SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

Rms value of supply voltage V = 50 V
The internal resistance of diode r = 20 Ω
The load resistance RL = 980 Ω
Maximum voltage on the secondary side ${\mathrm{V}}_{\mathrm{m}}$ $=\sqrt{2}\mathrm{V}=\sqrt{2}\times 50=70.7\mathrm{V}$
Average or DC output voltage ${\mathrm{V}}_0=\frac{2{\mathrm{V}}_{\mathrm{m}}}{\mathrm{\pi }}$ $=\frac{\left(2\times 70.7\right)}{\mathrm{\pi }}=45\mathrm{V}$
Average or mean load current is ${\mathrm{I}}_0=\frac{{\mathrm{V}}_0}{\left({\mathrm{R}}_{\mathrm{L}}+\mathit{r}\right)}$ $=\frac{45}{\left(980+20\right)}=45\mathrm{mA}$