1.

If the emf of an AC source is given by 6 sin omegat + 4 sin2omegat, V then rms values of the emf is

Answer»

`SQRT(10)`V
`sqrt(26)`V
`sqrt(32)`V
`sqrt(20)`V

Solution :Given, emf of an AC source
E = 6 sin `omega t + 4 sin 2 omega t `
Now, `E^(2) = (6 sinomega t + 4 sin 2 omega t )^(2)`
`therefore E^(2) = 36 sin^(2) 2 omega t +16 sin^(2) 2 omega t + 48 sin omega t . Sin 2 omega t `
`= 36 sin^(2) omega t+ 16 sin^(2) 2 omega t + 48 ( cos"" (omega t)/(2) - cos ""(3 omega t)/(2) )`
Now, THEAVERAGE value of `E^(2)`,
`therefore ""E_("avg") = (1)/(T) int_(0)^(T) [ E^(2) dt ] `
` = (1)/(T) int_(0)^(T) [ 36 sin^(2) "" omega t + 16 sin^(2) 2 omega t + 48 ( (cos omega t)/(2) - (cos 3 omega t)/(2) ) ] `
` = (1)/(T) [ 36. (T)/(2)+ 16 xx (T)/(2) + 48 (0-0) ] `
`therefore "" E_("avg") = 18 + 8 = 26 `V
`therefore "" E_("rms") = sqrt( E_("avg")) = sqrt(26) V `.


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