For the diagram shown, the resistance between points A and B when the ideal diode ' $D$ ' is forward biased…
For the diagram shown, the resistance between points A and B when the ideal diode ' $D$ ' is forward biased is ' $R_1$ ' and that when reverse biased is ' $R_2$ '. The ratio $\frac{R_1}{R_2}$ is
$\frac{2}{3}$
$\frac{2}{5}$
$\frac{3}{2}$
$\frac{5}{2}$
Solution
When the diode is forward biased, current will flow through both the arms.
$\therefore \quad$ The effective resistance is
$\mathrm{R}_1=\frac{40 \times 60}{100}=\frac{2400}{100}=24 \Omega$ When the diode is reverse biased, current will flow only through the bottom arm.
$\therefore \quad$ The effective resistance $\mathrm{R}_2$ is $60 \Omega$.
$\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{24}{60}=\frac{2}{5}$
