Physics - Semiconductors Question with Solution | TestHub

In the given circuit diode${\mathrm{D}}_2$is reverse biased, so it will not conduct. Diode${\mathrm{D}}_1$and${\mathrm{D}}_3$are forward biased, so they will conduct, Thus the equivalent circuit is



$\therefore {\mathrm{R}}_{\mathrm{net}}=\frac{(4+4)\times 16}{(4+4)+16}=\frac{8\times 16}{24}=\frac{16}{3} \mathrm{\Omega }$

$\therefore$current through the battery,

$\mathrm{I}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{net}}}=\frac{16}{\frac{16}{3}}=3 \mathrm{A}$