Physics Circle home Student Login

Home›Chapters›Semiconductor Electronics PYQs›Special Purpose Diodes›Question 31

MHT CET 2024 (4 May, Shift 2) · Special Purpose Diodes

Special Purpose Diodes question from MHT CET 2024 (4 May, Shift 2)

Q31MHT CET 2024 · 4 MAY · SHIFT 2Medium

Special Purpose Diodes

In the circuit diagram shown in figure, the current through the zener diode is

Figure

Show answer and solution

Answer: (b)

  1. Output held at 1515 V, so drop across RSR_S is 24−15=924 - 15 = 9 V.
  2. I=9300=30I = \frac{9}{300} = 30 mA
  3. IL=151200=12.5I_L = \frac{15}{1200} = 12.5 mA
  4. IZ=30−12.5=17.5I_Z = 30 - 12.5 = 17.5 mA, so the answer is (b).
Report a mistake

More questions on Special Purpose Diodes