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

Options

A. `(1)/(sqrt(2))A`
B. `sqrt(2)A`
C. `(1)/(2sqrt(2))A`
D. `(1)/(4)A`

Answer

C

Explanation

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`