Question

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

Answer 1

Correct Answer - A
Secondary voltage, `V_(S) = 24V`
Power associated with seconday, `P_(S) = 12W`
`I_(S) = P_(S)/V_(S) = 12/24=1/2A = 0.5A`
Peak value of the current in the secondary
`I_(0)=I_(s)sqrt(2)`
`(0.5)(1.414) = 0.707=1/sqrt(2)A`