The output of a step-down transformer is measured to be `24 V` when connected to a 12 watt light bulb. The value of the peak current is

Options

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

Answer

A

Explanation

Power `(P_S)` associated with the secondary of the transfomer = 12 W and `V_(S) = 24 V`
`:.` Current in the secondary `I_(S) = (P_S)/(V_S) = 12/24 = 1/2 A`
This is the r.m.s. current.
`:.` Peak value of the current in the secondary
`=sqrt(2) I_(rms) =sqrt(2) xx 1/2 = 1/(sqrt(2))A`.