In the circuit shown in figure (1), the V₀ ,I₁ , ${\mathrm{I}}_{{\mathrm{D}}_1}$ , and  ${\mathrm{I}}_{{\mathrm{D}}_3}$ are respectively

Since applied voltage is greater than 0.7 V and direction of current is same as direction of arrow head, both the diodes are in “on” state (see above figure(2)). Now V0is parallel to D₁ and D₂. Hence V₀=0.7 V
Current ${\mathrm{I}}_1=\frac{{\mathrm{V}}_{\mathrm{R}}}{\mathrm{R}}=\frac{\mathrm{E}-{\mathrm{V}}_{\mathrm{D}}}{\mathrm{R}}=\frac{10\mathrm{V}-0.7\mathrm{V}}{0.33\mathrm{k\Omega }}=$
28.18 mA
Bothe the silecon diodes have similar characteristics.
Hence  ${\mathrm{I}}_{{\mathrm{D}}_1}={\mathrm{I}}_{{\mathrm{D}}_2}=\frac{{\mathrm{I}}_1}{2}=14.09\mathrm{mA}$