An electric bulb is designed to operate at `12 v` DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source?

Explanation

Here `E=12 volts`
`i^2RT = i_(rms)^2RT`
`E^2/R^2= E_(rms)^2/R^2`
implies ` E^2 = E_(0)^2/2`
implies `E_(0)^2 = 2E^2`
implies `E_(0)^2 = 2 xx (12)^2 = 2 xx 144`
`E_0 = sqrt(2xx 144)`
`=16.97 = 17 V`