Rms value of rectified output from a half wave rectifier The rms value of an alternating quantity which is the input to a half wave rectifier is $\frac{I_{max}}{\sqrt2}$.  
Then the rms value of output should be $\frac{I_{max}}{2} \sqrt 2$. But it is given every where that it is $\frac{I_{max}}{2}$. Please tell me where am I wrong.
 A: You can simply use the definition of rms value of a periodic signal. If the signal at the input of your half-wave rectfier is $v_{in}(t)=V_{max}sin(\omega t)$ for $t\in [0,+\infty[$ the output will be $v_{out}=V_{max}sin(\omega t)$ for $t\in [kT,(k+1/2)T]$ and $v_{out}=0$ for $t\in [(k+1/2)T,(k+1)T]$. Then the definition of the rms value states that
$$\tag{1}
v_{rms}(t)=\left[\frac{1}{T}\int_0^T v^2_{out}(t)dt\right]^{1/2}=\left[\frac{V_{max}^2}{T}\int_0^{T/2} \sin^2(\omega t)dt\right]^{1/2}=\left[\frac{V_{max}^2}{2T}\int_0^{T/2} (1-\cos(2\omega t))dt\right]^{1/2}=\frac{V_{max}}{2}
$$
Clearly I've supposed the half-wave rectifier was ideal, otherwise the amplitude of the output signal is different from $V_{max}$.
A: In a half wave rectifier note that the only a half cycle is transmitted In other half cycle, current is (approximately) zero. How will it affect the rms value of output?
$Hint :$ Consider that the rms value in the first half cycle is same but in next half cycle it is zero. How do you find the rms value of current? Will it be the same case if you double the time there?  
