MHT CET · Physics · Semiconductors
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

- A \(2: 1\)
- B \(1: 1\)
- C \(1: 2\)
- D \(1: 4\)
Answer & Solution
Correct Answer
(C) \(1: 2\)
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
\(\mathrm{R}_1=\frac{40 \times 40}{80}=\frac{1600}{80}=20 \Omega\)
When the diode is reverse biased, current will flow through the bottom arm only
\(\therefore \quad\) The effective resistance \(\mathrm{R}_2\) is \(40 \Omega\).
\(\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{20}{40}=\frac{1}{2}\)
\(\therefore \quad\) 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 \(40 \Omega\).
\(\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{20}{40}=\frac{1}{2}\)
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