
A zener diode, having breakdown voltage 15 V is used in a voltage regulator circuit as shown. The current…

- 20 mA
- 5 mA
- 10 mA
- 15 mA
Solution
The voltage drop across $250 \Omega=20-15=5 \mathrm{~V}$ $\therefore \quad$ the current through it $I=\frac{V}{R}=\frac{5}{250}=20 \times 10^{-3} \mathrm{~A}$
The current through zener diode is, $I_z=\left(20 \times 10^{-3}\right)-\left(15 \times 10^{-3}\right)=5 \mathrm{~mA}$ *
Asked in: MHT CET 2024 (04 May Shift 1)