1.

The output of a step down transfomer is measured to be 24V when connected to a 12W light bulb. The value of the peak current is……..A

Answer»

`1/SQRT2`
`sqrt2`
2
`2sqrt2`

Solution :`V_2I_2=12W` (For IDEAL TRANSFORMER)
`therefore I_2=12/V_2=12/24=1/2A`
Now according to formula,
`I_(RMS)=I_m/sqrt2`
`therefore I_m=sqrt2 I_(rms)=sqrt2 times 1/2=1/sqrt2 A`


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