A Zener diode is connected to a battery and a load as show below: The currents, $\mathrm{I},…

A Zener diode is connected to a battery and a load as show below:
The currents, $\mathrm{I}, \mathrm{I}_{\mathrm{Z}}$ and $\mathrm{I}_{\mathrm{L}}$ are respectively.
  1. $15 \mathrm{~mA}, 5 \mathrm{~mA}, 10 \mathrm{~mA}$
  2. $15 \mathrm{~mA}, 7.5 \mathrm{~mA}, 7.5 \mathrm{~mA}$
  3. $12.5 \mathrm{~mA}, 5 \mathrm{~mA}, 7.5 \mathrm{~mA}$
  4. $12.5 \mathrm{~mA}, 7.5 \mathrm{~mA}, 5 \mathrm{~mA}$

Solution

Here, $R=4 \mathrm{k} \Omega=4 \times 10^3 \Omega$ $ V_i=60 \mathrm{~V} $ Zener voltage $V_z=10 \mathrm{~V}$ $R_L=2 \mathrm{k} \Omega=2 \times 10^3 \Omega$ Load current, $I_{\mathrm{L}}=\frac{V_Z}{R_L}=\frac{10}{2 \times 10^3}=5 \mathrm{~mA}$ Current through $R, I=\frac{V_i-V_Z}{R}$ $=\frac{60-10}{4 \times 10^3}=\frac{50}{4 \times 10^3}=12.5 \mathrm{~mA}$ Fom circuit diagram, $ \begin{aligned} & I=I_Z+I_L \\ \Rightarrow & 12.5=I_Z+5 \\ \Rightarrow & I_7=12.5-5=7.5 \mathrm{~mA} \end{aligned} $

Asked in: JEE Main 2014 (11 Apr Online)

Practice more Semiconductors questions on Aicharya