1.

An electric bulb is rated to give correct brightness at 24 V DC. It is connected to an AC source and its brightness is found to be one fourth of rated brightness. What is peak voltage across AC source ?

Answer»

`12 sqrt(2)`
`12//sqrt(2)`
`24 sqrt(2)`
`24 // sqrt(2)`

Solution :(a) If V is the rms voltage across bulb in case of AC source then we can write the FOLLOWING :
`(V^(2))/(R ) = (1)/(4) ((24^(2))/( R))`
`implies V = 12 ` volts
Peak voltage is `sqrt(2)` times the rms voltage . HENCE peak voltage across AC SUPPLY is `12SQRT(2)` volts.


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