Home
Class 12
PHYSICS
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

Verified by Experts

`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)`
Promotional Banner

Similar Questions

Explore conceptually related problems

The average value of current i=I_(m) sin omega t from t=(pi)/(2 omega ) to t=(3 pi)/(2 omega) si how many times of (I_m) ?

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

Find the rms value of current from t=0 to t= (2pi)/(omega) if the current avries as i=I_(m)sin omegat .

Find the rms value of current from t = 0 to t=(2pi)/omega if the current varies as i=I_(m)sin omegat .

The r.m.s. value of I = I_(1) sin omega t + I_(2) cos omega t is

The r.m.s. value of I = I_(1) sin omega t + I_(2) cos omega t is

What is the average value of atlernating current, I = I_(0) sin omega t over time interval t = pi//omega to t = pi//omega ?

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

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