Physics Circle home Student Login

Home›Chapters›Semiconductor Electronics PYQs›Special Purpose Diodes›Question 67

JEE Main 2021 (26 Aug, Shift 2) · Special Purpose Diodes

Special Purpose Diodes question from JEE Main 2021 (26 Aug, Shift 2)

Q67JEE Main 2021 · 26 AUG · SHIFT 2Medium

Special Purpose Diodes

For the given circuit, the power across Zener diode is .............. mW.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 120

  1. I=24−101 kΩ=14I = \frac{24 - 10}{1\text{ k}\Omega} = 14 mA
  2. IL=105 kΩ=2I_L = \frac{10}{5\text{ k}\Omega} = 2 mA, so IZ=14−2=12I_Z = 14 - 2 = 12 mA
  3. PZ=VZIZ=10×12=120P_Z = V_Z I_Z = 10 \times 12 = 120 mW
Report a mistake

More questions on Special Purpose Diodes