Physics - Semiconductors Question with Solution | TestHub

In circuit$\mathrm{A}$, both (p-n) junction diode act as forward biasing.

Hence, current flow in circuit$\mathrm{A}$.

Total resistance$\mathit{\text{R}}$is given by

$\genfrac{}{}{0.1ex}{}{1}{R}=\genfrac{}{}{0.1ex}{}{1}{4}+\genfrac{}{}{0.1ex}{}{1}{4}$

Or$\genfrac{}{}{0.1ex}{}{1}{R}=\genfrac{}{}{0.1ex}{}{2}{4}$

Or$R=2\text{Ω}$

According to Ohm's law

$I_A=\frac{8}{2}=\mathrm{4 }\mathrm{A}$

In circuit$B$, lower p-n-junction diode is reverse biased. Hence, no current will flow but upper diode is forward biased so current, can flow through it

$I_B=\frac{8}{4}=\mathrm{2 }\mathrm{A}$