A zener diode of breakdown voltage $V_Z=10V$  is used in the circuit given below. The current through the Zener diode is

Current through load resistance is 

${\mathrm{i}}_{\mathrm{L}}=\frac{10}{1000}=10\mathrm{mA}$

Voltage across the series resistance is (30 - 10)V = 20V.

Current passing through the series resistance is 

$\mathrm{i}=\frac{20}{500}=40\mathrm{mA}$

Therefore, 

${\mathrm{i}}_{\mathrm{Z}}=\mathrm{i}-{\mathrm{i}}_{\mathrm{L}}=40\mathrm{mA}-10\mathrm{mA}=30\mathrm{mA}$