In the given circuit, the value of current $I_L$ will be _____ $\mathrm{mA}.$ (When $R_L=1 \mathrm{k\Omega }$)

Potential drop across the branch containing load resistance will be $5 \mathrm{V}$ due to Zener diode. Therefore, potential drop across $800 \mathrm{\Omega }$ will be, $10 \mathrm{V}-5 \mathrm{V}=5 \mathrm{V}$.

Now, current through battery will be,

$I=\frac{5}{800}\times 1000$ $=6.25 \mathrm{mA}$

Current through load will be,

$I_L=\frac{5}{1000} \mathrm{A}$

$=5 \mathrm{mA}$