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.)