Input AC voltage

Here is the information extracted from the image:

Input AC voltage
If the voltage across the secondary coil of the transformer is:

$$V = V_p \sin \omega t \quad (4.1)$$

where $V_p$ is peak voltage and $\omega t$ is angular displacement of the rotating coil in the alternator or input signal generator.

Output current
Output current is given by:

$$\text{Output current} = \frac{\text{output voltage}}{\text{equivalent resistance}}$$ $$= \frac{V_p \sin \omega t}{R_D + R_L} \quad (4.2)$$

where $R_D$ and $R_L$ are resistances of diode and load respectively.

Output current taken over half a cycle
Output current taken over half a cycle as illustrated in Figure 4.5a is given as:

$$= \int_0^\pi \frac{V_p \sin \omega t}{R_D + R_L} d(\omega t) \quad (4.3)$$

Average current taken over a full cycle
Average current taken over a full cycle and illustrated in Fig. 4.5b is given as:
(The formula for this section is cut off in the image and cannot be fully transcribed.)