Physics - Semiconductors Question with Solution | TestHub

Here${\mathrm{D}}_2$and${\mathrm{D}}_4$are in reverse-bias.

Using KVL in the active loop:$5-2\mathrm{i}-0.7-20\mathrm{i}-2\mathrm{i}-0.7=0\Rightarrow 3.6=24\mathrm{i}\Rightarrow \mathrm{i}=\frac{3.6}{24}\mathrm{A}$

P.D. across${\mathrm{D}}_4$= P.D. between A and B points:${\mathrm{V}}_{\mathrm{A}}-20\mathrm{i}-2\mathrm{i}-0.7={\mathrm{V}}_{\mathrm{B}}\Rightarrow {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=22\mathrm{i}+0.7$

$=22\times \frac{3.6}{24}+0.7=4\mathrm{volt}$Answer: 4