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Find the rms value of current i=I(m)sin ...

Find the rms value of current `i=I_(m)sin omegat` from `(i) t=0 "to" t=(pi)/(omega)` (ii) `t=(pi)/(2omega) "to" t=(3pi)/(2omega)`

Text Solution

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`i_(rms) =sqrt((underset((pi)/(2omega))overset((pi)/(omega))int I_(m)^(2) sin omegatdt)/((2pi)/(omega)))= sqrt((i_(m)^(2))/(2))=(I_(m))/(sqrt2)`
`(ii) i=sqrt((underset((pi)/(2omega))overset((3pi)/(2omega))int I_(m)^(2) sin omegatdt)/((pi)/(omega)))= sqrt((I_(m)^(2))/(2))=(I_(m))/(sqrt2)`
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Knowledge Check

  • r.m.s. value of current i=3+4 sin (omega t+pi//3) is:

    A
    5A
    B
    `sqrt17A`
    C
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  • The voltage over a cycle varies as V=V_(0)sin omega t for 0 le t le (pi)/(omega) =-V_(0)sin omega t for (pi)/(omega)le t le (2pi)/(omega) The average value of the voltage one cycle is

    A
    `(V_(0))/(sqrt(2))`
    B
    `(V_(0))/(2)`
    C
    zero
    D
    `(2V_(0))/(pi)`

  • The average value of alternating current I=I_(0) sin omegat in time interval [0, pi/omega] is

    A
    `(2I_(0))/pi`
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