a current of 5A is flowing at 220V in the primary of a transformer. if the voltage produced in the secondary coil is 2200V and 50% poer is lost , then the current in the secondawill be
1) 0.25A2) 0.5A3) 2.5A
4) 5A

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Solution

$E f f i c i e n c y o f t h e t r a n s f r o m e r = \frac{p o w e r o u t p u t}{p o w e r i n p u t} \times 100 \frac{50}{100} = \frac{V_{2} I_{2}}{V_{1} I_{1}} \frac{50}{100} = \frac{2200 \times I_{2}}{5 \times 220} I_{2} = 0.25 A$

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a current of 5A is flowing at 220V in the primary of a transformer. if the voltage produced in the secondary coil is 2200V and 50% poer is lost , then the current in the secondawill be1) 0.25A2) 0.5A3) 2.5A4) 5A

Efficiency of the transfromer = power out putpower input×10050100 = V2I2V1I150100 = 2200×I25×220I2 =0.25 A

a current of 5A is flowing at 220V in the primary of a transformer. if the voltage produced in the secondary coil is 2200V and 50% poer is lost , then the current in the secondawill be
1) 0.25A2) 0.5A3) 2.5A
4) 5A

$E f f i c i e n c y o f t h e t r a n s f r o m e r = \frac{p o w e r o u t p u t}{p o w e r i n p u t} \times 100 \frac{50}{100} = \frac{V_{2} I_{2}}{V_{1} I_{1}} \frac{50}{100} = \frac{2200 \times I_{2}}{5 \times 220} I_{2} = 0.25 A$