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 ?

Options

A. `12 sqrt(2)`
B. `12//sqrt(2)`
C. `24 sqrt(2)`
D. `24 // sqrt(2)`

Answer

A

Explanation

(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.