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 is

Answer»

`1/sqrt2`A
`sqrt2`A
`2A`
`2sqrt2A`

Solution :Secondary VOLTAGE `V_S`=24 V
POWER associated with secondary `P_S`=12 W
Now `P_S=I_S V_S`
`therefore I_S=P_S/V_S=12/25`=0.5 A
`therefore` Peak value of the current in the secondary
`I_m=I_Sxxsqrt2` [ `because` Here, `(I_(rms))_S=I_S`]
`therefore I_m=0.5xx1.414`=0.707
`therefore I_m=1/sqrt2A [ because 0.707 = 1/sqrt2]`


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