Physics Circle home Student Login

Home›Chapters›Semiconductor Electronics PYQs›Special Purpose Diodes›Question 90

JEE Main 2014 (11 Apr, Shift 1) · Special Purpose Diodes

Special Purpose Diodes question from JEE Main 2014 (11 Apr, Shift 1)

Q90JEE Main 2014 · 11 APR · SHIFT 1Medium

Special Purpose Diodes

A Zener diode is connected to a battery and a load as shown below:

The currents I, IZ and IL are respectively

Figure

Show answer and solution

Answer: (d)

  1. Without the Zener the load gets 60×26=2060 \times \frac{2}{6} = 20 V >10> 10 V, so the Zener holds 1010 V.
  2. I=60−104 kΩ=12.5I = \frac{60 - 10}{4\text{ k}\Omega} = 12.5 mA
  3. IL=102 kΩ=5I_L = \frac{10}{2\text{ k}\Omega} = 5 mA, IZ=12.5−5=7.5I_Z = 12.5 - 5 = 7.5 mA
  4. So the answer is (d) 12.512.5 mA, 7.57.5 mA, 55 mA.
Report a mistake

More questions on Special Purpose Diodes