1.

A light bulb is rated at 100W for a 220 V supply. Find the peak voltage of the source

Answer»

Solution :We know, `V_("RMS") = (V_(m))/(SQRT(2))`
`RARR`Peak voltage `V_(m) = sqrt( 2) V_(rms)`
`:. V_(m) = ( 1.414) (220)``( :.` Here `V_(rms) = V =220 ` VOLT)


Discussion

No Comment Found

Related InterviewSolutions