1.

The output of a stepdown transformer is measured to be 48 V when connected to a 12 W bulb. The value of peak current is

Answer»

`(1)/(sqrt(2))A`
`sqrt(2)A`
`(1)/(2sqrt(2))A`
`(1)/(4)A`

SOLUTION :Given,
Output of STEP down transformer `(V_(s))=48 V`
Power associated with secondary coil `(P_(s))=12W`
For secondary coil,
`I_(s)=(P_(s))/(V_(s))=(12)/(48)=(1)/(4)=0.25 A`
Amplitude of current `(I_(o))=I_(s)sqrt(2)`
`= 0.25 (sqrt(2))`
`= (sqrt(2))/(2)`
`= (1)/(2sqrt(2))A`


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