Average current taken over a full cycle

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Average current taken over a full cycle
Average current taken over a full cycle and illustrated in Fig. 4.5b is given as:

$$I_{av} = \frac{1}{2\pi} \int_0^{2\pi} \frac{V_p \sin \omega t}{R_D + R_L} d(\omega t) + \int_\pi^{2\pi} 0 d(\omega t) \quad (4.4)$$ $$I_{av} = \frac{V_p}{2\pi(R_D + R_L)} \int_0^\pi \sin \omega t \, d(\omega t) \quad (4.5)$$ $$= \frac{V_p}{2\pi(R_D + R_L)} \int_0^\pi \sin \omega t \, d(\omega t)$$

Recall that $\theta = \omega t$.
Integration is with respect to $\omega t$ and not $t$.

$$= \frac{V_p}{2\pi(R_D + R_L)} [-\cos \omega t]_0^\pi$$ $$= \frac{V_p}{2\pi(R_D + R_L)} [-\cos \pi - (-\cos 0)]$$ $$= \frac{V_p}{2\pi(R_D + R_L)} [-(-1) + (1)]$$ $$= \frac{V_p}{2\pi(R_D + R_L)} [1 + 1]$$ $$= \frac{V_p}{2\pi(R_D + R_L)} [2]$$ $$I_{av} = \frac{V_p}{\pi(R_D + R_L)}$$

But maximum current $I_p = \frac{V_p}{R_D + R_L}$.
Therefore,

$$I_{av} = \frac{I_p}{\pi} \quad (4.6)$$

Output dc power
Output dc power is given by:

$$P_{dc} = I_{av}^2 \times R_L$$ $$P_{dc} = \left[\frac{I_p}{\pi}\right]^2 \times R_L \quad (4.7)$$

Observe that output power is taken from resistance of load.
The power in input ac signal (input ac power) is given as:

$$P_{in(ac)} = I_{r.m.s}^2 \times [R_D + R_L] = V_{in} I_{in} \quad (4.8)$$

where $I_{r.m.s} = \frac{I_p}{\sqrt{2}}$ and $I_{r.m.s}^2 = \frac{I_p^2}{2}$.
For a rectified half wave,

$$I_{r.m.s} = \frac{I_p}{2} = \frac{\text{maximum current through load}}{2} \quad (4.9)$$

Mehta and Mehta (2010)
Electronics for Beginners