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

Answer

B

Explanation

Here , power = 12 W, output voltage = 48 V
`i_("rms")= (P)/(V)= (12)/(48)= (1)/(4)A`
`"Value of Peak current"= sqrt(2)i_("rms")= (sqrt(2))/(4)= (1)/(2sqrt(2))A`