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