `eta=("Output POWER")/("Input power")=(E_(s)I_(s))/(E_(p)I_(p)) or E_(s)I_(s)=eta xxE_(p)I_(p)` `=(9)/(10)xx2000=1800W` `therefore E_(s)=(1800)/(I_(s))=(1800)/(5)="360 volt"` |
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