In the given circuit, the breakdown voltage of the Zener diode is 3 . 0 V . What is the value of I z ?

Given that, $V_z=3 \mathrm{V}$

Let potential at $B=0 \mathrm{V}$
Potential at $E\left(V_E\right)=10 \mathrm{V}$

Now, $V_C=V_A=3 \mathrm{V}$ and,

$I_z+I_1=I ...\left(1\right)$

The current through $1 \mathrm{k\Omega }$ resistor is given by

$\begin{array}{rcl}I& =& \frac{10-3}{1000} \mathrm{A}\\ & =& \frac{7}{1000} \mathrm{A}\end{array}$

From the diagram, it follows that

$I_1=\frac{3}{2000} \mathrm{A}$

Therefore, from equation (1), it follows that

$\begin{array}{rcl}{I}_z& =& \frac{7-1.5}{1000} \mathrm{A}\\ & =& 5.5 \mathrm{mA}\end{array}$