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/(sqrt2 A)`
B. `sqrt2A`
C. 2A
D. `2sqrt2A`

Answer

A

Explanation

As given that, Secondary voltage `(V_(s)) = 24` Volt
Power associated with secondary is :
`P_(s) = 12 watt`
As we know that `P_(s) = V_(s) I_(s)`
`I_(s) = (P_(s))/(V_(s)) = (12)/(24) = (1)/(2) A = 0.5 Amp`
Peak value of the current in the secondary
`I_(0) = I_(s) sqrt2 = 0.5 sqrt2`
`=(5)/(10) sqrt2 [I_(0) = (1)/(sqrt2) Amp]`