1.

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 isA. `1/sqrt(2)A`B. `sqrt(2)A`C. 2AD. `2sqrt(2)A`

Answer» 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`


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