Physics Circle home Student Login

Home›Chapters›Semiconductor Electronics PYQs›P-N Junction Diode›Question 46

MHT CET 2024 (9 May, Shift 1) · P-N Junction Diode

P-N Junction Diode question from MHT CET 2024 (9 May, Shift 1)

Q46MHT CET 2024 · 9 MAY · SHIFT 1Medium

P-N Junction Diode

For the diagram shown, the resistance between points A and B when the ideal diode ' DD ' is forward biased is ' R1R_1 ' and that when reverse biased is ' R2R_2 '. The ratio R1R2\frac{R_1}{R_2} is

Figure

Show answer and solution

Answer: (b)

  1. Forward bias: diode shorted, 40 Ω∥60 Ω40\ \Omega \parallel 60\ \Omega: R1=40×60100=24 ΩR_1 = \dfrac{40\times 60}{100} = 24\ \Omega.
  2. Reverse bias: diode branch open, R2=60 ΩR_2 = 60\ \Omega.
  3. R1R2=2460=25\dfrac{R_1}{R_2} = \dfrac{24}{60} = \dfrac{2}{5}
  4. So the answer is (b) 25\frac{2}{5}.
Report a mistake

More questions on P-N Junction Diode