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

Question

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 R 2 ​ R 1 ​ ​ is

TestHub · Accepted Answer

When the ideal diode is forward biased, it acts as a short circuit (zero resistance). The $40 Ω$ and $60 Ω$ resistors are in parallel.

$R_{1} = \frac{40 \times 60}{40 + 60} = \frac{2400}{100} = 24 Ω$

When the ideal diode is reverse biased, it acts as an open circuit (infinite resistance). No current flows through the top branch, so the equivalent resistance is simply the resistance of the bottom branch.

$R_{2} = 60 Ω$

The required ratio is:

$\frac{R _{1}}{R _{2}} = \frac{24}{60} = \frac{2}{5}$

Answer: $\frac{2}{5}$

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