Q148 / 150 · MHT CET 2024 (09 May Shift 1) · Physics

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 Figure

  • A
    $\frac{2}{3}$
  • Correct answer
    B
    $\frac{2}{5}$
  • C
    $\frac{3}{2}$
  • D
    $\frac{5}{2}$

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 $60 \Omega$. $\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{24}{60}=\frac{2}{5}$

Hence, the correct answer is option: 'B'.

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