Physics - Semiconductors Question with Solution | TestHub

When the diode is forward biased, current will flow through both the arms. $\therefore$ The effective resistance is ${\mathrm{R}}_1=\frac{40\times 40}{80}=\frac{1600}{80}=20\Omega$ When the diode is reverse biased, current will flow through the bottom arm only $\therefore$ The effective resistance ${\mathrm{R}}_2$ is $40\Omega$. $\therefore \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{20}{40}=\frac{1}{2}$