Find the rms value of the function i = 15 ( 1- e^1/2 . t from t= 0, t= 4?

Answer 1

Limits a and b

[integral] [sqrt(1/(b-a) * (di/dt)^2]

b = 4 and a =0 and di/dt = -7.5e^0.5 and 1/(b-a) = 0.25

[integral] [sqrt(0.25*7.5^2*(e^0.5)^2]

I = sqrt(14.0625e)

note I is constant

I=(14.0625e)^0.5 * t

I=14.0625e^0.5 * 4 - 0 = 24.73

RMS = 24.73A

Answer 2

First take the period. Then divide by the size of the period. ?15{a million - e^(-t/2)}dt = 15?dt - 15?e^(-t/2)dt = 15t - {15e^(-t/2)}/(-a million/2) = 15t + 3e355e4dab36951a7a989d4d54d2e01ce^(-t/2) eval over [e355e4dab36951a7a989d4d54d2e01ce355e4dab36951a7a989d4d54d2e01ce355e4dab36951a7a989d4d54d2e01c] = {15*0,4 + 3e355e4dab36951a7a989d4d54d2e01ce^(-0,4/2)} - {15*0,4 + 3e355e4dab36951a7a989d4d54d2e01ce^(-0,4/2)} = 6e355e4dab36951a7a989d4d54d2e01c + 3e355e4dab36951a7a989d4d54d2e01ce^(-2) - 3e355e4dab36951a7a989d4d54d2e01c = 3e355e4dab36951a7a989d4d54d2e01c + 3e355e4dab36951a7a989d4d54d2e01ce^(-2) = 3e355e4dab36951a7a989d4d54d2e01c{a million + e^(-2)} the final cost over the area is: 3e355e4dab36951a7a989d4d54d2e01c{a million + e^(-2)}/(0,4 - 0,4) = 7.5{a million + e^(-2)}

Answer 3

times root 2 divided by 2 (aka 0.707)?