MHT CET · Physics · Semiconductors
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
- A \(\frac{2}{3}\)
- B \(\frac{2}{5}\)
- C \(\frac{3}{2}\)
- D \(\frac{5}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{5}\)
Step-by-step Solution
Detailed explanation
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 \(60 \Omega\).
\(\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{24}{60}=\frac{2}{5}\)
\(\therefore \quad\) 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 \(60 \Omega\).
\(\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{24}{60}=\frac{2}{5}\)
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