For the diagram shown, the resistances between points $\mathrm{A}$ and $\mathrm{B}$ when ideal diode…
For the diagram shown, the resistances between points $\mathrm{A}$ and $\mathrm{B}$ when ideal diode $\mathrm{D}$ is forward biased is ' $R_1$ ' and that when reverse biased is ' $R_2$ '. The ratio $R_1: R_2$ is
$2: 1$
$1: 1$
$1: 2$
$1: 4$
Solution
When the diode is forward biased, current will flow through both the arms.
$\therefore \quad$ 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
$\therefore \quad$ The effective resistance $\mathrm{R}_2$ is $40 \Omega$.
$\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{20}{40}=\frac{1}{2}$
